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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

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 $x + y = 1$ .

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

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

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Payal Gupta | Contributor-Level 10

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This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

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) $y \neq 0$

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Payal Gupta | Contributor-Level 10

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This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

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) $- 1 + 2 i$

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Payal Gupta | Contributor-Level 10

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This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

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

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Payal Gupta | Contributor-Level 10

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This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

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

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Payal Gupta | Contributor-Level 10

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This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

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 $n \in ?$

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Payal Gupta | Contributor-Level 10

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This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

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

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Payal Gupta | Contributor-Level 10

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This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

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) $y > x > 0$

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Payal Gupta | Contributor-Level 10

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This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

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 $n \in ?$

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Payal Gupta | Contributor-Level 10

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This is an Objective Type Questions as classified in NCERT Exemplar

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

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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) $- 1 + 2 i$

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 $n \in ?$