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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

$s i n x + i c o s 2 x$ 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 $x$

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Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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Exemplar Solution Complex Numbers and Quadratic Equations

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

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Payal Gupta | Contributor-Level 10

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This is Other Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

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

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Asked by Payal Gupta

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Payal Gupta | Contributor-Level 10

a year ago

This is Other Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

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

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Payal Gupta | Contributor-Level 10

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This is Other Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

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

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Asked by Payal Gupta

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Payal Gupta | Contributor-Level 10

a year ago

This is Other Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

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

3 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is Other Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

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

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Payal Gupta | Contributor-Level 10

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This is Other Questions as classified in NCERT Exemplar

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

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Exemplar Solution Complex Numbers and Quadratic Equations

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

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is Other Questions as classified in NCERT Exemplar

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Five

If $\frac{{(a^{2} + 1)}^{2}}{2 a - i} = x + i y,$ what is the value of $x^{2} + y^{2}$ ?

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Payal Gupta | Contributor-Level 10

a year ago

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : \frac{{(a^{2} + 1)}^{2}}{2 a - i} = x + y i \dots (i) \\ T a k i n g c o n j u g a t e o n b o t h s i d e s \\ \frac{{(a^{2} + 1)}^{2}}{2 a + i} = x - y i \dots (i i) \\ M u l t i p l y i n g e q n . (i) a n d (i i) w e h a v e \\ \frac{{(a^{2} + 1)}^{2} {(a^{2} + 1)}^{2}}{(2 a - i) (2 a + i)} = x^{2} + y^{2} \\ \Rightarrow \frac{{(a^{2} + 1)}^{4}}{4 a^{2} - i^{2}} = x^{2} + y^{2} \\ \Rightarrow \frac{{(a^{2} + 1)}^{4}}{4 a^{2} + 1} = x^{2} + y^{2} \\ H e n c e, t h e v a l u e o f x^{2} + y^{2} = \frac{{(a^{2} + 1)}^{4}}{4 a^{2} + 1} . \end{array}$

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Exemplar Solution Complex Numbers and Quadratic Equations

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

3 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is Other Questions as classified in NCERT Exemplar

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

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$s i n x + i c o s 2 x$ 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 $x$

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

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

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

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

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

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

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

If $\frac{{(a^{2} + 1)}^{2}}{2 a - i} = x + i y,$ what is the value of $x^{2} + y^{2}$ ?

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