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Complex Numbers and Quadratic Equations

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New answer posted

3 months ago

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

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Payal Gupta

Contributor-Level 10

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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New answer posted

3 months ago

Complex Numbers and Quadratic Equations 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

This is an Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted

3 months ago

Complex Numbers and Quadratic Equations 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

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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New answer posted

3 months ago

Complex Numbers and Quadratic Equations 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

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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New answer posted

3 months ago

Complex Numbers and Quadratic Equations 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

This is an Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted

3 months ago

Complex Numbers and Quadratic Equations 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

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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New answer posted

3 months ago

Complex Numbers and Quadratic Equations Exemplar Solution

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

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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New answer posted

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Complex Numbers and Quadratic Equations Exemplar Solution

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

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Payal Gupta

Contributor-Level 10

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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New answer posted

3 months ago

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

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

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Payal Gupta

Contributor-Level 10

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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New answer posted

3 months ago

Complex Numbers and Quadratic Equations 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

This is Other Questions as classified in NCERT Exemplar

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

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