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Complex Numbers and Quadratic Equations Maths NCERT Exemplar Solutions Class 11th Chapter Five

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

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Payal Gupta | Contributor-Level 10

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This is an Objective Type Questions as classified in NCERT Exemplar

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

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Complex Numbers and Quadratic Equations Maths NCERT Exemplar Solutions Class 11th Chapter Five

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

(a) $0$  

(b) $\frac{π}{2}$  

(c) $π$  

(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

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

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Complex Numbers and Quadratic Equations Maths NCERT Exemplar Solutions Class 11th Chapter Five

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

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Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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Complex Numbers and Quadratic Equations Exemplar Solution

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

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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Complex Numbers and Quadratic Equations Maths NCERT Exemplar Solutions Class 11th Chapter Five

If $z$ is a complex number, then

(a) $∣ z ∣^{2} > ∣ z ∣^{2}$

(b) $∣ z ∣^{2} = ∣ z ∣^{2}$

(c) $∣ z ∣^{2} < ∣ z ∣^{2}$

(d) $∣ z ∣^{2} \geq ∣ z ∣^{2}$

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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Complex Numbers and Quadratic Equations 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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Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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Complex Numbers and Quadratic Equations 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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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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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Complex Numbers and Quadratic Equations 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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Complex Numbers and Quadratic Equations 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

a year ago

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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Complex Numbers and Quadratic Equations 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) $? z_{1} + z_{2} ? \geq ? ? z_{1} ? - ? z_{2} ? ?$

4 Views|Posted a year ago

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

If $z$ is a complex number, then

(a) $∣ z ∣^{2} > ∣ z ∣^{2}$

(b) $∣ z ∣^{2} = ∣ z ∣^{2}$

(c) $∣ z ∣^{2} < ∣ z ∣^{2}$

(d) $∣ z ∣^{2} \geq ∣ z ∣^{2}$

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) $- 1 + 2 i$