Let L be a common tangent line to the curves Then the square of the slope of the line L is………….

Given curves x 2 9 + y 4 4 = 1 . . . . . . . . . ( i ) & x 2 + y 2 = 3 1 4 . . . . . . . . . . . ( i i ) Equation of any tangent to (i) be y = mx + 9 m 2 + 4 . . . . . . . . . . . . ( i i i ) For common tangent (iii) also should be tangent to (ii) so by condition of common tangency 9 m 2 + 4 = 3 1 4 ( 1 + m 2 ) OR 36m2 + 16 = 31 + 31m2 =>m2 = 3

If the arithmetic mean and geometric mean of the pth and qth terms of the sequence -16,8,-4,2,…. satisfy the equation 4x2 – 9x + 5 = 0, then p + q is equation to……….

Given sequence is -16, 8, -4, 2, ..... are in G.P. with first term a = -16 & common ratio r = − 1 2 Now t p = a r P − 1 = − 1 6 ( − 1 2 ) p − 1 & t q = a r q − 1 = − 1 6 ( − 1 2 ) q − 1 So A.M. = -8 [ ( − 1 2 ) p − 1 + ( − 1 2 ) q − 1 ] = α ( l e t ) Given equation is 4x2 – 9x + 5 = 0 gives x = 1 , 5 4 From roots we get possible value of b = 1 so 1 6 ( − 1 2 ) P + q − 2 2 = 1 O R ( − 1 2 ) p + q − 2 2 = 1 1 6 − ( − 1 2 ) 4 ⇒ p + q − 2 2 = 4 ⇒ p + q = 1 0

The total number of 4-digit numbers whose greatest common divisor with 18 is 3, is…….

18 = 32 * 2 For G.C.D to be 3. no. of four digits should be only multiple of 3, but not multiple of 9 & also should not be even. As we know no. of the form 9 k -> 1000 9 k + 1 -> 1000 9 k + 2 -> 1000 9 k + 3 -> 1000 -> Total no. = 2000 9 k + 4 -> 1000 9 k + 5 -> 1000 9 k + 6 ->1000 9 k + 7 -> 1000 9 k + 8 -> 1000 In which half will be even & half be odd so Required no. = 1000

Let a and b be two real numbers such that α + β = 1 and αβ = -1. Let P n = ( α ) n + ( β ) n , Pn-1 = 11 and Pn+1 = 29 for some integer n ≥ 1 . Then, the value of p n 2 is……

Given P n = α n + β n , P n − 1 = 1 1 & P n + 1 = 2 9 P n = α n − 2 . α 2 + β n − 2 . β 2 . . . . . . . . . ( i ) Now quadratic equation having roots α & β will be x2 – (α + β)x + αβ = 0 i.e. x2 – x – 1 = 0, put x = α and put x = β So α2 = α + 1 & β2 = β + 1 (i) P n = α n − 2 ( α + 1 ) + β n − 2 ( β + 1 ) P n = P n − 1 + P n − 2 = > P n 2 = 2 3 4

Let z be those complex numbers which satisfy | z + 5 | ≤ 4 a n d z ( 1 + i ) + z ¯ ( 1 − i ) ≥ − 1 0 , i = − 1 . If the maximum value of | z + 1 | 2 i s α + β 2 , then the value of (α + β) is…….

G i v e n | z + 5 | ≤ 4 . . . . . . . . . . ( i ) ->Represent a circle ( x + 5 ) 2 + y 2 ≤ 1 6 = z ( 1 + i ) + z ¯ ( 1 − i ) ≥ − 1 0 . . . . . . . . . . ( i i ) ->Represent a line X – y ≥ − 5 So max |z + 1|2 = AQ2 = ( − 4 − 2 2 ) 2 + 8 = 3 2 + 1 6 2 = α + β 2 Hence α + β) = 48

The real value of α for which the expression 1−isin α1+isin α is purely real is: (a) ( n + 1 ) + π 2 (b) ( 2 n + 1 ) π 2 (c) n π (d) None of these, where n∈?

This is an Objective Type Questions as classified in NCERT Exemplar L e t z = 1 − i s i n α 1 + 2 i s i n α = ( 1 − i s i n α ) ( 1 − 2 i s i n α ) ( 1 + 2 i s i n α ) ( 1 − 2 i s i n α ) = 1 − 2 i s i n α − i s i n α + 2 i 2 s i n 2 α ( 1 ) 2 − ( 2 i s i n α ) 2 = 1 − 3 i s i n α − 2 s i n 2 α 1 − 4 i 2 s i n 2 α = ( 1 − 2 s i n 2 α ) − 3 i s i n α 1 + 4 s i n 2 α = 1 − 2 s i n 2 α 1 + 4 s i n 2 α − 3 s i n α 1 + 4 s i n 2 α . i Since, z is purely real, then 3 s i n α 1 + 4 s i n 2 α = 0 ⇒ s i n α = 0 S o , α = n π , n ∈ N . H e n c e , t h e c o r r e c t o p t i o n i s ( c ) .

If z = x + i y lies in the third quadrant, then z ¯ z ? also lies in the third quadrant if: (a) x > y > 0 (b) x (c) y (d) y > x > 0

This is an Objective Type Questions as classified in NCERT Exemplar G i v e n t h a t : z = x + i y I f z l i e s i n t h i r d q u a d r a n t . S o x < 0 a n d y < 0 . z ¯ = x − i y z ¯ z = x − i y x + i y = x − i y x + i y × x − i y x − i y = x 2 + i 2 y 2 − 2 x y i x 2 − i 2 y 2 = x 2 − y 2 − 2 x y i x 2 + y 2 = x 2 − y 2 x 2 + y 2 − 2 x y x 2 + y 2 i W h e n z l i e s i n t h i r d q u a d r a n t t h e n z ¯ z w i l l a l s o b e l i e i n t h i r d q u a d r a n t ∴ x 2 − y 2 x 2 + y 2 0 ⇒ x 2 0 S o , x < y < 0 . H e n c e , t h e c o r r e c t o p t i o n i s ( b ) .

The value of ? ( z + 3 ) ( z ¯ ? + 3 ) ? is equivalent to: (a) ? z + 3 ? 2 (b) z − 3 (c) z 2 + 3 (d) None of these

This is an Objective Type Questions as classified in NCERT Exemplar G i v e n t h a t : ( z + 3 ) ( z ¯ + 3 ) L e t z = x + y i S o ( z + 3 ) ( z ¯ + 3 ) = ( x + y i + 3 ) ( x − y i + 3 ) = [ ( x + 3 ) + y i ] [ ( x + 3 ) − y i ] = ( x + 3 ) 2 − y 2 i 2 = ( x + 3 ) 2 + y 2 = | x + 3 + i y | 2 = | z + 3 | 2 H e n c e , t h e c o r r e c t o p t i o n i s ( a ) .

What is the conjugate of 2−i(1−2i) z

This is Other Questions as classified in NCERT Exemplar G i v e n t h a t z = 2 − i ( 1 − 2 i ) 2 = 2 − i 1 + 4 i 2 − 4 i = 2 − i 1 − 4 − 4 i = 2 − i − 3 − 4 i = 2 − i − 3 − 4 i × − 3 + 4 i − 3 + 4 i = − 6 + 8 i + 3 i − 4 i 2 ( − 3 ) 2 − ( 4 i ) 2 = − 6 + 1 1 i + 4 9 − 1 6 i 2 = − 2 + 1 1 i 9 + 1 6 = − 2 2 5 + 1 1 2 5 i ∴ z ¯ = − 2 2 5 − 1 1 2 5 i H e n c e , z ¯ = − 2 2 5 − 1 1 2 5 i

Find z if ?z?=4 and (z)=5π6 .

This is Other Questions as classified in NCERT Exemplar G i v e n t h a t : | z | = 4 a n d a r g ( z ) = 5 π 6 ⇒ θ = 5 π 6 | z | = 4 ⇒ r = 4 S o P o l a r f o r m o f z = r [ c o s θ + i s i n θ ] = 4 [ c o s 5 π 6 + i s i n 5 π 6 ] = 4 [ c o s ( π − π 6 ) + i s i n ( π − π 6 ) ] = 4 [ − c o s π 6 + i s i n π 6 ] = 4 [ 3 2 + i . 1 2 ] = − 2 3 + 2 i H e n c e , z = − 2 3 + 2 i .

Find |(1+i)(2+i)(3+i)|

This is Other Questions as classified in NCERT Exemplar G i v e n t h a t : | ( 1 + i ) ( 2 + i ) ( 3 + i ) | | ( 1 + i ) ( 2 + i ) ( 3 + i ) × 3 − i 3 − i | = | ( 1 + i ) . 6 − 2 i + 3 i − i 2 9 − i 2 | = | ( 1 + i ) . ( 7 + i ) 9 + 1 | = | 7 + i + 7 i + i 2 1 0 | = | 7 + 8 i − 1 1 0 | = | 6 + 8 i 1 0 | = | 3 5 + 4 5 i | = ( 3 5 ) 2 + ( 4 5 ) 2 = 9 2 5 + 1 6 2 5 = 2 5 2 5 = 1 H e n c e , | ( 1 + i ) ( 2 + i ) ( 3 + i ) | = 1

For a positive integer n , find the value of ( 1 − i ) n 1 1 − i n .

This is a short answer type question as classified in NCERT Exemplar W e h a v e ( 1 − i ) n ( 1 − 1 i ) n = [ ( 1 − i ) ( 1 − 1 i ) ] n = [ ( 1 − i ) ( 1 − 1 i × i i ) ] n = [ ( 1 − i ) ( 1 − i i 2 ) ] n = [ ( 1 − i ) ( 1 + i ) ] n [ ? i 2 = − 1 ] = [ 1 − i 2 ] n = [ 1 + 1 ] n = 2 n H e n c e , ( 1 − i ) n ( 1 − 1 i ) n = 2 n .

If ( 1 + i ) 3 1 − i 3 − ( 1 − i ) 3 1 + i 3 = x + i y , then find ( x , y ) .

This is a short answer type question as classified in NCERT Exemplar W e h a v e ( 1 + i 1 − i ) 3 − ( 1 − i 1 + i ) 3 = x + i y ⇒ [ 1 + i 1 − i × 1 + i 1 + i ] 3 − [ 1 − i 1 + i × 1 − i 1 − i ] 3 = x + i y ⇒ [ 1 + i 2 + 2 i 1 − i 2 ] 3 − [ 1 + i 2 − 2 i 1 − i 2 ] 3 = x + i y ⇒ [ 1 − 1 + 2 i 1 + 1 ] 3 − [ 1 − 1 − 2 i 1 + 1 ] 3 = x + i y ⇒ [ 2 i 2 ] 3 − [ 2 i 2 ] 3 = x + i y ⇒ ( i ) 3 − ( − i ) 3 = x + i y ⇒ i 2 . i + i 2 . i = x + i y ⇒ − i − i = x + i y ⇒ 0 − 2 i = x + i y C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s , w e g e t x = 0 , y = − 2 . H e n c e , ( x , y ) = ( 0 , − 2 ) .

If 1 + i 1 − i = x + i y , then find the value of x + y .

This is a short answer type question as classified in NCERT Exemplar W e h a v e ( 1 + i ) 2 2 − i = x + i y ⇒ 1 + i 2 + 2 i 2 − i = x + i y ⇒ 1 − 1 + 2 i 2 − i = x + i y ⇒ 2 i 2 − i = x + i y ⇒ 2 i 2 − i × 2 + i 2 + i = x + i y ⇒ 4 i + 2 i 2 4 − i 2 = x + i y ⇒ 4 i − 2 4 + 1 = x + i y [ ? i 2 = − 1 ] ⇒ − 2 + 4 i 5 = x + i y ⇒ − 2 5 + 4 5 i = x + i y C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s , w e g e t x = − 2 5 , y = 4 5 . H e n c e , x + y = − 2 5 + 4 5 = 2 5 .

If 1 − i 1 0 0 1 + i = a + i b , then find ( a , b ) .

This is a short answer type question as classified in NCERT Exemplar W e h a v e ( 1 − i 1 + i ) 1 0 0 = a + b i ⇒ ( 1 − i 1 + i × 1 − i 1 − i ) 1 0 0 = a + b i ⇒ ( 1 + i 2 − 2 i 1 − i 2 ) 1 0 0 = a + b i ⇒ ( 1 − 1 − 2 i 1 + 1 ) 1 0 0 = a + b i ⇒ ( − 2 i 2 ) 1 0 0 = a + b i ⇒ ( − i ) 1 0 0 = a + b i ⇒ i 1 0 0 = a + b i ⇒ ( i 4 ) 2 5 = a + b i ⇒ ( 1 ) 2 5 = a + b i ⇒ 1 = a + b i ⇒ 1 + 0 i = a + b i C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s , w e g e t a = 1 , b = 0 . H e n c e , ( a , b ) = ( 1 , 0 ) .

Show that ? z − 2 z + 3 ? = 2 represents a circle. Find its centre and radius.

This is a long answer type question as classified in NCERT Exemplar G i v e n t h a t : | z − 2 z − 3 | = 2 L e t z = x + i y ∴ | x + i y − 2 x + i y − 3 | = 2 ⇒ | ( x − 2 ) + i y ( x − 3 ) + i y | = 2 ⇒ | ( x − 2 ) + i y | | ( x − 3 ) + i y | = 2 [ ? | a b | = | a | | b | ] ⇒ | ( x − 2 ) + i y | = 2 | ( x − 3 ) + i y | ⇒ ( x − 2 ) 2 + y 2 = 2 ( x − 3 ) 2 + y 2 [ ? | x + i y | = x 2 + y 2 ] S q u a r i n g b o t h s i d e s ( x − 2 ) 2 + y 2 = 4 [ ( x − 3 ) 2 + y 2 ] ⇒ x 2 + 4 − 4 x + y 2 = 4 [ x 2 + 9 − 6 x + y 2 ] ⇒ x 2 + y 2 − 4 x + 4 = 4 x 2 + 4 y 2 − 2 4 x + 3 6 ⇒ 3 x 2 + 3 y 2 − 2 0 x + 3 2 = 0 ⇒ x 2 + y 2 − 2 0 3 x + 3 2 3 = 0 H e r e , g = − 1 0 3 , f = 0 r = g 2 + f 2 − c = 1 0 0 9 − 0 − 3 2 3 = 4 9 = 2 3 H e n c e , t h e r e q u i r e d e q u a t i o n o f a c i r c l e i s x 2 + y 2 − 2 0 3 x + 3 2 3 = 0 C e n t r e = ( − g , − f ) = ( 1 0 3 , 0 ) a n d r = 2 3 .

If z − 1 z + 1 is a purely imaginary number ( z ≠ − 1 ), then find the value of ? z ? .

This is a long answer type question as classified in NCERT Exemplar G i v e n t h a t : z − 1 z + 1 i s p u r e l y i m a g i n a r y n u m b e r . L e t z = x + y i ∴ x + y i − 1 x + y i + 1 = ( x − 1 ) + i y ( x + 1 ) + i y = ( x − 1 ) + i y ( x + 1 ) + i y × ( x + 1 ) − i y ( x + 1 ) − i y ⇒ ( x − 1 ) ( x + 1 ) − i y ( x − 1 ) + ( x + 1 ) i y − i 2 y 2 ( x + 1 ) 2 − i 2 y 2 ⇒ x 2 − 1 + i y ( x + 1 − x + 1 ) + y 2 x 2 + 1 + 2 x + y 2 = x 2 + y 2 − 1 + 2 y i x 2 + y 2 + 2 x + 1 ⇒ x 2 + y 2 − 1 x 2 + y 2 + 2 x + 1 + 2 y x 2 + y 2 + 2 x + 1 i Since, the number is purely imaginary, then real part=0 ∴ x 2 + y 2 − 1 x 2 + y 2 + 2 x + 1 = 0 ⇒ x 2 + y 2 − 1 = 0 ⇒ x 2 + y 2 = 1 ⇒ x 2 + y 2 = 1 ∴ | z | = 1

l i m n → ∞ t a n { ∑ r = 1 n t a n − 1 ( 1 1 + r + r 2 ) } is equal to___________.

∑ r = 1 n t a n − 1 1 1 + r + r 2 = ∑ r = 1 n t a n − 1 ( r + 1 ) 1 + r ( r + 1 ) = t a n − 1 2 − t a n − 1 + . . . . . . + t a n − 1 ( n + 1 ) − t a n − 1 n For n ∞ value = π 2 − π 4 = π 4 ∴ t a n ( π 4 ) = 1

If the least and the largest real values of a, for which the equation z + |z−1|+2i=0(z∈C and i=−1) a has a solution, are p and q respectively, then 4(p2+q2) is equal to

z + α | z − 1 | + 2 i = 0 Let z = x + iy ⇒ x + i ( y + 2 ) = − α ( x − 1 ) 2 + y 2 for α ∈ R y + 2 = 0 ⇒ y = − 2 x 2 = α 2 [ ( x − 1 ) 2 + 4 ] = 0 1 α 2 ≥ 4 5 ⇒ α 2 ≤ 5 4 ⇒ − 5 2 ≤ α ≤ 5 2 4 ( p 2 + q 2 ) = 4 ( 5 4 + 5 4 ) = 1 0

The minimum values of a for which the equation 4sinx+11−sinx=α has at least one solution in (0,π2)is_____.

4 s i n x + 1 1 − s i n x = a ∀ x ∈ ( 0 , π 2 ) Let sin x = t, t ∈ (0, 1) g ( t ) = 4 t + 1 1 − t g ' ( t ) = 0 ⇒ t = 2 3 g ' ' ( 2 3 ) > 0 ∴ g ( t ) m i n i m u m = 4 2 / 3 + 1 1 − 2 / 3 = 9 Minimum value of a for which solution exist = 9

l i m n → ∞ t a n { ∑ r = 1 n t a n − 1 ( 1 1 + r + r 2 ) } is equal to___________.

∑ r = 1 n t a n − 1 1 1 + r + r 2 = ∑ r = 1 n t a n − 1 ( r + 1 ) 1 + r ( r + 1 ) = t a n − 1 2 − t a n − 1 + . . . . . . + t a n − 1 ( n + 1 ) − t a n − 1 n For n ∞ value = π 2 − π 4 = π 4 ∴ t a n ( π 4 ) = 1

The number of the real roots of the equation (x + 1)2 + | x − 5 | = 2 7 4 i s . . . . . . . . . . . . . . . . .

( x + 1 ) 2 + | x − 5 | = 2 7 4 Case l x ( 2 x + 3 ) ( 2 x − 1 ) = 0 ⇒ x = − 3 2 , 1 2 Case II ≥ x 5 x 2 + 2 x + 1 + x − 5 = 2 7 4 x = − 1 2 ± 1 4 4 + 4 3 × 1 6 2 × 4 = − 1 2 ± 8 3 2 8 ( r e j e c t e d ) because x > 5

Consider the following system of equations: x + 2y – 3z = a 2x + 6y – 11z = b x – 2y + 7z = c, where a, b and c are real constants. Then the system of equations:

Has infinite number of solution when 5a = 2b +c

Has no solution for all a, b, and c

Has a unique solution when 5a = 2b + c

Has a unique solution for all a, b and c

The sum of the series ∑ n = 1 ∞ n 2 + 6 n + 1 0 ( 2 n + 1 ) !  is equal to:

4 1 8 e − 1 9 8 e − 1 − 1 0

4 1 8 e + 1 9 8 e − 1 − 1 0

− 4 1 8 e + 1 9 8 e − 1 − 1 0

The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30°. If the jet plane is flying at a constant height, then its height is:

3600  3 m

2 4 0 0 3 m

The equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes 3x + y – 2z = 5 and

3 x − 1 0 y − 2 z + 1 1 = 0

Let F 1 (A,B,C) = ( A ∧ ∼ B ) ∨ [ ∼ C ∧ ( A ∨ B ) ] ∨ ∼ A a n d F 2 ( A , B ) = ( A ∨ B ) ∨ ( B → ∼ A )   be two logical expressions. Then:

F 1 is not a tautology but F 2 is a tautology

Both F 1 and F 2 are not tautologies

F 1 and F 2 both are tautologies

F 1 a tautology but F 2 is not a tautology

The area of the region : R =  { ( x , y ) : 5 x 2 ≤ y ≤ 2 x 2 + 9 } i s :

1 2 3 square units

6 3 square units

1 1 3 square units

Maths NCERT Exemplar Solutions Class 11th Chapter Five: Overview, Questions, Preparation

Updated on Aug 28, 2025 14:22 IST

By Raj Pandey

Table of content

Complex Numbers and Quadratic Equations Short Answer Type Questions
Complex Numbers and Quadratic Equations Long Answer Type Questions
Complex Numbers and Quadratic Equations Other Questions
Complex Numbers and Quadratic Equations Objective Type Questions
JEE MAINS 2 April 2025
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JEE Mains 2021
JEE Mains 2020

Complex Numbers and Quadratic Equations Short Answer Type Questions

1. For a positive integer $n$ , find the value of ${(1 - i)}^{n} \frac{1}{1 - i^{n}}$ .

Sol.

\begin{array}{l} W e h a v e {(1 - i)}^{n} {(1 - \frac{1}{i})}^{n} \\ = {[(1 - i) (1 - \frac{1}{i})]}^{n} = {[(1 - i) (1 - \frac{1}{i} \times \frac{i}{i})]}^{n} = {[(1 - i) (1 - \frac{i}{i^{2}})]}^{n} \\ = {[(1 - i) (1 + i)]}^{n} [∵ i^{2} = - 1] \\ = {[1 - i^{2}]}^{n} = {[1 + 1]}^{n} = 2^{n} \\ H e n c e, {(1 - i)}^{n} {(1 - \frac{1}{i})}^{n} = 2^{n} . \end{array}

2. Evaluate $\sum_{n = 1} {(1 + i)}^{3}$ , where $n \in ℕ$ .

Sol.

\begin{array}{l} W e h a v e \sum_{n = 1}^{13} (i^{n} + i^{n + 1}) \\ = (i + i^{2}) + (i^{2} + i^{3}) + (i^{3} + i^{4}) + (i^{4} + i^{5}) + (i^{5} + i^{6}) + (i^{6} + i^{7}) + (i^{7} + i^{8}) + (i^{8} + i^{9}) + (i^{9} + i^{10}) \\ (i^{10} + i^{11}) + (i^{11} + i^{12}) + (i^{12} + i^{13}) + (i^{13} + i^{14}) \\ = i + 2 (i^{2} + i^{3} + i^{4} + i^{5} + i^{6} + i^{7} + i^{8} + i^{9} + i^{10} + i^{11} + i^{12} + i^{13}) + i^{14} \\ = i + 2 [- 1 - i + 1 + i - 1 - i + 1 + i - 1 - i + 1 + i] + (- 1) \\ = i + 2 (0) - 1 \Rightarrow - 1 + i \\ H e n c e, \sum_{n = 1}^{13} (i^{n} + i^{n + 1}) = - 1 + i . \end{array}

Commonly asked questions

For a positive integer $n$ , find the value of ${(1 - i)}^{n} \frac{1}{1 - i^{n}}$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e {(1 - i)}^{n} {(1 - \frac{1}{i})}^{n} \\ = {[(1 - i) (1 - \frac{1}{i})]}^{n} = {[(1 - i) (1 - \frac{1}{i} \times \frac{i}{i})]}^{n} = {[(1 - i) (1 - \frac{i}{i^{2}})]}^{n} \\ = {[(1 - i) (1 + i)]}^{n} [? i^{2} = - 1] \\ = {[1 - i^{2}]}^{n} = {[1 + 1]}^{n} = 2^{n} \\ H e n c e, {(1 - i)}^{n} {(1 - \frac{1}{i})}^{n} = 2^{n} . \end{array}$

Evaluate $\sum_{n = 1} {(1 + i)}^{3}$ , where $n \in ?$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e \sum_{n = 1}^{13} (i^{n} + i^{n + 1}) \\ = (i + i^{2}) + (i^{2} + i^{3}) + (i^{3} + i^{4}) + (i^{4} + i^{5}) + (i^{5} + i^{6}) + (i^{6} + i^{7}) + (i^{7} + i^{8}) + (i^{8} + i^{9}) + (i^{9} + i^{10}) \\ (i^{10} + i^{11}) + (i^{11} + i^{12}) + (i^{12} + i^{13}) + (i^{13} + i^{14}) \\ = i + 2 (i^{2} + i^{3} + i^{4} + i^{5} + i^{6} + i^{7} + i^{8} + i^{9} + i^{10} + i^{11} + i^{12} + i^{13}) + i^{14} \\ = i + 2 [- 1 - i + 1 + i - 1 - i + 1 + i - 1 - i + 1 + i] + (- 1) \\ = i + 2 (0) - 1 \Rightarrow - 1 + i \\ H e n c e, \sum_{n = 1}^{13} (i^{n} + i^{n + 1}) = - 1 + i . \end{array}$

If $\frac{{(1 + i)}^{3}}{1 - i^{3}} - \frac{{(1 - i)}^{3}}{1 + i^{3}} = x + i y$ , then find $(x, y)$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e {(\frac{1 + i}{1 - i})}^{3} - {(\frac{1 - i}{1 + i})}^{3} = x + i y \\ \Rightarrow {[\frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i}]}^{3} - {[\frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i}]}^{3} = x + i y \\ \Rightarrow {[\frac{1 + i^{2} + 2 i}{1 - i^{2}}]}^{3} - {[\frac{1 + i^{2} - 2 i}{1 - i^{2}}]}^{3} = x + i y \\ \Rightarrow {[\frac{1 - 1 + 2 i}{1 + 1}]}^{3} - {[\frac{1 - 1 - 2 i}{1 + 1}]}^{3} = x + i y \\ \Rightarrow {[\frac{2 i}{2}]}^{3} - {[\frac{2 i}{2}]}^{3} = x + i y \\ \Rightarrow {(i)}^{3} - {(- i)}^{3} = x + i y \Rightarrow i^{2} . i + i^{2} . i = x + i y \\ \Rightarrow - i - i = x + i y \Rightarrow 0 - 2 i = x + i y \\ C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s, w e g e t x = 0, y = - 2 . \\ H e n c e, (x, y) = (0, - 2) . \end{array}$

If $\frac{1 + i}{1 - i} = x + i y$ , then find the value of $x + y$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e \frac{{(1 + i)}^{2}}{2 - i} = x + i y \\ \Rightarrow \frac{1 + i^{2} + 2 i}{2 - i} = x + i y \Rightarrow \frac{1 - 1 + 2 i}{2 - i} = x + i y \\ \Rightarrow \frac{2 i}{2 - i} = x + i y \Rightarrow \frac{2 i}{2 - i} \times \frac{2 + i}{2 + i} = x + i y \\ \Rightarrow \frac{4 i + 2 i^{2}}{4 - i^{2}} = x + i y \Rightarrow \frac{4 i - 2}{4 + 1} = x + i y [? i^{2} = - 1] \\ \Rightarrow \frac{- 2 + 4 i}{5} = x + i y \Rightarrow \frac{- 2}{5} + \frac{4}{5} i = x + i y \\ C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s, w e g e t x = \frac{- 2}{5}, y = \frac{4}{5} . \\ H e n c e, x + y = \frac{- 2}{5} + \frac{4}{5} = \frac{2}{5} . \end{array}$

If $\frac{1 - i^{100}}{1 + i^{}} = a + i b$ , then find $(a, b)$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e {(\frac{1 - i}{1 + i})}^{100} = a + b i \\ \Rightarrow {(\frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i})}^{100} = a + b i \Rightarrow {(\frac{1 + i^{2} - 2 i}{1 - i^{2}})}^{100} = a + b i \\ \Rightarrow {(\frac{1 - 1 - 2 i}{1 + 1})}^{100} = a + b i \Rightarrow {(\frac{- 2 i}{2})}^{100} = a + b i \\ \Rightarrow {(- i)}^{100} = a + b i \Rightarrow i^{100} = a + b i \\ \Rightarrow {(i^{4})}^{25} = a + b i \Rightarrow {(1)}^{25} = a + b i \Rightarrow 1 = a + b i \\ \Rightarrow 1 + 0 i = a + b i \\ C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s, w e g e t a = 1, b = 0 . \\ H e n c e, (a, b) = (1, 0) . \end{array}$

If $a = c o s θ + i s i n θ$ , find the value of $\frac{1 + a}{1 - a}$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : a = c o s θ + i s i n θ \\ ∴ \frac{1 + a}{1 - a} = \frac{1 + c o s θ + i s i n θ}{1 - c o s θ - i s i n θ} \\ = \frac{1 + c o s θ + i s i n θ}{1 - c o s θ - i s i n θ} \times \frac{1 - c o s θ + i s i n θ}{1 - c o s θ + i s i n θ} \\ = \frac{1 - c o s θ + i s i n θ + c o s θ - c o s^{2} θ + i s i n θ c o s θ + i s i n θ - i s i n θ c o s θ + i^{2} s i n^{2} θ}{{(1 - c o s θ)}^{2} - i^{2} s i n^{2} θ} \\ = \frac{1 + i s i n θ - c o s^{2} θ + i s i n θ - s i n^{2} θ}{1 + c o s^{2} θ - 2 c o s θ + s i n^{2} θ} \\ = \frac{s i n^{2} θ + 2 i s i n θ - s i n^{2} θ}{1 + 1 - 2 c o s θ} = \frac{2 i s i n θ}{2 - 2 c o s θ} = \frac{2 i s i n θ}{2 (1 - c o s θ)} = \frac{i s i n θ}{1 - c o s θ} \\ = \frac{2 s i n \frac{θ}{2} c o s \frac{θ}{2} . i}{2 s i n^{2} \frac{θ}{2}} = c o t \frac{θ}{2} . i \\ H e n c e, \frac{1 + a}{1 - a} = i . c o t \frac{θ}{2} . \end{array}$

If $(1 + i) z = (1 - i) z$ , then show that $z = - i z$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : (1 + i) z = (1 - i) \bar{z} \\ \Rightarrow \frac{z}{\bar{z}} = \frac{1 - i}{1 + i} = \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{1 + i^{2} - 2 i}{1 - i^{2}} \\ = \frac{1 - 1 - 2 i}{1 + 1} = \frac{- 2 i}{2} = - i \\ \Rightarrow \frac{z}{\bar{z}} = - i \\ ∴ z = - i . \bar{z} H e n c e p r o v e d . \end{array}$

If $z = x + i y$ , then show that $? z ?^{2} + 2 (z + z) + b = 0$ , where $b \in ?$ , represents a circle.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : z = x + i y \\ T o p r o v e : z \bar{z} + 2 (z + \bar{z}) + b = 0 \\ \Rightarrow (x + i y) (x - i y) + 2 (x + i y + x - i y) + b = 0 \\ \Rightarrow x^{2} + y^{2} - 2 (x + x) + b = 0 \\ \Rightarrow x^{2} + y^{2} - 4 x + b = 0 \\ W h i c h r e p r e s e n t s a c i r c l e . H e n c e p r o v e d . \end{array}$

If the real part of $\frac{z + 2}{z - 1}$ is 4, then show that the locus of the point representing $z$ in the complex plane is a circle.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t z = x + i y \\ ∴ \bar{z} = x - i y \\ S o \frac{\bar{z} + 2}{\bar{z} - 1} = \frac{x - i y + 2}{x - i y - 1} \\ = \frac{(x + 2) - i y}{(x - 1) - i y} \times \frac{(x - 1) + i y}{(x - 1) + i y} \\ = \frac{(x + 2) (x - 1) + (x + 2) y i - (x - 1) y i - i^{2} y^{2}}{{(x - 1)}^{2} - i^{2} y^{2}} \\ = \frac{x^{2} + 2 x - x - 2 + (x + 2 - x + 1) y i + y^{2}}{{(x - 1)}^{2} + y^{2}} \\ = \frac{x^{2} + y^{2} + x - 2}{{(x - 1)}^{2} + y^{2}} + \frac{3 y}{{(x - 1)}^{2} + y^{2}} i \\ Real p a r t = 4 \\ ∴ \frac{x^{2} + y^{2} + x - 2}{{(x - 1)}^{2} + y^{2}} = 4 \\ \Rightarrow x^{2} + y^{2} + x - 2 = 4 [{(x - 1)}^{2} + y^{2}] \\ \Rightarrow x^{2} + y^{2} + x - 2 = 4 [x^{2} + 1 - 2 x + y^{2}] \\ \Rightarrow x^{2} + y^{2} + x - 2 = 4 x^{2} + 4 - 8 x + 4 y^{2} \\ \Rightarrow x^{2} - 4 x^{2} + y^{2} - 4 y^{2} + x + 8 x - 2 - 4 = 0 \\ \Rightarrow - 3 x^{2} - 3 y^{2} + 9 x - 6 = 0 \\ \Rightarrow x^{2} + y^{2} - 3 x + 2 = 0 \\ W h i c h r e p r e s e n t s a c i r c l e . H e n c e, z l i e s o n a c i r c l e . \end{array}$

Show that the complex number $z$ , satisfying the condition $arg (\frac{1 - z}{1 + z}) = \frac{π}{4}$ , lies on a circle.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t z = x + i y \\ G i v e n t h a t : a r g (\frac{z - 1}{z + 1}) = \frac{π}{4} \\ \Rightarrow a r g (z - 1) - a r g (z + 1) = \frac{π}{4} [? a r g (z_{1}) - a r g (z_{2}) = a r g \frac{z_{1}}{z_{2}}] \\ \Rightarrow a r g (x + i y - 1) - a r g (x + i y + 1) = \frac{π}{4} \\ \Rightarrow a r g [(x - 1) + i y] - a r g [(x + 1) + i y] = \frac{π}{4} \\ \Rightarrow t a n^{- 1} \frac{y}{x - 1} - t a n^{- 1} \frac{y}{x + 1} = \frac{π}{4} [? a r g (x + y i) = t a n^{- 1} \frac{y}{x}] \\ \Rightarrow t a n^{- 1} [\frac{\frac{y}{x - 1} - \frac{y}{x + 1}}{1 + \frac{y}{x - 1} \times \frac{y}{x + 1}}] = \frac{π}{4} \\ \Rightarrow \frac{x y + y - x y + y}{x^{2} - 1 + y^{2}} = t a n \frac{π}{4} \\ \Rightarrow \frac{2 y}{x^{2} + y^{2} - 1} = 1 \\ \Rightarrow x^{2} + y^{2} - 1 = 2 y \\ \Rightarrow x^{2} + y^{2} - 2 y - 1 = 0 w h i c h i s a c i r c l e . \\ H e n c e, z l i e s o n a c i r c l e . \end{array}$

Solve the equation $z = \overset{?}{z} + 1 + 2 i$ .

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | z | = z + 1 + 2 i \\ L e t z = x + y i \\ | z | = (z + 1) + 2 i \\ S q u a r i n g b o t h s i d e s \\ {| z |}^{2} = {(z + 1)}^{2} + 4 i^{2} + 4 (z + 1) i \\ \Rightarrow {| z |}^{2} = {| z |}^{2} + 1 + 2 z - 4 + 4 (z + 1) i \\ \Rightarrow 0 = - 3 + 2 z + 4 (z + 1) i \\ \Rightarrow 3 - 2 z - 4 (z + 1) i = 0 \\ \Rightarrow 3 - 2 (x + y i) - 4 (x + y i + 1) i = 0 \\ \Rightarrow 3 - 2 x - 2 y i - 4 x i - 4 y i^{2} - 4 i = 0 \\ \Rightarrow 3 - 2 x + 4 y - 2 y i - 4 x i - 4 i = 0 \\ \Rightarrow (3 - 2 x + 4 y) - i (2 y + 4 x + 4) = 0 \\ \Rightarrow 3 - 2 x + 4 y = 0 \Rightarrow 2 x - 4 y = 3 \dots (i) \\ \Rightarrow 2 y + 4 x + 4 = 0 \Rightarrow 2 x + y = - 2 \dots (i i) \\ S o l v i n g e q n . (i) a n d (i i), w e g e t y = - 1 a n d x = - \frac{1}{2} \\ H e n c e, t h e v a l u e o f z = x + y i = (- \frac{1}{2} - i) . \end{array}$

Complex Numbers and Quadratic Equations Long Answer Type Questions

1. If $\overset{ˉ}{∣ z} + 1 ∣ = z + 2 (1 + i)$ , then find $z$ .

Sol.

\begin{array}{l} G i v e n t h a t : | z + 1 | = z + 2 (1 + i) \\ L e t z = x + i y \\ S o, | x + i y + 1 | = (x + i y) + 2 (1 + i) \\ \Rightarrow | (x + 1) + i y | = x + i y + 2 + 2 i \\ \Rightarrow | (x + 1) + i y | = (x + 2) + (y + 2) i \\ \Rightarrow \sqrt[]{{(x + 1)}^{2} + y^{2}} = (x + 2) + (y + 2) i [∵ | x + i y | = \sqrt[]{x^{2} + y^{2}}] \\ S q u a r i n g b o t h s i d e s \\ {(x + 1)}^{2} + y^{2} = {(x + 2)}^{2} + {(y + 2)}^{2} . i^{2} + 2 (x + 2) (y + 2) i \\ \Rightarrow x^{2} + 1 + 2 x + y^{2} = x^{2} + 4 + 4 x - y^{2} - 4 y - 4 + 2 (x + 2) (y + 2) i \\ C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s, w e g e t \\ x^{2} + 1 + 2 x + y^{2} = x^{2} + 4 x - y^{2} - 4 y a n d 2 (x + 2) (y + 2) = 0 \\ \Rightarrow 2 y^{2} - 2 x + 4 y + 1 = 0 \dots (i) \\ a n d (x + 2) (y + 2) = 0 \dots (i i) \\ x + 2 = 0 o r y + 2 = 0 \\ ∴ x = - 2 o r y = - 2 \\ N o w p u t x = - 2 i n e q n . (i) \\ \Rightarrow 2 y^{2} - 2 \times (- 2) + 4 y + 1 = 0 \\ \Rightarrow 2 y^{2} + 4 + 4 y + 1 = 0 \\ \Rightarrow 2 y^{2} + 4 y + 5 = 0 \\ b^{2} - 4 a c = {(4)}^{2} - 4 \times 2 \times 5 = 16 - 40 = - 24 < 0 n o r e a l r o o t s . \\ N o w p u t y = - 2 i n e q n . (i) \\ \Rightarrow 2 {(- 2)}^{2} - 2 x + 4 (- 2) + 1 = 0 \\ \Rightarrow 8 - 2 x - 8 + 1 = 0 \Rightarrow x = \frac{1}{2} a n d y = - 2 \\ H e n c e, t h e v a l u e o f z = x + i y = (\frac{1}{2} - 2 i) . \end{array}

2. If $arg (z - 1) = arg (z + 3 i)$ , then find $x - 1 : y$ , where $z = x + i y$ .

Sol.

\begin{array}{l} G i v e n t h a t : a r g (z - 1) = a r g (z + 3 i) \\ \Rightarrow a r g (x + y i - 1) = a r g (x + y i + 3 i) \\ \Rightarrow a r g [(x - 1) + y i] = a r g [x + (y + 3) i] \\ \Rightarrow t a n^{- 1} \frac{y}{x - 1} = t a n^{- 1} \frac{y + 3}{x} \\ \Rightarrow \frac{y}{x - 1} = \frac{y + 3}{x} \\ \Rightarrow x y = (x - 1) (y + 3) \Rightarrow x y = x y + 3 x - y - 3 \\ \Rightarrow 3 x - y = 3 \Rightarrow 3 x - 3 = y \\ \Rightarrow 3 (x - 1) = y \Rightarrow \frac{(x - 1)}{y} = \frac{1}{3} \Rightarrow x - 1 : y = 1 : 3 \\ H e n c e, x - 1 : y = 1 : 3 . \end{array}

Commonly asked questions

If $arg (z - 1) = arg (z + 3 i)$ , then find $x - 1 : y$ , where $z = x + i y$ .

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : a r g (z - 1) = a r g (z + 3 i) \\ \Rightarrow a r g (x + y i - 1) = a r g (x + y i + 3 i) \\ \Rightarrow a r g [(x - 1) + y i] = a r g [x + (y + 3) i] \\ \Rightarrow t a n^{- 1} \frac{y}{x - 1} = t a n^{- 1} \frac{y + 3}{x} \\ \Rightarrow \frac{y}{x - 1} = \frac{y + 3}{x} \\ \Rightarrow x y = (x - 1) (y + 3) \Rightarrow x y = x y + 3 x - y - 3 \\ \Rightarrow 3 x - y = 3 \Rightarrow 3 x - 3 = y \\ \Rightarrow 3 (x - 1) = y \Rightarrow \frac{(x - 1)}{y} = \frac{1}{3} \Rightarrow x - 1 : y = 1 : 3 \\ H e n c e, x - 1 : y = 1 : 3 . \end{array}$

If $\overset{?}{? z} + 1 ? = z + 2 (1 + i)$ , then find $z$ .

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | z + 1 | = z + 2 (1 + i) \\ L e t z = x + i y \\ S o, | x + i y + 1 | = (x + i y) + 2 (1 + i) \\ \Rightarrow | (x + 1) + i y | = x + i y + 2 + 2 i \\ \Rightarrow | (x + 1) + i y | = (x + 2) + (y + 2) i \\ \Rightarrow \sqrt[]{{(x + 1)}^{2} + y^{2}} = (x + 2) + (y + 2) i [? | x + i y | = \sqrt[]{x^{2} + y^{2}}] \\ S q u a r i n g b o t h s i d e s \\ {(x + 1)}^{2} + y^{2} = {(x + 2)}^{2} + {(y + 2)}^{2} . i^{2} + 2 (x + 2) (y + 2) i \\ \Rightarrow x^{2} + 1 + 2 x + y^{2} = x^{2} + 4 + 4 x - y^{2} - 4 y - 4 + 2 (x + 2) (y + 2) i \\ C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s, w e g e t \\ x^{2} + 1 + 2 x + y^{2} = x^{2} + 4 x - y^{2} - 4 y a n d 2 (x + 2) (y + 2) = 0 \\ \Rightarrow 2 y^{2} - 2 x + 4 y + 1 = 0 \dots (i) \\ a n d (x + 2) (y + 2) = 0 \dots (i i) \\ x + 2 = 0 o r y + 2 = 0 \\ ∴ x = - 2 o r y = - 2 \\ N o w p u t x = - 2 i n e q n . (i) \\ \Rightarrow 2 y^{2} - 2 \times (- 2) + 4 y + 1 = 0 \\ \Rightarrow 2 y^{2} + 4 + 4 y + 1 = 0 \\ \Rightarrow 2 y^{2} + 4 y + 5 = 0 \\ b^{2} - 4 a c = {(4)}^{2} - 4 \times 2 \times 5 = 16 - 40 = - 24 < 0 n o r e a l r o o t s . \\ N o w p u t y = - 2 i n e q n . (i) \\ \Rightarrow 2 {(- 2)}^{2} - 2 x + 4 (- 2) + 1 = 0 \\ \Rightarrow 8 - 2 x - 8 + 1 = \end{array}$

Show that $? \frac{z - 2}{z + 3} ? = 2$ represents a circle. Find its centre and radius.

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | \frac{z - 2}{z - 3} | = 2 \\ L e t z = x + i y \\ ∴ | \frac{x + i y - 2}{x + i y - 3} | = 2 \\ \Rightarrow | \frac{(x - 2) + i y}{(x - 3) + i y} | = 2 \\ \Rightarrow \frac{| (x - 2) + i y |}{| (x - 3) + i y |} = 2 [? | \frac{a}{b} | = \frac{| a |}{| b |}] \\ \Rightarrow | (x - 2) + i y | = 2 | (x - 3) + i y | \\ \Rightarrow \sqrt[]{{(x - 2)}^{2} + y^{2}} = 2 \sqrt[]{{(x - 3)}^{2} + y^{2}} [? | x + i y | = \sqrt[]{x^{2} + y^{2}}] \\ S q u a r i n g b o t h s i d e s \\ {(x - 2)}^{2} + y^{2} = 4 [{(x - 3)}^{2} + y^{2}] \\ \Rightarrow x^{2} + 4 - 4 x + y^{2} = 4 [x^{2} + 9 - 6 x + y^{2}] \\ \Rightarrow x^{2} + y^{2} - 4 x + 4 = 4 x^{2} + 4 y^{2} - 24 x + 36 \\ \Rightarrow 3 x^{2} + 3 y^{2} - 20 x + 32 = 0 \\ \Rightarrow x^{2} + y^{2} - \frac{20}{3} x + \frac{32}{3} = 0 \\ H e r e, g = - \frac{10}{3}, f = 0 \\ r = \sqrt[]{g^{2} + f^{2} - c} = \sqrt[]{\frac{100}{9} - 0 - \frac{32}{3}} = \sqrt[]{\frac{4}{9}} = \frac{2}{3} \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f a c i r c l e i s \\ x^{2} + y^{2} - \frac{20}{3} x + \frac{32}{3} = 0 \\ C e n t r e = (- g, - f) = (\frac{10}{3}, 0) a n d r = \frac{2}{3} . \end{array}$

If $\frac{z - 1}{z + 1}$ is a purely imaginary number ( $z \neq - 1$ ), then find the value of $? z ?$ .

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : \frac{z - 1}{z + 1} i s p u r e l y i m a g i n a r y n u m b e r . \\ L e t z = x + y i \\ ∴ \frac{x + y i - 1}{x + y i + 1} = \frac{(x - 1) + i y}{(x + 1) + i y} = \frac{(x - 1) + i y}{(x + 1) + i y} \times \frac{(x + 1) - i y}{(x + 1) - i y} \\ \Rightarrow \frac{(x - 1) (x + 1) - i y (x - 1) + (x + 1) i y - i^{2} y^{2}}{{(x + 1)}^{2} - i^{2} y^{2}} \\ \Rightarrow \frac{x^{2} - 1 + i y (x + 1 - x + 1) + y^{2}}{x^{2} + 1 + 2 x + y^{2}} = \frac{x^{2} + y^{2} - 1 + 2 y i}{x^{2} + y^{2} + 2 x + 1} \\ \Rightarrow \frac{x^{2} + y^{2} - 1}{x^{2} + y^{2} + 2 x + 1} + \frac{2 y}{x^{2} + y^{2} + 2 x + 1} i \\ Since, t h e n u m b e r i s p u r e l y i m a g i n a r y, t h e n r e a l p a r t = 0 \\ ∴ \frac{x^{2} + y^{2} - 1}{x^{2} + y^{2} + 2 x + 1} = 0 \\ \Rightarrow x^{2} + y^{2} - 1 = 0 \Rightarrow x^{2} + y^{2} = 1 \\ \Rightarrow \sqrt[]{x^{2} + y^{2}} = 1 ∴ | z | = 1 \end{array}$

$z_{1}$ and $z_{2}$ are two complex numbers such that $? z_{1} ? = ? z_{2} ? = 1$ and $arg (z_{1}) + arg (z_{2}) = π$ , then show that $z_{1} = - \frac{}{z_{2}}$ .

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t z_{1} = r_{1} (c o s θ_{1} + i s i n θ_{1}) a n d z_{2} = r_{2} (c o s θ_{2} + i s i n θ_{2}) a r e p o l a r f o r m o f t w o c o m p l e x \\ n u m b e r s z_{1} a n d z_{2} . \\ G i v e n t h a t : | z_{1} | = | z_{2} | \Rightarrow r_{1} = r_{2} \\ a n d a r g (z_{1}) + a r g (z_{2}) = π \\ \Rightarrow θ_{1} + θ_{2} = π \\ \Rightarrow θ_{1} = π - θ_{2} \\ N o w z_{1} = r_{1} [c o s (π - θ_{2}) + i s i n (π - θ_{2})] \\ \Rightarrow z_{1} = r_{1} [- c o s θ_{2} + i s i n θ_{2}] \\ \Rightarrow z_{1} = - r_{1} [c o s θ_{2} - i s i n θ_{2}] \dots (i) \\ \Rightarrow z_{2} = r_{2} [c o s θ_{2} + i s i n θ_{2}] \\ \Rightarrow {\bar{z}}_{2} = r_{1} [c o s θ_{2} - i s i n θ_{2}] [? r_{1} = r_{2}] \dots (i i) \\ F r o m e q n . (i) a n d (i i) w e g e t, \\ z_{1} = - {\bar{z}}_{2} . H e n c e p r o v e d . \end{array}$

If $z_{1} = 1$ ( $z_{1} \neq - 1$ ) and $z_{2} = \frac{z_{1} - 1}{z_{1} + 1}$ , then show that the real part of $z_{2}$ is zero.

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t z_{1} = x + y i \\ | z_{1} | = \sqrt[]{x^{2} + y^{2}} = 1 [G i v e n t h a t | z_{1} | = 1] \\ \Rightarrow x^{2} + y^{2} = 1 \dots (i) \\ N o w z_{2} = \frac{z_{1} - 1}{z_{1} + 1} = \frac{x + y i - 1}{x + y i + 1} \\ = \frac{(x - 1) + y i}{(x + 1) + y i} = \frac{(x - 1) + y i}{(x + 1) + y i} \times \frac{x + 1 - y i}{x + 1 - y i} \\ = \frac{(x - 1) (x + 1) - y (x - 1) i + y (x + 1) i - y^{2} i^{2}}{{(x + 1)}^{2} - y^{2} i^{2}} \\ = \frac{x^{2} - 1 + y i (x + 1 - x + 1) + y^{2}}{x^{2} + 1 + 2 x + y^{2}} \\ = \frac{(x^{2} + y^{2} - 1) + 2 y i}{x^{2} + y^{2} + 2 x + 1} \\ = \frac{(1 - 1)}{x^{2} + y^{2} + 2 x + 1} + \frac{2 y}{x^{2} + y^{2} + 2 x + 1} i \\ = 0 + \frac{2 y}{x^{2} + y^{2} + 2 x + 1} i \\ H e n c e, t h e r e a l p a r t o f z_{2} i s 0 . \end{array}$

If $z_{1}, z_{2}$ and $z_{3}, z_{4}$ are two pairs of conjugate complex numbers, then find $arg (\frac{z_{1}}{z_{4}}) + arg (\frac{z_{2}}{z_{3}})$ .

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t t h e p o l a r f o r m o f z_{1} = r_{1} (c o s θ_{1} + i s i n θ_{1}) \\ ∴ z_{2} = {\bar{z}}_{1} = r_{1} (c o s θ_{1} - i s i n θ_{1}) = r_{1} [c o s (- θ_{1}) + i s i n (- θ_{1})] \\ S i m i l a r l y, z_{3} = r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ ∴ z_{4} = {\bar{z}}_{3} = r_{1} (c o s θ_{2} - i s i n θ_{2}) = r_{2} [c o s (- θ_{2}) + i s i n (- θ_{2})] \\ a r g (\frac{z_{1}}{z_{4}}) + a r g (\frac{z_{2}}{z_{3}}) = a r g (z_{1}) - a r g (z_{4}) + a r g (z_{2}) - a r g (z_{3}) \\ = θ_{1} - (- θ_{2}) + (- θ_{1}) - θ_{2} \\ = θ_{1} + θ_{2} - θ_{1} - θ_{2} = 0 \\ H e n c e, a r g (\frac{z_{1}}{z_{4}}) + a r g (\frac{z_{2}}{z_{3}}) = 0 \end{array}$

If $? z_{1} ? = ? z_{2} ? = . . . = ? z_{n} ? = 1$ , then show that $? z_{1} + z_{2} + z_{3} + . . . + \frac{1_{}}{z_{1}} = \frac{z_{2}}{z_{2}} + \frac{z_{3}}{z_{3}} + \frac{z_{4}}{z_{3}} + . . . + \frac{z_{1}}{z_{n}}$ .

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e | z_{1} | = | z_{2} | = \dots = | z_{n} | = 1 \\ \Rightarrow {| z_{1} |}^{2} = {| z_{2} |}^{2} = \dots = {| z_{n} |}^{2} = 1 \dots (i) \\ \Rightarrow z_{1} {\bar{z}}_{1} = z_{2} {\bar{z}}_{2} = \dots = z_{n} {\bar{z}}_{n} = 1 [? z \bar{z} = {| z |}^{2}] \\ \Rightarrow z_{1} = \frac{1}{{\bar{z}}_{1}}, z_{2} = \frac{1}{{\bar{z}}_{2}} = \dots = z_{n} = \frac{1}{{\bar{z}}_{n}} \\ L . H . S . | z_{1} + z_{2} + z_{3} + \dots + z_{n} | \\ = | \frac{z_{1} {\bar{z}}_{1}}{{\bar{z}}_{1}} + \frac{z_{2} {\bar{z}}_{2}}{{\bar{z}}_{2}} + \frac{z_{3} {\bar{z}}_{3}}{{\bar{z}}_{3}} + \dots + \frac{z_{n} {\bar{z}}_{n}}{{\bar{z}}_{n}} | \\ = | \frac{{| z_{1} |}^{2}}{{\bar{z}}_{1}} + \frac{{| z_{2} |}^{2}}{{\bar{z}}_{2}} + \frac{{| z_{3} |}^{2}}{{\bar{z}}_{3}} + \dots + \frac{{| z_{n} |}^{2}}{{\bar{z}}_{n}} | [? z \bar{z} = {| z |}^{2}] \\ = | \frac{1}{{\bar{z}}_{1}} + \frac{1}{{\bar{z}}_{2}} + \frac{1}{{\bar{z}}_{3}} + \dots + \frac{1}{{\bar{z}}_{n}} | [Using (i)] \\ = | \bar{\frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} + \dots + \frac{1}{z_{n}}} | [? {\bar{z}}_{1} + {\bar{z}}_{2} = \bar{z_{1} + z_{2}}] \\ = | \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} + \dots + \frac{1}{z_{n}} | [? | z | = | {\bar{z}}_{1} |] \\ L . H . S . - R . H . S . H e n c e p r o v e d . \end{array}$

If for complex numbers $z_{1}$ and $z_{2}$ , $arg (z_{1}) - arg (z_{2}) = 0$ , then show that $? z_{1} - z_{2} ? = ? z_{1} ? ? z_{2} ?$

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t f o r z_{1} a n d z_{2}, a r g (z_{1}) - a r g (z_{2}) = 0 \\ L e t u s r e p r e s e n t z_{1} a n d z_{2} i n p o l a r f o r m \\ z_{1} = r_{1} (c o s θ_{1} + i s i n θ_{1}) a n d z_{2} = r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ a r g (z_{1}) = θ_{1} a n d a r g (z_{2}) = θ_{2} \\ Since a r g (z_{1}) - a r g (z_{2}) = 0 \\ \Rightarrow θ_{1} - θ_{2} = 0 \Rightarrow θ_{1} = θ_{2} \\ N o w z_{1} - z_{2} = r_{1} (c o s θ_{1} - i s i n θ_{1}) - r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ = r_{1} c o s θ_{1} - i r_{1} s i n θ_{1} - r_{2} c o s θ_{1} + i r_{2} s i n θ_{1} [? θ_{1} = θ_{2}] \\ = (r_{1} c o s θ_{1} - r_{2} c o s θ_{1}) + i (r_{1} s i n θ_{1} - r_{2} s i n θ_{1}) \\ ∴ | z_{1} - z_{2} | = \sqrt[]{{(r_{1} c o s θ_{1} - r_{2} c o s θ_{1})}^{2} + {(r_{1} s i n θ_{1} - r_{2} s i n θ_{1})}^{2}} \\ = \sqrt[]{(r_{1}^{2} c o s^{2} θ_{1} + r_{2}^{2} c o s^{2} θ_{1} - 2 r_{1} r_{2} c o s^{2} θ_{1} + r_{1}^{2} s i n^{2} θ_{1} + r_{2}^{2} s i n^{2} θ_{1} - 2 r_{1} r_{2} s i n^{2} θ_{1})} \\ = \sqrt[]{r_{1}^{2} (c o s^{2} θ_{1} + s i n^{2} θ_{1}) + r_{2}^{2} (c o s^{2} θ_{1} + s i n^{2} θ_{1}) - 2 r_{1} r_{2} (c o s^{2} θ_{1} + s i n^{2} θ_{1})} \\ = \sqrt[]{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2}} = \sqrt[]{{(r_{1} - r_{2})}^{2}} = r_{1} - r_{2} = | z_{1} | - | z_{2} | \\ H e n c e, | z_{1} - z_{2} | = | z_{1} | - | z_{2} | \end{array}$

Solve the system of equations $Re (z^{2}) = 0, z = 2$ .

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : R e (z^{2}) = 0 a n d | z | = 2 \\ L e t z = x + y i \\ ∴ | z | = \sqrt[]{x^{2} + y^{2}} \\ \Rightarrow \sqrt[]{x^{2} + y^{2}} = 2 \Rightarrow x^{2} + y^{2} = 4 \dots (i) \\ Since z = x + y i \\ z^{2} = x^{2} + y^{2} i^{2} + 2 x y i \Rightarrow z^{2} = x^{2} - y^{2} + 2 x y i \\ ∴ R e (z^{2}) = x^{2} - y^{2} \\ \Rightarrow x^{2} - y^{2} = 0 \dots (i i) \\ F r o m e q n . (i) a n d (i i), w e g e t \\ x^{2} + y^{2} = 4 \Rightarrow 2 x^{2} = 4 \Rightarrow x^{2} = 2 \Rightarrow x = \pm \sqrt[]{2} a n d y = \pm \sqrt[]{2} \\ H e n c e, z = \pm \sqrt[]{2} \pm i \sqrt[]{2}, - \sqrt[]{2} \pm i \sqrt[]{2} . \end{array}$

Find the complex number satisfying the equation $z + \sqrt 2 ? z + 1 ? + i = 0$ .

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : z + \sqrt[]{2} | (z + 1) | + i = 0 \\ L e t z = x + y i \\ ∴ (x + y i) + \sqrt[]{2} | (x + y i + 1) | + i = 0 \\ \Rightarrow x + (y + 1) i + \sqrt[]{2} | (x + 1) + y i | = 0 \\ \Rightarrow x + (y + 1) i + \sqrt[]{2} \sqrt[]{{(x + 1)}^{2} + y^{2}} = 0 \\ \Rightarrow x + (y + 1) i + \sqrt[]{2} \sqrt[]{x^{2} + 2 x + 1 + y^{2}} = 0 + 0 i \\ \Rightarrow x + \sqrt[]{2} \sqrt[]{x^{2} + 2 x + 1 + y^{2}} = 0, y + 1 = 0 \\ \Rightarrow x = - \sqrt[]{2} \sqrt[]{x^{2} + 2 x + 1 + y^{2}} a n d y = - 1 \\ \Rightarrow x^{2} = 2 (x^{2} + 2 x + 1 + y^{2}) \\ \Rightarrow x^{2} = 2 x^{2} + 4 x + 2 + 2 y^{2} \\ \Rightarrow x^{2} + 4 x + 2 + 2 y^{2} = 0 \\ \Rightarrow x^{2} + 4 x + 2 + 2 {(- 1)}^{2} = 0 [? y = - 1] \\ \Rightarrow x^{2} + 4 x + 4 = 0 \\ \Rightarrow {(x + 2)}^{2} = 0 \\ \Rightarrow x + 2 = 0 \Rightarrow x = - 2 \\ H e n c e, z = x + y i = - 2 - i . \end{array}$

Write the complex number $z = \frac{1 - i}{c o s \frac{π}{3} + i s i n \frac{π}{3}}$ in polar form.

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : \\ z = \frac{1 - i}{\frac{1}{2} + i \frac{\sqrt[]{3}}{2}} = \frac{2 - 2 i}{1 + i \sqrt[]{3}} = \frac{2 - 2 i}{1 + i \sqrt[]{3}} \times \frac{1 - i \sqrt[]{3}}{1 - i \sqrt[]{3}} \\ \Rightarrow z = \frac{2 - 2 \sqrt[]{3} i - 2 i + 2 \sqrt[]{3} i^{2}}{{(1)}^{2} - {(i \sqrt[]{3})}^{2}} = \frac{2 - 2 \sqrt[]{3} i - 2 i - 2 \sqrt[]{3}}{1 - 3 i^{2}} \\ = \frac{(2 - 2 \sqrt[]{3}) - (2 + 2 \sqrt[]{3}) i}{4} = \frac{1 - \sqrt[]{3}}{2} - \frac{1 + \sqrt[]{3}}{2} i \\ r = \sqrt[]{{(\frac{1 - \sqrt[]{3}}{2})}^{2} + {(- \frac{1 + \sqrt[]{3}}{2})}^{2}} = \sqrt[]{\frac{1 + 3 - 2 \sqrt[]{3}}{4} + \frac{1 + 3 + 2 \sqrt[]{3}}{4}} \\ = \sqrt[]{\frac{4 - 2 \sqrt[]{3} + 4 + 2 \sqrt[]{3}}{4}} = \sqrt[]{\frac{8}{4}} = \sqrt[]{2} \\ S o r = \sqrt[]{2} \\ N o w a r g (z) = t a n^{- 1} \frac{y}{x} \\ \Rightarrow θ = t a n^{- 1} \frac{- (\frac{1 + \sqrt[]{3}}{2})}{(\frac{1 - \sqrt[]{3}}{2})} = t a n^{- 1} [- (\frac{1 + \sqrt[]{3}}{1 - \sqrt[]{3}})] = t a n^{- 1} \frac{\sqrt[]{3} + 1}{\sqrt[]{3} - 1} \\ \Rightarrow θ = t a n^{- 1} [t a n (\frac{π}{4} + \frac{π}{6})] [? t a n (\frac{π}{4} + \frac{π}{6}) = \frac{t a n \frac{π}{4} + t a n \frac{π}{6}}{1 - t a n \frac{π}{4} t a n \frac{π}{6}}] \\ \Rightarrow θ = \frac{5 π}{12} \\ H e n c e, t h e p o l a r i s z = \sqrt[]{2} [c o s (\frac{5 π}{12}) + i s i n (\frac{5 π}{12})] . \end{array}$

If $z$ and $w$ are two complex numbers such that $? z w ? = 1$ and $arg (z) - arg (w) = \frac{π}{2}$ , then show that $z^{} w = - i$ .

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t z = r_{1} (c o s θ_{1} + i s i n θ_{1}) a n d w = r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ z w = r_{1} r_{2} [(c o s θ_{1} - i s i n θ_{1})] [(c o s θ_{2} + i s i n θ_{2})] \\ | z w | = r_{1} r_{2} = 1 [G i v e n] \\ N o w a r g (z) - a r g (w) = \frac{π}{2} \\ \Rightarrow θ_{1} - θ_{2} = \frac{π}{2} \Rightarrow a r g (\frac{z}{w}) = \frac{π}{2} \\ \bar{z} w = r_{1} (c o s θ_{1} - i s i n θ_{1}) r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ = r_{1} r_{2} [c o s θ_{1} c o s θ_{2} + i c o s θ_{1} s i n θ_{2} - i s i n θ_{1} c o s θ_{2} - i^{2} s i n θ_{1} s i n θ_{2}] \\ = r_{1} r_{2} [(c o s θ_{1} c o s θ_{2} + s i n θ_{1} s i n θ_{2}) + i (c o s θ_{1} s i n θ_{2} - s i n θ_{1} c o s θ_{2})] \\ = r_{1} r_{2} [c o s (θ_{2} - θ_{1}) + i s i n (θ_{2} - θ_{1})] \\ = r_{1} r_{2} [c o s (- \frac{π}{2}) + i s i n (- \frac{π}{2})] \\ = r_{1} r_{2} [c o s \frac{π}{2} - i s i n \frac{π}{2}] = 1 . [0 - i] = - i \\ H e r e \bar{z} w = - i . H e n c e p r o v e d . \end{array}$

Complex Numbers and Quadratic Equations Other Questions

1. What is the conjugate of $\frac{2 - i}{(1 - 2 i)}$ z

Sol.

\begin{array}{l} G i v e n t h a t z = \frac{2 - i}{{(1 - 2 i)}^{2}} = \frac{2 - i}{1 + 4 i^{2} - 4 i} = \frac{2 - i}{1 - 4 - 4 i} \\ = \frac{2 - i}{- 3 - 4 i} = \frac{2 - i}{- 3 - 4 i} \times \frac{- 3 + 4 i}{- 3 + 4 i} \\ = \frac{- 6 + 8 i + 3 i - 4 i^{2}}{{(- 3)}^{2} - {(4 i)}^{2}} = \frac{- 6 + 11 i + 4}{9 - 16 i^{2}} \\ = \frac{- 2 + 11 i}{9 + 16} = \frac{- 2}{25} + \frac{11}{25} i \\ ∴ \bar{z} = \frac{- 2}{25} - \frac{11}{25} i \\ H e n c e, \bar{z} = \frac{- 2}{25} - \frac{11}{25} i \end{array}

2. If $∣ z_{1} ∣ = ∣ z_{2} ∣ = 1$ , is it necessary that $z_{1} = z_{2}$ ?

Sol.

\begin{array}{l} L e t z_{1} = x_{1} + y_{1} i a n d z_{2} = x_{2} + y_{2} i \\ ∴ | x_{1} + y_{1} i | = | x_{2} + y_{2} i | \\ \Rightarrow \sqrt[]{x_{1}^{2} + y_{1}^{2}} = \sqrt[]{x_{2}^{2} + y_{2}^{2}} \\ \Rightarrow x_{1}^{2} + y_{1}^{2} = x_{2}^{2} + y_{2}^{2} \Rightarrow x_{1}^{2} = x_{2}^{2} a n d y_{1}^{2} = y_{2}^{2} \\ \Rightarrow x_{1} = \pm x_{2} a n d y_{1} = \pm y_{2} \\ S o, z_{1} = x_{1} + y_{1} i a n d z_{2} = \pm x_{2} \pm y_{2} i \\ ∴ z_{1} \neq z_{2} \\ H e n c e, i t i s n o t n e c e s s a r y t h a t z_{1} = z_{2} . \end{array}

Commonly asked questions

What is the conjugate of $\frac{2 - i}{(1 - 2 i)}$ z

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t z = \frac{2 - i}{{(1 - 2 i)}^{2}} = \frac{2 - i}{1 + 4 i^{2} - 4 i} = \frac{2 - i}{1 - 4 - 4 i} \\ = \frac{2 - i}{- 3 - 4 i} = \frac{2 - i}{- 3 - 4 i} \times \frac{- 3 + 4 i}{- 3 + 4 i} \\ = \frac{- 6 + 8 i + 3 i - 4 i^{2}}{{(- 3)}^{2} - {(4 i)}^{2}} = \frac{- 6 + 11 i + 4}{9 - 16 i^{2}} \\ = \frac{- 2 + 11 i}{9 + 16} = \frac{- 2}{25} + \frac{11}{25} i \\ ∴ \bar{z} = \frac{- 2}{25} - \frac{11}{25} i \\ H e n c e, \bar{z} = \frac{- 2}{25} - \frac{11}{25} i \end{array}$

If $? z_{1} ? = ? z_{2} ? = 1$ , is it necessary that $z_{1} = z_{2}$ ?

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z_{1} = x_{1} + y_{1} i a n d z_{2} = x_{2} + y_{2} i \\ ∴ | x_{1} + y_{1} i | = | x_{2} + y_{2} i | \\ \Rightarrow \sqrt[]{x_{1}^{2} + y_{1}^{2}} = \sqrt[]{x_{2}^{2} + y_{2}^{2}} \\ \Rightarrow x_{1}^{2} + y_{1}^{2} = x_{2}^{2} + y_{2}^{2} \Rightarrow x_{1}^{2} = x_{2}^{2} a n d y_{1}^{2} = y_{2}^{2} \\ \Rightarrow x_{1} = \pm x_{2} a n d y_{1} = \pm y_{2} \\ S o, z_{1} = x_{1} + y_{1} i a n d z_{2} = \pm x_{2} \pm y_{2} i \\ ∴ z_{1} \neq z_{2} \\ H e n c e, i t i s n o t n e c e s s a r y t h a t z_{1} = z_{2} . \end{array}$

If $? z_{1} ? = ? z_{2} ? = 1$ , is it necessary that $z_{1} = z_{2}$ ?

This is Other Questions as classified in NCERT Exemplar

Find $z$ if $? z ? = 4$ and $(z) = \frac{5 π}{6}$ .

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | z | = 4 a n d a r g (z) = \frac{5 π}{6} \Rightarrow θ = \frac{5 π}{6} \\ | z | = 4 \Rightarrow r = 4 \\ S o P o l a r f o r m o f z = r [c o s θ + i s i n θ] \\ = 4 [c o s \frac{5 π}{6} + i s i n \frac{5 π}{6}] \\ = 4 [c o s (π - \frac{π}{6}) + i s i n (π - \frac{π}{6})] \\ = 4 [- c o s \frac{π}{6} + i s i n \frac{π}{6}] = 4 [\frac{\sqrt[]{3}}{2} + i . \frac{1}{2}] \\ = - 2 \sqrt[]{3} + 2 i \\ H e n c e, z = - 2 \sqrt[]{3} + 2 i . \end{array}$

Find $| (1 + i) \frac{(2 + i)}{(3 + i)} |$

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | (1 + i) \frac{(2 + i)}{(3 + i)} | \\ | (1 + i) \frac{(2 + i)}{(3 + i)} \times \frac{3 - i}{3 - i} | = | (1 + i) . \frac{6 - 2 i + 3 i - i^{2}}{9 - i^{2}} | \\ = | \frac{(1 + i) . (7 + i)}{9 + 1} | = | \frac{7 + i + 7 i + i^{2}}{10} | = | \frac{7 + 8 i - 1}{10} | \\ = | \frac{6 + 8 i}{10} | = | \frac{3}{5} + \frac{4}{5} i | = \sqrt[]{{(\frac{3}{5})}^{2} + {(\frac{4}{5})}^{2}} = \sqrt[]{\frac{9}{25} + \frac{16}{25}} = \sqrt[]{\frac{25}{25}} = 1 \\ H e n c e, | (1 + i) \frac{(2 + i)}{(3 + i)} | = 1 \end{array}$

Find the principal argument of ${(1 + i \sqrt[]{3})}^{2}$ .

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : {(1 + i \sqrt[]{3})}^{2} = 1 + i^{2} . 3 + 2 \sqrt[]{3} i \\ = 1 - 3 + 2 \sqrt[]{3} i = - 2 + 2 \sqrt[]{3} i \\ t a n α = | \frac{2 \sqrt[]{3}}{- 2} | [? t a n α = | \frac{I m g (z)}{R e (z)} |] \\ \Rightarrow t a n α = | - \sqrt[]{3} | = \sqrt[]{3} \\ \Rightarrow t a n α = t a n \frac{π}{3} ∴ α = \frac{π}{3} \\ N o w R e (z) < 0 a n d i m a g e (z) > 0 \\ ∴ a r g (z) = π - α = π - \frac{π}{3} = \frac{2 π}{3} \\ H e n c e, t h e p r i n c i p l e a r g = \frac{2 π}{3} . \end{array}$

Where does $z$ lie, if $\frac{z - 5 i}{z + 5 i} = \frac{1}{5} ?$

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | \frac{z - 5 i}{z + 5 i} | = 1 \\ L e t z = x + y i \\ ∴ | \frac{x + y i - 5 i}{x + y i + 5 i} | = 1 \Rightarrow | \frac{x + (y - 5) i}{x + (y + 5) i} | = 1 \\ \Rightarrow | x + (y - 5) i | = | x + (y + 5) i | \\ \Rightarrow x^{2} + {(y - 5)}^{2} = x^{2} + {(y + 5)}^{2} \\ \Rightarrow {(y - 5)}^{2} = {(y + 5)}^{2} \\ \Rightarrow y^{2} + 25 - 10 y = y^{2} + 25 + 10 y \\ \Rightarrow 20 y = 0 \Rightarrow y = 0 \\ H e n c e, z l i e s o n x - a x i s i . e ., r e a l a x i s . \end{array}$

Complex Numbers and Quadratic Equations Objective Type Questions

1. $s i n x + i c o s 2 x$ and $c o s x - i s i n 2 x$ are conjugate to each other for:

(a) $x = n π$

(b) $x = n + \frac{π}{2} + \frac{π}{2}$

(c) $x = 0$

(d) no value of $x$

Sol.

\begin{array}{l} L e t z = s i n x + i c o s 2 x \\ \bar{z} = s i n x - i c o s 2 x \\ B u t w e a r e g i v e n t h a t \bar{z} = c o s x - i s i n 2 x \\ ∴ s i n x - i c o s 2 x = c o s x - i s i n 2 x \\ C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s, w e g e t \\ s i n x = c o s x a n d c o s 2 x = s i n 2 x \\ \Rightarrow t a n x = 1 a n d t a n 2 x = 1 \\ \Rightarrow t a n x = t a n \frac{π}{4} a n d t a n 2 x = t a n \frac{π}{4} \\ ∴ x = n π + \frac{π}{4}, n \in I a n d 2 x = n π + \frac{π}{4} \\ \Rightarrow x = 2 x \Rightarrow 2 x - x = 0 \Rightarrow x = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}

2. The real value of $α$ for which the expression $\frac{1 - i s i n α}{1 + i s i n α}$ is purely real is:

(a) $(n + 1) + \frac{π}{2}$

(b) $\frac{(2 n + 1) π}{2}$

(c) $n π$

(d) None of these, where $n \in ℕ$

Sol.

\begin{array}{l} L e t z = \frac{1 - i s i n α}{1 + 2 i s i n α} = \frac{(1 - i s i n α) (1 - 2 i s i n α)}{(1 + 2 i s i n α) (1 - 2 i s i n α)} \\ = \frac{1 - 2 i s i n α - i s i n α + 2 i^{2} s i n^{2} α}{{(1)}^{2} - {(2 i s i n α)}^{2}} \\ = \frac{1 - 3 i s i n α - 2 s i n^{2} α}{1 - 4 i^{2} s i n^{2} α} = \frac{(1 - 2 s i n^{2} α) - 3 i s i n α}{1 + 4 s i n^{2} α} \\ = \frac{1 - 2 s i n^{2} α}{1 + 4 s i n^{2} α} - \frac{3 s i n α}{1 + 4 s i n^{2} α} . i \\ Since, z i s p u r e l y r e a l, t h e n \\ \frac{3 s i n α}{1 + 4 s i n^{2} α} = 0 \Rightarrow s i n α = 0 \\ S o, α = n π, n \in N . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}

Commonly asked questions

If $z$ is a complex number, then

(a) $∣ z ∣^{2} > ∣ z ∣^{2}$

(b) $∣ z ∣^{2} = ∣ z ∣^{2}$

(c) $∣ z ∣^{2} < ∣ z ∣^{2}$

(d) $∣ z ∣^{2} \geq ∣ z ∣^{2}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z = x + y i \\ | z | = | x + y i | a n d {| z |}^{2} = {| x + y i |}^{2} \\ \Rightarrow {| z |}^{2} = x^{2} + y^{2} \dots (i) \\ N o w z^{2} = x^{2} + y^{2} i^{2} + 2 x y i \\ \Rightarrow z^{2} = x^{2} - y^{2} + 2 x y i \\ | z^{2} | = \sqrt[]{{(x^{2} - y^{2})}^{2} + {(2 x y)}^{2}} = \sqrt[]{x^{4} + y^{4} - 2 x^{2} y^{2} + 4 x^{2} y^{2}} \\ = \sqrt[]{x^{4} + y^{4} + 2 x^{2} y^{2}} = \sqrt[]{{(x^{2} + y^{2})}^{2}} \\ S o | z^{2} | = x^{2} + y^{2} = {| z |}^{2} \\ S o | z^{2} | = {| z |}^{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

$s i n x + i c o s 2 x$ and $c o s x - i s i n 2 x$ are conjugate to each other for:

(a) $x = n π$

(b) $x = n + \frac{π}{2} + \frac{π}{2}$

(c) $x = 0$

(d) no value of $x$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z = s i n x + i c o s 2 x \\ \bar{z} = s i n x - i c o s 2 x \\ B u t w e a r e g i v e n t h a t \bar{z} = c o s x - i s i n 2 x \\ ∴ s i n x - i c o s 2 x = c o s x - i s i n 2 x \\ C o m p a r i n g t h e r e a l a n d i m a g i n a r y p a r t s, w e g e t \\ s i n x = c o s x a n d c o s 2 x = s i n 2 x \\ \Rightarrow t a n x = 1 a n d t a n 2 x = 1 \\ \Rightarrow t a n x = t a n \frac{π}{4} a n d t a n 2 x = t a n \frac{π}{4} \\ ∴ x = n π + \frac{π}{4}, n \in I a n d 2 x = n π + \frac{π}{4} \\ \Rightarrow x = 2 x \Rightarrow 2 x - x = 0 \Rightarrow x = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

The real value of $α$ for which the expression $\frac{1 - i s i n α}{1 + i s i n α}$ is purely real is:

(a) $(n + 1) + \frac{π}{2}$

(b) $\frac{(2 n + 1) π}{2}$

(c) $n π$

(d) None of these, where $n \in ?$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z = \frac{1 - i s i n α}{1 + 2 i s i n α} = \frac{(1 - i s i n α) (1 - 2 i s i n α)}{(1 + 2 i s i n α) (1 - 2 i s i n α)} \\ = \frac{1 - 2 i s i n α - i s i n α + 2 i^{2} s i n^{2} α}{{(1)}^{2} - {(2 i s i n α)}^{2}} \\ = \frac{1 - 3 i s i n α - 2 s i n^{2} α}{1 - 4 i^{2} s i n^{2} α} = \frac{(1 - 2 s i n^{2} α) - 3 i s i n α}{1 + 4 s i n^{2} α} \\ = \frac{1 - 2 s i n^{2} α}{1 + 4 s i n^{2} α} - \frac{3 s i n α}{1 + 4 s i n^{2} α} . i \\ Since, z i s p u r e l y r e a l, t h e n \\ \frac{3 s i n α}{1 + 4 s i n^{2} α} = 0 \Rightarrow s i n α = 0 \\ S o, α = n π, n \in N . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

If $z = x + i y$ lies in the third quadrant, then $\frac{\bar{z}}{\overset{?}{z}}$ also lies in the third quadrant if:

(a) $x > y > 0$

(b) $x < y < 0$

(c) $y < x < 0$

(d) $y > x > 0$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : z = x + i y \\ I f z l i e s i n t h i r d q u a d r a n t . \\ S o x < 0 a n d y < 0 . \\ \bar{z} = x - i y \\ \frac{\bar{z}}{z} = \frac{x - i y}{x + i y} = \frac{x - i y}{x + i y} \times \frac{x - i y}{x - i y} \\ = \frac{x^{2} + i^{2} y^{2} - 2 x y i}{x^{2} - i^{2} y^{2}} = \frac{x^{2} - y^{2} - 2 x y i}{x^{2} + y^{2}} = \frac{x^{2} - y^{2}}{x^{2} + y^{2}} - \frac{2 x y}{x^{2} + y^{2}} i \\ W h e n z l i e s i n t h i r d q u a d r a n t t h e n \frac{\bar{z}}{z} w i l l a l s o b e l i e i n t h i r d q u a d r a n t \\ ∴ \frac{x^{2} - y^{2}}{x^{2} + y^{2}} < 0 a n d \frac{2 x y}{x^{2} + y^{2}} < 0 \\ \Rightarrow x^{2} - y^{2} < 0 a n d 2 x y > 0 \\ \Rightarrow x^{2} < y^{2} a n d x y > 0 \\ S o, x < y < 0 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

The value of $? (z + 3) (\overset{?}{\bar{z}} + 3) ?$ is equivalent to:

(a) $? z + 3 ?^{2}$

(b) $z - 3$

(c) $z^{2} + 3$

(d) None of these

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : (z + 3) (\bar{z} + 3) \\ L e t z = x + y i \\ S o (z + 3) (\bar{z} + 3) = (x + y i + 3) (x - y i + 3) \\ = [(x + 3) + y i] [(x + 3) - y i] \\ = {(x + 3)}^{2} - y^{2} i^{2} = {(x + 3)}^{2} + y^{2} \\ = {| x + 3 + i y |}^{2} = {| z + 3 |}^{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

If $\frac{1 + i}{1 - i^{}} = 1,$ then

(a) $x = 2 n + 1$

(b) $x = 4 n$

(c) $x = 2 n$

(d) $x = 4 n + 1$ , where $n \in ?$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : {(\frac{1 + i}{1 - i})}^{x} = 1 \\ \Rightarrow {(\frac{(1 + i) (1 + i)}{(1 - i) (1 + i)})}^{x} = 1 \Rightarrow {(\frac{1 + i^{2} + 2 i}{1 - i^{2}})}^{x} = 1 \\ \Rightarrow {(\frac{1 - 1 + 2 i}{1 + 1})}^{x} = 1 \Rightarrow {(\frac{2 i}{2})}^{x} = 1 \\ \Rightarrow {(i)}^{x} = {(i)}^{4 n} \Rightarrow x = 4 n, n \in N \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

A real value of $x$ satisfies the equation $(\frac{3 - 4 i x}{3 + 4 i x}) = α - i β (α, β \in R)$ if $α^{2} + β^{2} =$

(a) 1

(b) -1

(c) 2

(d) -2

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : (\frac{3 - 4 i x}{3 + 4 i x}) = α - i β \\ \Rightarrow (\frac{3 - 4 i x}{3 + 4 i x} \times \frac{3 - 4 i x}{3 - 4 i x}) = α - i β \\ \Rightarrow (\frac{9 - 12 i x - 12 i x + 16 i^{2} x^{2}}{9 - 16 i^{2} x^{2}}) = α - i β \\ \Rightarrow \frac{9 - 24 i x - 16 x^{2}}{9 + 16 x^{2}} = α - i β \\ \Rightarrow \frac{9 - 16 x^{2}}{9 + 16 x^{2}} - \frac{24 x}{9 + 16 x^{2}} . i = α - i β \dots (i) \\ \Rightarrow \frac{9 - 16 x^{2}}{9 + 16 x^{2}} + \frac{24 x}{9 + 16 x^{2}} . i = α + i β \dots (i i) \\ M u l t i p l y i n g e q n . (i) a n d (i i) w e g e t \\ \Rightarrow {(\frac{9 - 16 x^{2}}{9 + 16 x^{2}})}^{2} + {(\frac{24 x}{9 + 16 x^{2}})}^{2} = α^{2} + β^{2} \\ \Rightarrow \frac{{(9 - 16 x^{2})}^{2} + {(24 x)}^{2}}{{(9 + 16 x^{2})}^{2}} = α^{2} + β^{2} \\ \Rightarrow \frac{81 + 256 x^{4} - 288 x^{2} + 576 x^{2}}{{(9 + 16 x^{2})}^{2}} = α^{2} + β^{2} \\ \Rightarrow \frac{81 + 256 x^{4} + 288 x^{2}}{{(9 + 16 x^{2})}^{2}} = α^{2} + β^{2} \\ \Rightarrow \frac{{(9 + 16 x^{2})}^{2}}{{(9 + 16 x^{2})}^{2}} = α^{2} + β^{2} \\ S o, α^{2} + β^{2} = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Which of the following is correct for any two complex numbers $z_{1}$ and $z_{2}$ ?

(a) $? z_{1} z_{2} ? = ? z_{1} ? ? z_{2} ?$

(b) $arg (z_{1} z_{2}) = arg (z_{1}) \cdot arg (z_{2})$

(c) $? z_{1} + z_{2} ? = ? z_{1} ? + ? z_{2} ?$

(d) $? z_{1} + z_{2} ? \geq ? ? z_{1} ? - ? z_{2} ? ?$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z_{1} = r_{1} (c o s θ_{1} + i s i n θ_{1}) \\ ∴ | z_{1} | = r_{1} \\ a n d z_{2} = r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ ∴ | z_{2} | = r_{2} \\ z_{1} z_{2} = r_{1} (c o s θ_{1} + i s i n θ_{1}) . r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ = r_{1} r_{2} (c o s θ_{1} + i s i n θ_{1}) . (c o s θ_{2} + i s i n θ_{2}) \\ = r_{1} r_{2} (c o s θ_{1} c o s θ_{2} + i s i n θ_{2} c o s θ_{1} + i s i n θ_{1} c o s θ_{2} + i^{2} s i n θ_{1} s i n θ_{2}) \\ = r_{1} r_{2} [(c o s θ_{1} c o s θ_{2} - s i n θ_{1} s i n θ_{2}) + i (s i n θ_{1} c o s θ_{2} + c o s θ_{1} s i n θ_{2})] \\ = r_{1} r_{2} [c o s (θ_{1} + θ_{2}) + i s i n (θ_{1} + θ_{2})] \\ ∴ | z_{1} z_{2} | = | z_{1} | | z_{2} | \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

The point represented by the complex number $2 - i$ is rotated about the origin through an angle $\frac{π}{2}$ in the clockwise direction, the new position of the point is:

(a) $1 + 2 i$

(b) $- 1 - 2 i$

(c) $2 + i$

(d) $- 1 + 2 i$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : z = 2 - i \\ I f z r o t a t e d t h r o u g h a n a n g l e o f \frac{π}{2} a b o u t t h e o r i g i n i n c l o c k w i s e d i r e c t i o n . \\ T h e n t h e n e w p o s i t i o n = z . e^{- (π / 2)} \\ = (2 - i) . e^{- (π / 2)} \\ = (2 - i) [c o s (- \frac{π}{2}) + i s i n (- \frac{π}{2})] \\ = (2 - i) (0 - i) = - 1 - 2 i \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Let $x, y \in ?$ , then $x + i y$ is a non-real complex number if:

(a) $x = 0$

(b) $y = 0$

(c) $x \neq 0$

(d) $y \neq 0$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} x + y i i s a n o n - r e a l c o m p l e x n u m b e r i f y \neq 0 . I f x, y \in R . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

If $a + i b = c + i d$ , then

(a) $a^{2} + c^{2} = 0$

(b) $b^{2} + c^{2} = 0$

(c) $b^{2} + d^{2} = 0$

(d) $a^{2} + b^{2} = c^{2} + d^{2}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : a + i b = c + i d \\ \Rightarrow | a + i b | = | c + i d | \\ \Rightarrow \sqrt[]{a^{2} + b^{2}} = \sqrt[]{c^{2} + d^{2}} \\ S q u a r i n g b o t h s i d e s, w e g e t a^{2} + b^{2} = c^{2} + d^{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

The complex number $z$ which satisfies the condition $\frac{1 + i z}{1 - i z} = 1$ lies on Circle

(a) Circle $x^{2} + y^{2} = 1$

(b) The x-axis

(c) The y-axis

(d) The line $x + y = 1$ .

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : | \frac{i + z}{i - z} | = 1 \\ L e t z = x + y i \\ ∴ | \frac{i + x + y i}{i - x - y i} | = 1 \Rightarrow | \frac{x + (y + 1) i}{- x - (y - 1) i} | = 1 \\ \Rightarrow | x + (y + 1) i | = | - x - (y - 1) i | \\ \Rightarrow \sqrt[]{x^{2} + {(y + 1)}^{2}} = \sqrt[]{x^{2} + {(y - 1)}^{2}} \\ S q u a r i n g b o t h s i d e s, w e g e t \\ \Rightarrow x^{2} + {(y + 1)}^{2} = x^{2} + {(y - 1)}^{2} \\ \Rightarrow {(y + 1)}^{2} = {(y - 1)}^{2} \\ \Rightarrow y^{2} + 2 y + 1 = y^{2} - 2 y + 1 \Rightarrow 2 y = - 2 y \\ \Rightarrow 4 y = 0 \Rightarrow y = 0 \Rightarrow x - a x i s \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

$? z_{1} + z_{2} ? = ? z_{1} ? + ? z_{2} ?$ is possible if:

(a) $z_{2} = \frac{1}{z_{1}}$

(b) $z_{2} = \frac{1}{{\overset{?}{z}}_{1}}$

(c) $arg (z_{1}) = arg (z_{2})$

(d) $z_{1} = z_{2}$ $(d)$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z_{1} = r_{1} (c o s θ_{1} + i s i n θ_{1}) a n d z_{2} = r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ Since | z_{1} + z_{2} | = | z_{1} | + | z_{2} | \\ z_{1} + z_{2} = r_{1} c o s θ_{1} + i r_{1} s i n θ_{1} + r_{2} c o s θ_{2} + i r_{2} s i n θ_{2} \\ | z_{1} + z_{2} | = \sqrt[]{\begin{array}{l} r_{1}^{2} c o s^{2} θ_{1} + r_{2}^{2} c o s^{2} θ_{2} + 2 r_{1} r_{2} c o s θ_{1} c o s θ_{2} \\ + r_{1}^{2} s i n^{2} θ_{1} + r_{2}^{2} s i n^{2} θ_{2} + 2 r_{1} r_{2} s i n θ_{1} s i n θ_{2} \end{array}} \\ = \sqrt[]{r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} c o s (θ_{1} - θ_{2})} \\ B u t | z_{1} + z_{2} | = | z_{1} | + | z_{2} | \\ S o \sqrt[]{r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} c o s (θ_{1} - θ_{2})} = r_{1} + r_{2} \\ S q u a r i n g b o t h s i d e s, w e g e t \\ r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} c o s (θ_{1} - θ_{2}) = r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} \\ \Rightarrow 2 r_{1} r_{2} - 2 r_{1} r_{2} c o s (θ_{1} - θ_{2}) = 0 \\ \Rightarrow 1 - c o s (θ_{1} - θ_{2}) = 0 \Rightarrow c o s (θ_{1} - θ_{2}) = 1 \\ \Rightarrow θ_{1} - θ_{2} = 0 \Rightarrow θ_{1} = θ_{2} \\ S o, a r g (z_{1}) = a r g (z_{2}) \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

The real value of $θ$ for which the expression $\frac{1 + i c o s θ}{1 - i c o s θ}$ is a real number is:

(a) $\frac{π}{4} + n π$

(b) $n π + \frac{(2 n + 1) π}{4}$

(c) $\pm \frac{π}{2} + n π$

(d) None of these

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z = \frac{1 + i c o s θ}{1 - 2 i c o s θ} = \frac{1 + i c o s θ}{1 - 2 i c o s θ} \times \frac{1 + 2 i c o s θ}{1 + 2 i c o s θ} \\ = \frac{1 + 2 i c o s θ + i c o s θ + 2 i^{2} c o s^{2} θ}{1 - 4 i^{2} c o s^{2} θ} \\ = \frac{1 + 3 i c o s θ - 2 c o s^{2} θ}{1 + 4 c o s^{2} θ} = \frac{1 - 2 c o s^{2} θ}{1 + 4 c o s^{2} θ} + \frac{3 c o s θ}{1 + 4 c o s^{2} θ} i \\ I f z i s a r e a l n u m b e r, t h e n \\ \Rightarrow \frac{3 c o s θ}{1 + 4 c o s^{2} θ} = 0 \\ \Rightarrow 3 c o s θ = 0 \Rightarrow c o s θ = 0 \\ ∴ θ = (2 n + 1) \frac{π}{2}, n \in N . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

The value of $arg (x)$ when $x < 0$ is:

(a) $0$

(b) $\frac{π}{2}$

(c) $π$

(d) None of these

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z = - x + 0 i a n d x < 0 \\ ∴ | z | = \sqrt[]{{(- 1)}^{2} + {(0)}^{2}} = 1, x < 0 \\ Since, t h e point (- x, 0) l i e s o n t h e n e g a t i v e s i d e o f t h e r e a l a x i s (? x < 0) \\ ∴ Principle argument (z) = π \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

If $f (z) = \frac{2 z - 7}{1 - z}$ , where $z = 1 + 2 i$ , then $f (z)$ is:

(a) $\frac{2}{z}$

(b) $z$

(c) $2 z$

(d) None of these

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t z = 1 + 2 i \\ ∴ | z | = \sqrt[]{{(1)}^{2} + {(2)}^{2}} = \sqrt[]{5} \\ N o w f (z) = \frac{7 - z}{1 - z^{2}} \\ = \frac{7 - (1 + 2 i)}{1 - {(1 + 2 i)}^{2}} = \frac{7 - 1 - 2 i}{1 - 1 - 4 i^{2} - 4 i} = \frac{6 - 2 i}{4 - 4 i} \\ = \frac{3 - i}{2 - 2 i} = \frac{3 - i}{2 - 2 i} \times \frac{2 + 2 i}{2 + 2 i} = \frac{6 + 6 i - 2 i - 2 i^{2}}{4 - 4 i^{2}} \\ = \frac{6 + 4 i + 2}{4 + 4} = \frac{8 + 4 i}{8} = 1 + \frac{1}{2} i \\ S o, | f (z) | = \sqrt[]{{(1)}^{2} + {(\frac{1}{2})}^{2}} = \sqrt[]{1 + \frac{1}{4}} = \frac{\sqrt[]{5}}{2} = \frac{| z |}{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Commonly asked questions

Let $λ$ be an integer. IF the shortest distance between the lines $x - λ = 2 y - 1 = - 2 z$ and $x = y + 2 λ = z - λ i s \frac{\sqrt[]{7}}{2 \sqrt[]{2}}$ , then the value of $| λ | i s . . . . . . . . . . . . . . . . . . .$

$L_{1} = \frac{x - λ}{1} = \frac{y - \frac{1}{2}}{\frac{1}{2}} = \frac{z}{- \frac{1}{2}}$

$∴ S D = \frac{| - 2 λ + 3 (2 λ + \frac{1}{2}) + λ |}{\sqrt[]{14}} = \frac{| 5 λ + \frac{3}{2} |}{\sqrt[]{14}}$

$5 λ + \frac{3}{2} = - \frac{7}{2} \Rightarrow 5 λ = - 5 \Rightarrow λ = - 1$

$∴ | λ | = 1$

For integers n and r, let

The maximum value of k for which the sum

$\sum_{i = 0}^{k} (\begin{matrix} 10 & i \end{matrix}) (\begin{matrix} 15 & k - i \end{matrix}) + \sum_{i = 0}^{k + 1} (\begin{matrix} 12 & i \end{matrix}) (\begin{matrix} 13 & k + 1 - i \end{matrix})$

$\sum_{i = 0}^{k} (\begin{array}{l} 10 \\ i \end{array}) (\begin{array}{l} 15 \\ K - i \end{array}) + \sum_{i = 0}^{k + 1} (\begin{array}{l} 12 \\ i \end{array}) (\begin{array}{l} 13 \\ k + 1 - i \end{array})$

= Equating the coefficient of $x^{k} i n {(1 + x)}^{25} =^{25} C_{k}$ ..............(i)

= Equating the coefficient of $x^{k + 1} i n {(1 + x)}^{12} {(x + 1)}^{13}$

From (i) and (ii)

Hence $26 \geq k + 1 \Rightarrow k \leq 25$

But maximum value of k is not defined bonus

Let $i = \sqrt[]{- 1} .$ If $\frac{{(- 1 + i \sqrt[]{3})}^{21}}{{(1 - i)}^{24}} + \frac{{(1 + i \sqrt[]{3})}^{21}}{{(1 + i)}^{24}} = k,$ and n = $[| k |]$ be the greatest integral part of $| k | .$ Then $\sum_{j = 0}^{n + 5} {(j + 5)}^{2} - \sum_{j = 0}^{n + 5} (j + 5)$ is equal to………………..

$\frac{2^{21} {(\frac{- 1 + i \sqrt[]{3}}{2})}^{21}}{{(\sqrt[]{2})}^{24} {(\frac{1 - i}{\sqrt[]{2}})}^{24}} + \frac{2^{21} {(\frac{1 + \sqrt[]{3} i}{2})}^{21}}{{(\sqrt[]{2})}^{24} {(\frac{1 + i}{\sqrt[]{2}})}^{24}} = k$

= $\sum_{j = 0}^{5} (j + 5) (j + 5 - 1) = \sum_{j = 0}^{5} (j + 5) (j + 4)$

= $\sum_{j = 0}^{5} (j^{2} + 9 \hat{j} + 20) = \frac{5 \times 6 \times 11}{6} + \frac{9 \times 5 \times 6}{2} + 20 \times 6$

55 + 135 + 120 = 310

The number of the real roots of the equation (x + 1)² + $| x - 5 | = \frac{27}{4} i s . . . . . . . . . . . . . . . . .$

${(x + 1)}^{2} + | x - 5 | = \frac{27}{4}$

Case l x < 5

$x^{2} + 2 x + 1 - x + 5 = \frac{27}{4}$

$(2 x + 3) (2 x - 1) = 0 \Rightarrow x = - \frac{3}{2}, \frac{1}{2}$

Case II $\geq$ x 5

$x^{2} + 2 x + 1 + x - 5 = \frac{27}{4}$

$x = \frac{- 12 \pm \sqrt[]{144 + 43 \times 16}}{2 \times 4} = \frac{- 12 \pm \sqrt[]{832}}{8} (r e j e c t e d)$

because x > 5

Let a point P be such that its distance from the point (5,0) is thrice the distance of P from the point (-5,0). If the locus of the point P is a circle of radius r, then 4r² is equal to………………….

Let P (h, k)

$\sqrt[]{{(h - 5)}^{2} + k^{2}} = 3 \sqrt[]{{(h + 5)}^{2} + k^{2}}$

$h^{2} - 10 h + 25 + k^{2} = 9 h^{2} + 90 h + 225 + 9 k^{2}$

$8 h^{2} + 8 k^{2} + 100 h + 200 = 0$

$x^{2} + y^{2} + \frac{25}{2} x + 25 = 0$

$r^{2} = \frac{625}{16} - 25 = \frac{625 - 400}{16} = \frac{225}{16}$

$4 r^{2} = \frac{225}{4} = 56.25$

The sum of first four terms of a geometric progression (G. P) is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18} .$ If the product of first three terms of the G.P is 1,and the third term is a, then 2a is………………

Let a, ar, ar², ar³ are in G.P.

a + ar + ar² + ar³ = $\frac{65}{12}$

$\frac{1}{a} + \frac{1}{a r} + \frac{1}{a r^{2}} + \frac{1}{a r^{3}} = \frac{65}{18}$

${(a r)}^{2} r = \frac{3}{2} \Rightarrow a = \frac{2}{3}, r = \frac{3}{2}$

$∴$ third term = ar² = $\frac{2}{3} \times \frac{9}{4} = \frac{3}{2}$

$∴ α = \frac{3}{2} \Rightarrow 2 α = 3$

The students S₁, S₂,…………….S₁₀ are to be divided into 3 groups A, B and C such that each group has at least one student and the group C has at most 3 students. Then the total number of possibilities of forming such groups is……………..

A B C

- - 1

- - 2

- - 3

Number of groups $=^{10} C_{1} (2^{9} - 2) = 5100$

Number of groups $=^{10} C_{2} (2^{8} - 2) = 11430$

Number of groups = $=^{10} C_{3} (2^{7} - 2) = 15120$

Total number of groups = 31650

If a + a = 1, b + b = 2 and af(x) + $α f (\frac{1}{x}) = b x + \frac{β}{x}, x \neq 0,$ then the value of the equation $\frac{f (x) + f (\frac{1}{x})}{x + \frac{1}{x}}$ is………………..

$a f (x) + α f (\frac{1}{x}) = b x + \frac{β}{x} . . . . . . . . . . . . (i)$

Replace x by $\frac{1}{x}$

$a f (\frac{1}{x}) + α f (x) = \frac{b}{x} + β x . . . . . . . . . . . . . . (i i)$

$a (f (x) + f (\frac{1}{x})) + α (f (x) + f (\frac{1}{x})) = b (x + \frac{1}{x}) + β (x + \frac{1}{x})$

$∴ \frac{f (x) + f (\frac{1}{x})}{x + \frac{1}{x}} = \frac{b + β}{a + α} = \frac{2}{1} = 2$

IF the variance of 10 natural numbers 1, 1, 1……… 1, k less than 10, then the maximum possible value of k is…………………

$σ^{2} = \frac{\sum x^{2}}{n} - {(\frac{\sum x}{n})}^{2} = \frac{9 + k^{2}}{10} - {(\frac{9 + k}{10})}^{2} < 10$

$\Rightarrow k < \frac{10 \sqrt[]{10}}{3} + 1 \Rightarrow k \leq 11$

$∴$ maximum value of k is 11

If the area of the triangle formed by the positive x-axis, the normal and the tangent to the circle (x - 2)² + (y – 3)² = 25 at the point (5, 7) is A, then 24A is equal to………………

Equation of normal is 4x – 3y + 1 = 0

Equation of tangent is 3x + 4y – 43 = 0

Area of triangle = $\frac{1}{2} (\frac{43}{3} + \frac{1}{4}) \times 7$

$= \frac{1}{2} \times (\frac{172 + 3}{12}) \times 7 = \frac{1225}{24}$

$∴ 24 A = 1225$

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Commonly asked questions

$\underset{n \to \infty}{l i m} t a n {\sum_{r = 1}^{n} t a n^{- 1} (\frac{1}{1 + r + r^{2}})}$ is equal to___________.

$\sum_{r = 1}^{n} t a n^{- 1} \frac{1}{1 + r + r^{2}} = \sum_{r = 1}^{n} t a n^{- 1} \frac{(r + 1)}{1 + r (r + 1)}$

$= t a n^{- 1} 2 - t a n^{- 1} + . . . . . . + t a n^{- 1} (n + 1) - t a n^{- 1} n$

For n $\infty$ value = $\frac{π}{2} - \frac{π}{4} = \frac{π}{4}$

$∴ t a n (\frac{π}{4}) = 1$

If the least and the largest real values of a, for which the equation z + $| z - 1 | + 2 i = 0 (z \in C a n d i = \sqrt[]{- 1})$ a has a solution, are p and q respectively, then $4 (p^{2} + q^{2})$ is equal to

$z + α | z - 1 | + 2 i = 0$

Let z = x + iy

$\Rightarrow x + i (y + 2) = - α \sqrt[]{{(x - 1)}^{2} + y^{2}}$

for $α \in R y + 2 = 0 \Rightarrow y = - 2$

$x^{2} = α^{2} [{(x - 1)}^{2} + 4]$ = 0

$\frac{1}{α^{2}} \geq \frac{4}{5} \Rightarrow α^{2} \leq \frac{5}{4} \Rightarrow \frac{- \sqrt[]{5}}{2} \leq α \leq \frac{\sqrt[]{5}}{2}$

$4 (p^{2} + q^{2}) = 4 (\frac{5}{4} + \frac{5}{4}) = 10$

Let M be any 3 × 3 matrix with entires from the set {0, 1, 2}. The maximum number of such matrices, for which the sum of diagonal elements M^TM is seven is____________.

$M = [\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & c_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}]$

$M^{T} M = [\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix}] [\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}]$

$T r (M^{T} M) = a_{1}^{2} + b_{1}^{2} + c_{1}^{2} + a_{2}^{2} + b_{2}^{2} + c_{2}^{2} + a_{3}^{2} + b_{3}^{2} + c_{3}^{2} = 7$

all $a_{i}, b_{i}, c_{i} \in {0, 1, 2} f o r$ i = 1, 2, 3

Case 1 7 one’s and two zeroes which can occur in ways

Case 2 One 2 three 1’s five zeroes =

total such matrices = 504 + 36 = 540

The minimum values of a for which the equation $\frac{4}{s i n x} + \frac{1}{1 - s i n x} = α$ has at least one solution in $(0, \frac{π}{2}) i s_____.$

$\frac{4}{s i n x} + \frac{1}{1 - s i n x} = a \forall x \in (0, \frac{π}{2})$

Let sin x = t, t $\in$ (0, 1)

$g (t) = \frac{4}{t} + \frac{1}{1 - t}$

$g' (t) = 0 \Rightarrow t = \frac{2}{3}$

$g'' (\frac{2}{3}) > 0$

$∴ g {(t)}_{m i n i m u m} = \frac{4}{2 / 3} + \frac{1}{1 - 2 / 3} = 9$

Minimum value of a for which solution exist = 9

$\underset{n \to \infty}{l i m} t a n {\sum_{r = 1}^{n} t a n^{- 1} (\frac{1}{1 + r + r^{2}})}$ is equal to___________.

$\sum_{r = 1}^{n} t a n^{- 1} \frac{1}{1 + r + r^{2}} = \sum_{r = 1}^{n} t a n^{- 1} \frac{(r + 1)}{1 + r (r + 1)}$

$= t a n^{- 1} 2 - t a n^{- 1} + . . . . . . + t a n^{- 1} (n + 1) - t a n^{- 1} n$

For n $\infty$ value = $\frac{π}{2} - \frac{π}{4} = \frac{π}{4}$

$∴ t a n (\frac{π}{4}) = 1$

$z + α | z - 1 | + 2 i = 0$

Let z = x + iy

$\Rightarrow x + i (y + 2) = - α \sqrt[]{{(x - 1)}^{2} + y^{2}}$

for $α \in R y + 2 = 0 \Rightarrow y = - 2$

$x^{2} = α^{2} [{(x - 1)}^{2} + 4]$ = 0

$\frac{1}{α^{2}} \geq \frac{4}{5} \Rightarrow α^{2} \leq \frac{5}{4} \Rightarrow \frac{- \sqrt[]{5}}{2} \leq α \leq \frac{\sqrt[]{5}}{2}$

$4 (p^{2} + q^{2}) = 4 (\frac{5}{4} + \frac{5}{4}) = 10$

Let M be any 3 × 3 matrix with entires from the set {0, 1, 2}. The maximum number of such matrices, for which the sum of diagonal elements M^TM is seven is____________.

$M = [\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & c_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}]$

$T r (M^{T} M) = a_{1}^{2} + b_{1}^{2} + c_{1}^{2} + a_{2}^{2} + b_{2}^{2} + c_{2}^{2} + a_{3}^{2} + b_{3}^{2} + c_{3}^{2} = 7$

all $a_{i}, b_{i}, c_{i} \in {0, 1, 2} f o r$ i = 1, 2, 3

Case 1 7 one’s and two zeroes which can occur in ways

Case 2 One 2 three 1’s five zeroes =

total such matrices = 504 + 36 = 540

The minimum values of a for which the equation $\frac{4}{s i n x} + \frac{1}{1 - s i n x} = α$ has at least one solution in $(0, \frac{π}{2}) i s_____.$

$\frac{4}{s i n x} + \frac{1}{1 - s i n x} = a \forall x \in (0, \frac{π}{2})$

Let sin x = t, t $\in$ (0, 1)

$g (t) = \frac{4}{t} + \frac{1}{1 - t}$

$g' (t) = 0 \Rightarrow t = \frac{2}{3}$

$g'' (\frac{2}{3}) > 0$

$∴ g {(t)}_{m i n i m u m} = \frac{4}{2 / 3} + \frac{1}{1 - 2 / 3} = 9$

Minimum value of a for which solution exist = 9

Let A = ${n \in N : n i s a 3 - d i g i t n u m b e r}$

B = ${9 k + 2 : k \in N}$

And C = ${9 k + l : k \in N}$ for some $l (0 < l < 9)$

If the sum of all the elements of the set $A \cap (B \cup C)$ is 274 × 400, then $l$ is equal to___________.

Sum of elements $A \cap (B \cup C) = 274 \times 400$

In set B numbers of the form 9k + 2 are {101, 109, ....992}

$∴ s u m = \frac{100}{2} (101 + 992) = \frac{100 \times 1093}{2} . . . . . . (i)$

Another possible number is 9k + 5 forms are {104, ..........995}

$∴ s u m = \frac{100}{2} (104 + 995) = \frac{100}{2} \times 1099 . . . . . . . . . (i i)$

$∴ T o t a l = \frac{100}{2} \times [1093 + 1099] = 100 \times 1096 = 274 \times 4 \times 100 = 274 \times 400$

$∴$ possible value of $l$ = 5

If one of the diameters of the circle x² + y² – 2x – 6y + 6 = 0 is a chord of another circle ‘C’, whose centre is at (2, 1), then its radius is_________

$x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ centre (1, 3)

$\begin{array}{l} r = \sqrt[]{1 + 9 - 6} = 2 \\ C M = \sqrt[]{1 + 4} = \sqrt[]{5} \end{array}$

$r = \sqrt[]{5 + 4} = 3$

Let B_i (i = 1, 2, 3) be three independent events in a sample space. The probability that only B₁ occurs is $γ .$ a, only B₂ occurs is b and B₃ occurs is Let p be the probability that none of the events B₁ occurs and these 4 probabilities satisfy the equations (a - 2b) = ab and (b - 3 $γ)$ p = $2 β γ$ (All the probabilities are assumed to lie in the integral (0, 1)). The $\frac{P (B_{1})}{P (B_{3})}$ is equal to_________.

Let P (B₁) = a P (B₂) = b P (B₃) = c

Given a (1 – b) (1 – c) = a . (i)

b (1 – a) (1 – c) = b . (ii)

c (1 – b) (1 – a) = $γ$ . (iii)

(1 – a) (1 – b) (1 – c) = p . (iv)

$(α - 2 β) p = α β$

->a – ab – 2b + 2ab = ab Þ a = 2b . (v)

Again $(β - 3 γ) p = 2 β γ$

->b – bc – 3c + 3bc = 2bc Þ b = 3c . (vi)

$\Rightarrow \frac{P (B_{1})}{P (B_{3})} = \frac{a}{c} = \frac{2 b}{b / 3} = 6$

Let three vectors $\vec{a}, \vec{b} a n d \vec{c}$ be such that $\vec{c}$ is coplanar with $\vec{a} a n d \vec{b}, \vec{a} . \vec{c} = 7 a n d \vec{b}$ is perpendicular to $\vec{c}, w h e r e \vec{a} = - \hat{i} + \hat{j} + \hat{k} a n d \hat{b} = 2 \hat{i} + \hat{k},$ then the value of 2 ${| \vec{a} + \vec{b} + \vec{c} |}^{2}$ is____________.

$\vec{c} = α \vec{a} + β \vec{b} . . . . . (i)$

$\vec{a} . \vec{c} = 7 \vec{b} . \vec{c} = 0$

$\vec{a} = - \hat{i} + \hat{j} + \hat{k} \Rightarrow | \vec{a} | = \sqrt[]{3}$

$\vec{b} = 2 \hat{i} + \hat{k} \Rightarrow | \vec{b} | = \sqrt[]{5} \vec{a} . \vec{b} = - 2 + 1 = - 1$

From $(i) \vec{a} . \vec{c} = α {| \vec{a} |}^{2} - β$

$3 α - β = 7 . . . . . . . . . . (i i)$

$\bar{b} . \bar{c} = α \bar{b} . \bar{a} + β {| \vec{b} |}^{2} \Rightarrow - α + 5 β = 0 . . . . . . . (i i i)$

Solving $α = \frac{5}{2} a n d β = \frac{1}{2}$

$\Rightarrow 2 {| \vec{a} + \vec{b} + \vec{c} |}^{2} = 75$

Let P = $[\begin{matrix} 3 & - 1 & - 2 \\ 2 & 0 & α \\ 3 & - 5 & 0 \end{matrix}] w h e r e α \in R .$ Suppose Q = [q_ij] is a matrix satisfying PQ = kI₃ for some non-zero k $\in$ R. If $q_{23} = \frac{- k}{8} a n d | Q | = \frac{k^{2}}{2}, t h e n α^{2} + k^{2}$ is equal to_________.

$P = [\begin{matrix} 3 & - 1 & - 2 \\ 2 & 0 & α \\ 3 & - 5 & 0 \end{matrix}] a n d Q = [q_{i j}] \Rightarrow P Q = k l_{3}$

$q_{23} = - \frac{k}{8} a n d | Q | = \frac{k^{2}}{2}$

$P Q = k l_{3} \Rightarrow P^{- 1} = \frac{Q}{k} = {(\begin{matrix} 3 & - 1 & - 2 \\ 2 & 0 & α \\ 3 & - 5 & 0 \end{matrix})}^{- 1}$

$| p | | Q | = (k l_{3}) \Rightarrow 8 . \frac{k^{2}}{2} = k^{3}$

$k \neq 0 \Rightarrow k = 4$

$∴ α^{2} + k^{2} = 1 + 16 = 17 .$

If $\int_{- a}^{a} (| x | + | x - 2 |) d x = 22, (a > 2) a n d [x]$ denotes the greatest integer $\leq x,$ then $\int_{a}^{- a} (x + [x]) d x$ is equal to__________.

$\int_{- a}^{a} (| x | + | x - 2 |) d x = 22, a > 2$

$\int_{- a}^{0} (- 2 x + 2) d x + \int_{0}^{2} (x - x + 2) d x + \int_{2}^{a} (2 x - 2) d x = 22$

$\Rightarrow 2 a^{2} + 2 = 20 \Rightarrow a^{2} = 9 \Rightarrow a = 3$

$∴ \int_{3}^{- 3} (x + [x]) d x = - \int_{- 3}^{3} (2 x - {x}) d x = - \int_{- 3}^{3} 2 x d x + 6 {\int_{0}^{1} x d x = 6 . \frac{x^{2}}{2} |}_{0}^{1} = 3$

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