Math Ncert Exemplar 2019 Solutions for Class 11 Science Maths Chapter 5 - Complex Numbers And Quadratic Equations

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Page No 91:

Question 1:

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

Answer:

We have, ${(1 - i)}^{n} {(1 - \frac{1}{i})}^{n}$ .
$= {[(1 - i) (1 - \frac{1}{i})]}^{n} = {[(1 - i) (1 - (- i))]}^{n} [∵ \frac{1}{i} = - i] = {[(1 - i) (1 + i)]}^{n} = {[1 - i^{2}]}^{n} = {[1 - (- 1)]}^{n} = 2^{n}$

Hence, ${(1 - i)}^{n} {(1 - \frac{1}{i})}^{n} = 2^{n}$ .

Page No 91:

Question 2:

Evaluate $\sum_{n = 1}^{13} (i^{n} + i^{n + 1})$ , where n ∈ N.

Answer:

We have, $\sum_{n = 1}^{13} (i^{n} + i^{n + 1})$
$= \sum_{i = 1}^{13} (1 + i) i^{n} = (1 + i) (i + i^{2} + i^{3} + i^{4} + . . . + i^{13}) = (1 + i) (i - 1 - i + 1 + i - 1 - i + 1 + i - 1 - i + 1 + i) = (1 + i) i = i + i^{2} = i - 1$

Hence, $\sum_{n = 1}^{13} (i^{n} + i^{n + 1}) = i - 1$ .

Page No 91:

Question 4:

If $\frac{{(1 + i)}^{2}}{2 - i} = x + i y$ , then find the value of x + y.

Answer:

We have,
$\begin{array}{rcl} x + i y & = & \frac{{(1 + i)}^{2}}{2 - i} \\ = & \frac{1 + i^{2} + 2 i}{2 - i} \\ = & \frac{1 - 1 + 2 i}{2 - i} \times \frac{2 + i}{2 + i} \\ = & \frac{2 i (2 + i)}{2^{2} - i^{2}} \\ = & \frac{4 i + 2 i^{2}}{4 + 1} \\ = & - \frac{2}{5} + \frac{4}{5} i \end{array}$
$\Rightarrow x = - \frac{2}{5}, y = \frac{4}{5} \Rightarrow x + y = - \frac{2}{5} + \frac{4}{5} = \frac{2}{5}$
Hence, the value of x + y is $\frac{2}{5}$ .

Page No 91:

Question 3:

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

Answer:

We have,
$\begin{array}{rcl} x + i y & = & {(\frac{1 + i}{1 - i})}^{3} - {(\frac{1 - i}{1 + i})}^{3} \\ = & {(\frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i})}^{3} - {(\frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i})}^{3} \\ = & {(\frac{1 + i^{2} + 2 i}{1 - i^{2}})}^{3} - {(\frac{1 + i^{2} - 2 i}{1 - i^{2}})}^{3} \\ = & {(\frac{1 - 1 + 2 i}{1 + 1})}^{3} - {(\frac{1 - 1 - 2 i}{1 + 1})}^{3} \\ = & {(i)}^{3} - {(- i)}^{3} \\ = & i^{3} + i^{3} \\ = & 2 i^{3} \\ = & 0 - 2 i \end{array}$
Thus, x = 0 and y = −2.
Hence, (x, y) = (0, −2).

Page No 91:

Question 5:

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

Answer:

We have,
$\begin{array}{rcl} a + i b & = & {(\frac{1 - i}{1 + i})}^{100} \\ = & {(\frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i})}^{100} \\ = & {(\frac{1 + i^{2} - 2 i}{1 - i^{2}})}^{100} \\ = & {(\frac{1 - 1 - 2 i}{1 + 1})}^{100} \\ = & {(- i)}^{100} \\ = & i^{100} \\ = & {(i^{4})}^{25} \\ = & 1^{25} (∵ i^{4} = 1) \\ = & 1 + 0 i \end{array}$
Thus, a = 1 and b = 0.
Hence, (a, b) = (1, 0).

Page No 91:

Question 6:

If a = cos θ + i sin θ, find the value of $\frac{1 + a}{1 - a} .$

Answer:

Given: a = cos θ + i sin θ

$\begin{array}{rcl} ∴ \frac{1 + a}{1 - a} & = & \frac{1 + \cos θ + i \sin θ}{1 - \cos θ - i \sin θ} \\ = & \frac{1 + \cos θ + i \sin θ}{1 - \cos θ - i \sin θ} \times \frac{1 - \cos θ + i \sin θ}{1 - \cos θ + i \sin θ} \\ = & \frac{{(1 + i \sin θ)}^{2} - {(\cos θ)}^{2}}{{(1 - \cos θ)}^{2} - {(i \sin θ)}^{2}} \\ = & \frac{1 + i^{2} \sin^{2} θ + 2 i \sin θ - \cos^{2} θ}{1 + \cos^{2} θ - 2 \cos θ - i^{2} \sin^{2} θ} \\ = & \frac{1 - \sin^{2} θ - \cos^{2} θ + 2 i \sin θ}{1 + \sin^{2} θ + \cos^{2} θ - 2 \cos θ} (∵ i^{2} = - 1) \\ = & \frac{1 - 1 + 2 i \sin θ}{1 + 1 - 2 \cos θ} (∵ \sin^{2} θ + \cos^{2} θ = 1) \\ = & \frac{i \sin θ}{1 - \cos θ} \\ = & \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} \frac{θ}{2}} i (∵ \begin{matrix} \sin 2 θ = 2 \sin θ \cos θ \\ 1 - \cos 2 θ = 2 \sin^{2} θ \end{matrix}) \\ = & \cot \frac{θ}{2} i \end{array}$
Hence, $\frac{1 + a}{1 - a} = i \cot \frac{θ}{2}$ .

Page No 91:

Question 7:

If (1 + i) z = (1 – i) $\bar{z}$ , then show that $z = - i z$ .

Answer:

Given: (1 + i) z = (1 – i) $\bar{z}$
$\begin{array}{rcl} \Rightarrow & \frac{z}{\bar{z}} = \frac{1 - i}{1 + i} \\ = & \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} \\ = & \frac{1 + i^{2} - 2 i}{1 - i^{2}} \\ = & \frac{1 - 1 - 2 i}{1 + 1} \\ = & - i \end{array}$
$\begin{array}{rcl} ∴ z & = & i \bar{z} \end{array}$
Hence proved.

Page No 91:

Question 8:

If z = x + iy, then show that z $\bar{z}$ + 2 (z + $\bar{z}$ ) + b = 0, where b ∈ R, represents a circle.

Answer:

Given that z = x + iy.
Consider: z $\bar{z}$ + 2 (z + $\bar{z}$ ) + b = 0
$\Rightarrow (x + i y) (x - i y) + 2 (x + i y + x - i y) + b = 0 \Rightarrow x^{2} + y^{2} + 2 (2 x) + b = 0 \Rightarrow x^{2} + 4 x + y^{2} + b = 0$
represents a circle.
Hence proved.

Page No 91:

Question 9:

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

Answer:

Let z = x + iy.
$\Rightarrow \bar{z} = x - i y$
$\begin{array}{rcl} ∴ \frac{\bar{z} + 2}{\bar{z} - 1} & = & \frac{x - i y + 2}{x - i y - 1} \\ = & \frac{(x + 2) - i y}{(x - 1) - i y} \\ = & \frac{(x + 2) - i y}{(x - 1) - i y} \times \frac{(x - 1) + i y}{(x - 1) + i y} \\ = & \frac{(x + 2) (x - 1) + i y (x + 2) - i y (x - 1) - i^{2} y^{2}}{{(x - 1)}^{2} - i^{2} y^{2}} \\ = & \frac{x^{2} + x - 2 + y^{2} + i y (x + 2 - x + 1)}{{(x - 1)}^{2} + y^{2}} \\ = & \frac{x^{2} + x - 2 + y^{2}}{{(x - 1)}^{2} + y^{2}} + i \frac{3 y}{{(x - 1)}^{2} + y^{2}} \end{array}$
Since, the real part of $\frac{\bar{z} + 2}{\bar{z} - 1}$ is 4.
$∴ \begin{array}{rcl} \frac{x^{2} + x - 2 + y^{2}}{{(x - 1)}^{2} + y^{2}} \end{array} = 4 \Rightarrow x^{2} + y^{2} + x - 2 = 4 [{(x - 1)}^{2} + y^{2}] \Rightarrow x^{2} + y^{2} + x - 2 = 4 x^{2} + 4 y^{2} - 8 x + 4 \Rightarrow 3 x^{2} + 3 y^{2} - 9 x + 6 = 0 \Rightarrow x^{2} + y^{2} - 3 x + 2 = 0 \Rightarrow x^{2} - 3 x + y^{2} = - 2 \Rightarrow x^{2} - 3 x + \frac{9}{4} + y^{2} = - 2 + \frac{9}{4} \Rightarrow {(x - \frac{3}{2})}^{2} + y^{2} = \frac{1}{4}$
which represents a circle.
Hence, the locus of z is a circle.

Page No 91:

Question 10:

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

Answer:

Let z = x + iy.
Given: arg $(\frac{z - 1}{z + 1}) = \frac{π}{4}$
$\Rightarrow \arg (z - 1) - \arg (z + 1) = \frac{π}{4} \Rightarrow \arg (x + i y - 1) - \arg (x + i y + 1) = \frac{π}{4} \Rightarrow \arg [(x - 1) + i y] - \arg [(x + 1) + i y] = \frac{π}{4} . . . . . (1)$
Let the arguments of $(x - 1) + i y be θ_{1} and (x + 1) + i y be θ_{2}$ .
From (1), $θ_{1} - θ_{2} = \frac{π}{4} . . . . . (2)$
$∴ \tan θ_{1} = \frac{y}{x - 1} and \tan θ_{2} = \frac{y}{x + 1}$
$\Rightarrow \tan (θ_{1} - θ_{2}) = \frac{\tan (θ_{1}) - \tan (θ_{2})}{1 + \tan (θ_{1}) \tan (θ_{2})} \Rightarrow \tan (\frac{π}{4}) = \frac{\frac{y}{x - 1} - \frac{y}{x + 1}}{1 + \frac{y}{x - 1} \times \frac{y}{x + 1}} [From (2)] \Rightarrow 1 = \frac{y (x + 1) - y (x - 1)}{(x^{2} - 1) + y^{2}} \Rightarrow 1 = \frac{y (x + 1 - x + 1)}{x^{2} + y^{2} - 1} \Rightarrow 1 = \frac{2 y}{x^{2} + y^{2} - 1} \Rightarrow x^{2} + y^{2} - 1 = 2 y \Rightarrow x^{2} + y^{2} - 2 y - 1 = 0$
which represents a circle.

Page No 91:

Question 11:

Solve the equation |z| = z + 1 + 2i.

Answer:

Given: $|z| = z + 1 + 2 i$
Let $z = x + i y$
Squaring both sides, we get

$\Rightarrow {|z|}^{2} = {|z + 1|}^{2} + 4 i^{2} + 4 (z + 1) i \Rightarrow {|z|}^{2} = {|z|}^{2} + 1 + 2 z - 4 + 4 (z + 1) i \Rightarrow 0 = - 3 + 2 z + 4 (z + 1) i \Rightarrow 3 - 2 z - 4 (z + 1) i = 0$
$\Rightarrow 3 - 2 (x + i y) - 4 (x + i y + 1) i = 0 \Rightarrow 3 - 2 x - 2 y i - 4 x i - 4 y i^{2} - 4 i = 0 \Rightarrow 3 - 2 x + 4 y - 2 y i - 4 i - 4 x i = 0 \Rightarrow (3 - 2 x + 4 y) - i (2 y + 4 x + 4) = 0$
So,
$3 - 2 x + 4 y = 0 \Rightarrow 2 x - 4 y = 3 . . . . . (1)$
$And, 2 y + 4 x + 4 = 0 \Rightarrow y + 2 x + 2 = 0 \Rightarrow 2 x + y = - 2 . . . . . (2)$
Solving (1) and (2), we get
$y = - 1$ and $x = - \frac{1}{2}$
Hence, the value of $z = x + i y = - \frac{1}{2} - i$ .

Page No 92:

Question 12:

If |z + 1| = z + 2(1 + i), then find z.

Answer:

Given: $|z + 1| = z + 2 (1 + i)$
Let $z = x + i y$
So, $|z + 1| = z + 2 (1 + i)$
$\Rightarrow |x + i y + 1| = (x + i y) + 2 (1 + i) \Rightarrow |x + i y + 1| = x + i y + 2 + 2 i \Rightarrow |(x + 1) + i y| = (x + 2) + (2 + y) i \Rightarrow \sqrt{{(x + 1)}^{2} + y^{2}} = (x + 2) + i (2 + y) [∵ |x + i y| = \sqrt{x^{2} + y^{2}}]$
Squaring on both sides, we get
$\Rightarrow {(x + 1)}^{2} + y^{2} = {(x + 2)}^{2} + {(y + 2)}^{2} \times i^{2} + 2 (x + 2) (y + 2) i \Rightarrow x^{2} + 1 + 2 x + y^{2} = x^{2} + 4 + 4 x - y^{2} - 4 - 4 y + i 2 (x + 2) (y + 2)$
Comparing the real and imaginary parts, we get
$x^{2} + 1 + 2 x + y^{2} = x^{2} + 4 x - y^{2} - 4 y$ and $2 (x + 2) (y + 2) = 0$

$\Rightarrow 2 y^{2} - 2 x + 4 y + 1 = 0$ $. . . . . (1)$
and $(x + 2) (y + 2) = 0$ $. . . . . (2)$
$x + 2 = 0$ or $y + 2 = 0$
$∴ x = - 2$ or $y = - 2$
Now substituting $x = - 2$ in (1), we get
$2 y^{2} - 2 \times (- 2) + 4 y + 1 = 0 \Rightarrow 2 y^{2} + 4 + 4 y + 1 = 0 \Rightarrow 2 y^{2} + 4 y + 5 = 0$
$\begin{array}{rcl} ∴ b^{2} - 4 a c & = & {(4)}^{2} - 4 \times 2 \times 5 \\ = & 16 - 40 \\ = & - 24 < 0 \end{array}$
i.e. y has no real roots.
Now substituting $y = - 2$ in (1), we get
$2 {(- 2)}^{2} - 2 x + 4 (- 2) + 1 = 0 \Rightarrow 8 - 2 x - 8 + 1 = 0 \Rightarrow x = \frac{1}{2}$
So, $x = \frac{1}{2}$ and $y = - 2$ .
Hence, $z = x + i y = \frac{1}{2} - 2 i$ .

Page No 92:

Question 13:

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

Answer:

We have, arg (z – 1) = arg (z + 3i), where z = x + iy
⇒ arg (x + iy – 1) = arg (x + iy + 3i)
⇒ arg (x – 1 + iy) = arg [x + i(y + 3)] .....(1)
Let the arguments of $(x - 1) + i y be θ_{1} and x + i (y + 3) be θ_{2}$ .
From (1), $θ_{1} = θ_{2} . . . . . (2)$
$∴ \tan θ_{1} = \tan θ_{2} \Rightarrow \frac{y}{x - 1} = \frac{y + 3}{x} \Rightarrow x y = (x - 1) (y + 3) \Rightarrow x y = x y + 3 x - y - 3 \Rightarrow 3 x - 3 = y \Rightarrow \frac{x - 1}{y} = \frac{1}{3} \Rightarrow (x - 1) : y = 1 : 3$
Hence, (x – 1) : y = 1 : 3.

Page No 92:

Question 14:

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

Answer:

Let z = x + iy.
Given, $|\frac{z - 2}{z - 3}| = 2$ .
$\Rightarrow |\frac{x + i y - 2}{x + i y - 3}| = 2 \Rightarrow |x - 2 + i y| = 2 |x - 3 + i y| \Rightarrow \sqrt{{(x - 2)}^{2} + y^{2}} = 2 \sqrt{{(x - 3)}^{2} + y^{2}} \Rightarrow x^{2} - 4 x + 4 + y^{2} = 4 (x^{2} - 6 x + 9 + y^{2}) \Rightarrow 3 x^{2} + 3 y^{2} - 20 x + 32 = 0 \Rightarrow x^{2} + y^{2} - \frac{20}{3} x + \frac{32}{3} = 0 \Rightarrow x^{2} - \frac{20}{3} x + {(\frac{10}{3})}^{2} + y^{2} = {(\frac{10}{3})}^{2} - \frac{32}{3} \Rightarrow {(x - \frac{10}{3})}^{2} + y^{2} = {(\frac{2}{3})}^{2}$
Hence, the centre of the circle is $(\frac{10}{3}, 0)$ and radius is $\frac{2}{3}$ units.

Page No 92:

Question 15:

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

Answer:

Let z = x + iy.
$\begin{array}{rcl} ∴ \frac{z - 1}{z + 1} & = & \frac{x + i y - 1}{x + i y + 1} \\ = & \frac{x - 1 + i y}{x + 1 + i y} \\ = & \frac{x - 1 + i y}{x + 1 + i y} \times \frac{x + 1 - i y}{x + 1 - i y} \\ = & \frac{(x^{2} - 1) - i y (x - 1) + i y (x + 1) - i^{2} y^{2}}{{(x + 1)}^{2} - {(i y)}^{2}} \\ = & \frac{(x^{2} - 1) + y^{2} + i y [(x + 1) - (x - 1)]}{{(x + 1)}^{2} + y^{2}} \\ = & \frac{(x^{2} - 1) + y^{2} + 2 i y}{{(x + 1)}^{2} + y^{2}} \end{array}$
Given that $\frac{z - 1}{z + 1}$ is a purely imaginary number.
$∴ \frac{(x^{2} - 1) + y^{2}}{{(x + 1)}^{2} + y^{2}} = 0 \Rightarrow x^{2} - 1 + y^{2} = 0 \Rightarrow x^{2} + y^{2} = 1 \Rightarrow \sqrt{x^{2} + y^{2}} = 1 \Rightarrow |z| = 1$

Hence, the value of |z| is 1.

Page No 92:

Question 16:

z₁ and z₂ are two complex numbers such that |z₁| = |z₂| and arg (z₁) + arg (z₂) = π, then show that z₁ = − $z_{2}$ .

Answer:

Let $z_{1} = |z_{1}| (\cos θ_{1} + i \sin θ_{1}) and z_{2} = |z_{2}| (\cos θ_{2} + i \sin θ_{2})$ .
Given that |z₁| = |z₂|.
And, arg(z₁) + arg(z₂) = π
$\Rightarrow θ_{1} + θ_{2} = π \Rightarrow θ_{1} = π - θ_{2}$
$\begin{array}{rcl} ∴ z_{1} & = & |z_{1}| (\cos θ_{1} + i \sin θ_{1}) \\ = & |z_{2}| [\cos (π - θ_{2}) + i \sin (π - θ_{2})] \\ = & |z_{2}| (- \cos θ_{2} + i \sin θ_{2}) \\ = & - |z_{2}| (\cos θ_{2} - i \sin θ_{2}) \\ = & - \bar{z_{2}} \end{array}$
Hence proved.

Page No 92:

Question 17:

If |z₁| = 1 (z₁ ≠ –1) and $z_{2} = \frac{z_{1} - 1}{z_{1} + 1}$ , then show that the real part of z₂is zero.

Answer:

Let z₁ = x + iy
$\Rightarrow |z_{1}| = \sqrt{x^{2} + y^{2}} = 1$
$\begin{array}{rcl} ∴ z_{2} & = & \frac{z_{1} - 1}{z_{1} + 1} \\ = & \frac{x + i y - 1}{x + i y + 1} \\ = & \frac{x - 1 + i y}{x + 1 + i y} \\ = & \frac{x - 1 + i y}{x + 1 + i y} \times \frac{x + 1 - i y}{x + 1 - i y} \\ = & \frac{(x^{2} - 1) - i y (x - 1) + i y (x + 1) - i^{2} y^{2}}{{(x + 1)}^{2} - {(i y)}^{2}} \\ = & \frac{(x^{2} - 1) + y^{2} + i y [(x + 1) - (x - 1)]}{{(x + 1)}^{2} + y^{2}} \\ = & \frac{(x^{2} - 1) + y^{2} + 2 i y}{{(x + 1)}^{2} + y^{2}} \\ = & \frac{x^{2} + y^{2} - 1 + 2 i y}{{(x + 1)}^{2} + y^{2}} \\ = & \frac{1 - 1 + 2 i y}{{(x + 1)}^{2} + y^{2}} [∵ x^{2} + y^{2} = 1] \\ = & 0 + i \frac{2 y}{{(x + 1)}^{2} + y^{2}} \end{array}$
Hence, the real part of z₂ is zero.

Page No 92:

Question 18:

If z₁, z₂ and z₃, z₄ are two pairs of conjugate complex numbers, then find arg $(\frac{z_{1}}{z_{4}}) + \arg (\frac{z_{2}}{z_{3}})$ .

Answer:

Given that z₁ and z₂ are conjugate complex numbers.
$∴ z_{2} = \bar{z_{1}}$
Also, z₃ and z₄ are conjugate complex numbers.
$∴ z_{4} = \bar{z_{3}}$
Thus,
$\begin{array}{rcl} \arg (\frac{z_{1}}{z_{4}}) + \arg (\frac{z_{2}}{z_{3}}) & = & \arg (\frac{z_{1}}{z_{4}}) (\frac{z_{2}}{z_{3}}) \\ = & \arg (\frac{z_{1}}{\bar{z_{3}}}) (\frac{\bar{z_{1}}}{z_{3}}) \\ = & \arg (\frac{z_{1} \bar{z_{1}}}{z_{3} \bar{z_{3}}}) \\ = & \arg (\frac{{|z_{1}|}^{2}}{{|z_{3}|}^{2}}) \\ = & 0 [∵ \frac{{|z_{1}|}^{2}}{{|z_{3}|}^{2}} is purely real] \end{array}$
Hence, arg $(\frac{z_{1}}{z_{4}}) + \arg (\frac{z_{2}}{z_{3}})$ = 0.

Page No 92:

Question 19:

If |z₁| = |z₂| = ... = |z_n| =1, then show that |z₁+ z₂ + z₃ + ... + z_n| = $|\frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} + . . . + \frac{1}{z_{n}}|$ .

Answer:

We have,
|z₁| = |z₂| = ... = |z_n| =1
⇒ |z₁|² = |z₂|² = ... = |z_n|² =1
$\Rightarrow z_{1} \bar{z_{1}} = z_{2} \bar{z_{2}} = . . . = z_{n} \bar{z_{n}} = 1 \Rightarrow z_{1} = \frac{1}{\bar{z_{1}}}, z_{2} = \frac{1}{\bar{z_{2}}}, . . ., z_{n} = \frac{1}{\bar{z_{n}}}$
$\begin{array}{rcl} ∴ |z_{1} + z_{2} + z_{3} + . . . + z_{n}| & = & |\frac{1}{\bar{z_{1}}} + \frac{1}{\bar{z_{2}}} + \frac{1}{\bar{z_{3}}} + . . . + \frac{1}{\bar{z_{n}}}| \\ = & |\bar{\frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} + . . . + \frac{1}{z_{n}}}| \\ = & |\frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} + . . . + \frac{1}{z_{n}}| \end{array}$

Hence proved.

Page No 92:

Question 20:

If for complex numbers z₁and z₂, arg (z₁) – arg (z₂) = 0, then show that |z₁ – z₂| = |z₁| – |z₂|

Answer:

Let z₁ = $|z_{1}| (\cos θ_{1} + ι \sin θ_{1})$ and z₂ = $|z_{2}| (\cos θ_{2} + ι \sin θ_{2})$ where arg(z₁) = $θ_{1}$ and arg(z₂) = $θ_{2}$ .
Then,
arg(z₁) − arg(z₂) = 0
⇒ $θ_{1}$ − $θ_{2}$ = 0
⇒ $θ_{1}$ = $θ_{2}$

Also,
z₁− z₂ = $|z_{1}| (\cos θ_{1} + ι \sin θ_{1})$ − $|z_{2}| (\cos θ_{2} + ι \sin θ_{2})$
⇒ z₁− z₂ = $(|z_{1}| \cos θ_{1} - |z_{2}| \cos θ_{1}) + ι (|z_{1}| \sin θ_{1} - |z_{2}| \sin θ_{1})$ [ $θ_{1}$ = $θ_{2}$ ]
= $\sqrt{{(|z_{1}| \cos θ_{1} - |z_{2}| \cos θ_{1})}^{2} + {(|z_{1}| \sin θ_{1} - |z_{2}| \sin θ_{1})}^{2}}$
= $\sqrt{{|z_{1}|}^{2} \cos^{2} θ_{1} + {|z_{2}|}^{2} \cos^{2} θ_{1} - 2 |z_{1}| |z_{2}| \cos^{2} θ_{1} + {|z_{1}|}^{2} \sin^{2} θ_{1} + {|z_{2}|}^{2} \sin^{2} θ_{1} - 2 |z_{1}| |z_{2}| \sin^{2} θ_{1}}$
= $\sqrt{{|z_{1}|}^{2} (\cos^{2} θ_{1} + \sin^{2} θ_{1}) + {|z_{2}|}^{2} (\cos^{2} θ_{1} + \sin^{2} θ_{1}) - 2 |z_{1}| |z_{2}| (\cos^{2} θ_{1} + \sin^{2} θ_{1})}$
= $\sqrt{{|z_{1}|}^{2} + {|z_{2}|}^{2} - 2 |z_{1}| |z_{2}|}$
= $\sqrt{{(|z_{1}| - |z_{2}|)}^{2}}$
= $|z_{1}| - |z_{2}|$

Hence proved.

Page No 92:

Question 21:

Solve the system of equations Re (z²) = 0, |z| = 2.

Answer:

$Let z = x + ι y . Then, |z| = \sqrt{x^{2} + y^{2}} .$
Now,
|z| = 2
$\Rightarrow \sqrt{x^{2} + y^{2}} = 2 \Rightarrow x^{2} + y^{2} = 4 . . . . . (1)$
Also,
$\begin{array}{rcl} z^{2} & = & {(x + ι y)}^{2} \\ = & x^{2} + {(i y)}^{2} + 2 x (i y) \\ = & x^{2} - y^{2} + 2 i x y \end{array}$
And,
Re (z²) = 0
$\Rightarrow x^{2} - y^{2} = 0 \Rightarrow x^{2} = y^{2} . . . . . (2)$

Equating (1) and (2),
$x^{2} + x^{2} = 4 \Rightarrow 2 x^{2} = 4 \Rightarrow x^{2} = 2 \Rightarrow x = \pm \sqrt{2}$

Similarly, $y = \pm \sqrt{2}$ .

So, $z = \pm \sqrt{2} \pm i \sqrt{2}$ .
Hence, there are 4 solutions.

Page No 92:

Question 22:

Find the complex number satisfying the equation $z + \sqrt{2} |(z + 1)| + i = 0 .$

Answer:

Let $z = x + i y$ . Then,
$z + \sqrt{2} |(z + 1)| + i = 0 \Rightarrow x + i y + \sqrt{2} |(x + i y + 1)| + i = 0 \Rightarrow x + i (y + 1) + \sqrt{2} [\sqrt{{(x + 1)}^{2} + y^{2}}] = 0 \Rightarrow x + i (y + 1) + \sqrt{2} \sqrt{x^{2} + 2 x + 1 + y^{2}} = 0$
Comparing real and imaginary parts to 0 we get,
$x + \sqrt{2} \sqrt{x^{2} + 2 x + 1 + y^{2}} = 0 . . . . . (1) y + 1 = 0 \Rightarrow y = - 1 . . . . . (2)$
Substituting (2) in (1) we get,
$x + \sqrt{2} \sqrt{x^{2} + 2 x + 1 + 1} = 0 \Rightarrow \sqrt{2} \sqrt{x^{2} + 2 x + 2} = - x \Rightarrow 2 (x^{2} + 2 x + 2) = x^{2} \Rightarrow 2 x^{2} + 4 x + 4 = x^{2} \Rightarrow x^{2} + 4 x + 4 = 0 \Rightarrow {(x + 2)}^{2} = 0 \Rightarrow x = - 2$

Hence, $z = - 2 - i$ .

Page No 92:

Question 23:

Write the complex number $z = \frac{1 - i}{\cos \frac{π}{3} + i \sin \frac{π}{3}}$ in polar form.

Answer:

$\begin{array}{rcl} z & = & \frac{1 - ι}{\cos \frac{π}{3} + ι \sin \frac{π}{3}} \\ = & \frac{\sqrt{2} (\frac{1}{\sqrt{2}} - \frac{ι}{\sqrt{2}})}{\cos \frac{π}{3} + ι \sin \frac{π}{3}} \\ = & \frac{\sqrt{2} (\cos \frac{π}{4} - i \sin \frac{π}{4})}{\cos \frac{π}{3} + ι \sin \frac{π}{3}} \\ = & \sqrt{2} [\cos (\frac{- π}{4} - \frac{π}{3}) + i \sin (\frac{- π}{4} - \frac{π}{3})] \\ = & \sqrt{2} [\cos (\frac{- 7 π}{12}) + i \sin (\frac{- 7 π}{12})] \\ = & \sqrt{2} [\cos (\frac{5 π}{12}) + i \sin (\frac{5 π}{12})] \end{array}$

Page No 92:

Question 24:

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

Answer:

Let z = (z) (cos θ₁ + i sin θ₁) and w = |w| (cos θ₂ + i sin θ₂)
Given, |zw| = |z| |w| = 1
Also, arg (z) – arg (w) = $\frac{π}{2}$
$\Rightarrow θ_{1} - θ_{2} = \frac{π}{2}$

Now, $\bar{z} w = | z | ({cosθ}_{1} - i {sinθ}_{1}) | w | ({cosθ}_{2} + i {sinθ}_{2})$
$= | z | | w | [\cos (- θ_{1}) + i \sin (- θ_{1})] ({cosθ}_{2} + i {sinθ}_{2}) = 1 [\cos (θ_{2} - θ_{1}) + i \sin (θ_{2} - θ_{1})] = \cos (\frac{- π}{2}) + i \sin (- \frac{π}{2})$

$= 1 (0 - i) = - i$

Hence, proved.

Page No 93:

Question 25:

Fill in the blanks of the following

(i) For any two complex numbers z₁, z₂ and any real numbers a, b, ${|a z_{1} - b z_{2}|}^{2} + {|b z_{1} + a z_{2}|}^{2} = . . . . . . . . . .$

(ii) The value of $\sqrt{- 25} \times \sqrt{- 9}$ is .....................

(iii) The number $\frac{{(1 - i)}^{3}}{1 - i^{3}}$ is equal to ...............

(iv) The sum of the series i + i² + i³ + ... upto 1000 terms is ..........

(v) Multiplicative inverse of 1 + i is ................

(vi) If z₁ and z₂ are complex numbers such that z₁ + z₂ is a real number, then z₂ = ...........

(vii) arg (z) + arg $z (z \neq 0)$ is ...............

(viii) If |z + 4| ≤ 3, then the greatest and least values of |z +1| are ............ and .............

(ix) If $|\frac{z - 2}{z + 2}| = \frac{π}{6}$ , then the locus of z is ............

(x) If |z| = 4 and arg (z) = $\frac{5 π}{6}$ , then z = .............

Answer:

(i) |az₁ – 6x₂|² + |bz₁ + az₂|²
= |az₁|² + |bz₂|² – 2 Re (az₁ × b ${\bar{z}}_{2}$ ) + |bz₁|² + |az₂|² + 2 Re (bz₁ × a ${\bar{z}}_{2}$ )
= (a² + b²) |z₁|² + (a² + b²) |z₂|²
= (a² + b²) ( |z₁|² + |z₂|²)

(ii) $\sqrt{- 25} \times \sqrt{- 9}$

$= i \sqrt{25} \times i \sqrt{9} i^{2} (5 \times 3) = - 15$

(iii) $\frac{{(1 - i)}^{3}}{1 - i^{3}}$

$= \frac{{(1 - i)}^{3}}{(1 - i) (1 + i + i^{2})} = \frac{{(1 - i)}^{2}}{i} = \frac{1 + i^{2} - 2 i}{i} = \frac{- 2 i}{i} = - 2$

(iv) i + i² + i³ + . . . upto 1000 terms
= (i + i² + i³ + i⁴) + (i⁵ + i⁶ + i⁷ + i⁸) + . . . 250 brackets
= 0 + 0 + 0 + . . . + 0        [âˆµ iⁿ + iⁿ⁺¹ + iⁿ⁺² + iⁿ⁺³ = 0 where nâˆŠN]

(v) Multiplicative inverse of 1 + i
   $= \frac{1}{1 + i} = \frac{1 - i}{1 - i^{2}} = \frac{1}{2} (1 - i)$

(vi) Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂z₁ + z₂ = (x₁ + x₂) + i (y₁ + y₂)
⇒ y₁ + y₂ = 0
⇒ y₁ = – y2
since, z₂ = x₁ – iy₁                [Assuming x₁ = x₂]
⇒ 2z = ${\bar{z}}_{1}$

(vii) arg (z) + arg $(\bar{z})$ = θ + (– θ)
                                  = 0

(viii) Given that |z + 4| ≤ 3
For the greatest value of |z + 1|, |z + 1| = |z + 4 – 3|
⇒ |z + 1| ≤ |z + 4| + |– 3|
⇒ |z + 1| ≤ 3 + 3
⇒ |z + 1| ≤ 6
So, greatest value of |z + 1| is 6. the least value of the modules of a complete number is zero. So, the least value of |z + 1| is zero.

(ix) $|\frac{z - 2}{z + 2}| = \frac{π}{6}$

$\Rightarrow |\frac{x + i y - z}{x + i y + z}| = \frac{π}{6} \Rightarrow |\frac{x - 2 + i y}{x + 2 + i y}| = \frac{π}{6} \Rightarrow 6 | x - 2 + i y | = π | x + 2 + iy |$
$\Rightarrow 6 \sqrt{{(x - 2)}^{2} + y^{2}} = π \sqrt{{(x + 2)}^{2} + y^{2}} \Rightarrow 36 (x^{2} + 4 - 4 \times + y^{2}) = π^{2} (x^{2} + 4 x + 4 + y^{2}) \Rightarrow (36 - π^{2}) x^{2} + (36 - π^{2}) y^{2} - (144 + 4 π^{2}) x + 144 + 4 π^{2} = 0$

This is a circle.

(x) Let z = x + iy = r (cosθ + i sinθ) |z| = r = 4 and arg (z) = 0

Since, tanθ = $\frac{5 π}{6}$

$\Rightarrow z = 4 (\cos \frac{π}{6} + i \sin \frac{π}{6})$

   $= 4 [\cos (π - \frac{π}{6}) + i \sin (π - \frac{π}{6})] = 4 [\cos (π - \frac{π}{6}) + i \sin (π - \frac{π}{6})] = 4 [- \cos \frac{π}{6} + i \sin \frac{π}{6}] = 4 (- \frac{\sqrt{3}}{2} + \frac{i}{2}) = - 2 \sqrt{3} + 2 i$

Page No 93:

Question 26:

State True or False for the following :

(i) The order relation is defined on the set of complex numbers.

(ii) Multiplication of a non zero complex number by – i rotates the point about origin through a right angle in the anti-clockwise direction.

(iii) For any complex number z the minimum value of |z| + |z −1| is 1.

(iv) The locus represented by |z −1| = |z − i| is a line perpendicular to the join of (1, 0) and (0, 1).

(v) If z is a complex number such that z ≠ 0 and Re (z) = 0, then Im (z²) = 0.

(vi) The inequality |z − 4| < |z − 2| represents the region given by x > 3.

(vii) Let z₁ and z₂ be two complex numbers such that |z₁ + z₂| = |z₁| + |z₂|, then arg (z₁ – z₂) = 0.

(viii) 2 is not a complex number.

Answer:

(i) False, we can compare two complex numbers when they are purely real, otherwise comparison of complex numbers is not possible.

(ii) False.
Let z = x + iy, where x, y > 0.
i.e., z or point A(x, y) lies in the first quadrant.
Now,
–iz = – i (x + iy)
= – ix – izy
= y – ix
Now, point B(y₁ – x) lies in the fourth quadrant.
Also, ∠AOB = 90°
Thus, B is obtained by rotating A in clockwise direction about origin.

(iii) True,
|z₁| + |z₂| ≥ |z₁ – z₂|
⇒ |z| + |z – 1| ≥ |z – (z – 1)|
⇒ |z| + |z – 1| ≥ 1
So, minimum value of |z| + |z – 1| is 1.

(iv) True,
|z – 1| = |z – i|
Putting z = x + iy, we get,
|x – 1 + xy| = |x – i (1 – y)|
⇒ (x – 1)² + y² = x² + (1 – y)²
⇒ x² – 2x + 1 + y² = x² + 1 + y² – 2y
⇒ –2x + 1 = 1 – 2y
⇒ –2x + 2y = 0
⇒ x – y = 0
Also, equation of a line long the points (1, 0) and (0, 1) is

y – 0 = $\frac{1 - 0}{0 - 1} (x - 1)$

⇒ x + y = 1
So, this line is perpendicular to the line x – y = 0

(v) False,
Let z = x + iy, z ≠ 0 and Re(z) = 0
⇒ x = 0
∴ z = iy
⇒ Im (z²) = i²y² = – y² = – y² ≠ 0

(vi) True,
|z – 4| < |z – 2|
Putting z = x + iy, we get
|x – 4 + iy| < x – 2 + iy|

$\Rightarrow \sqrt{{(x - 4)}^{2} + y^{2}} < \sqrt{{(x - 2)}^{2} + y^{2}} \Rightarrow {(x - 4)}^{2} + y^{2} < {(x - 2)}^{2} + y^{2} \Rightarrow x^{2} - 8 x + 16 + y^{2} < x^{2} - 9 x + 4 + y^{2}$
⇒ – 8x + 16 < – 4x + 4
⇒ – 8x < – 4x – 12
⇒ 4x > 12
⇒ x > 3

(vii) False,
Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂
Now,
|z₁ + z₂| = |z₁| + |z₂|
⇒ |x₁ + iy₁ + x₂ + iy₂| = |x₁ + iy₁| + |x₂ + iy₂|

$\Rightarrow \sqrt{{(x_{1} + x_{2})}^{2} + {(y_{1} + y_{2})}^{2}} = \sqrt{x_{1}^{2} + y_{1}^{2}} + \sqrt{x_{2}^{2} + y_{2}^{2}} \Rightarrow {(x_{1} + x_{2})}^{2} + {(y_{1} + y_{2})}^{2} = x_{2}^{1} + y_{1}^{2} + x_{2}^{2} + y_{2}^{2} + 2 \sqrt{(x_{1}^{2} + y_{1}^{2}) (x_{2}^{2} + y_{2}^{2})} \Rightarrow x_{1}^{2} + x_{2}^{2} + 2 x_{1} x_{2} + y_{1}^{2} + y_{2}^{2} + 2 y_{1} y_{2} = x_{1}^{2} + y_{1}^{2} + x_{2}^{2} + y_{2}^{2} + 2 \sqrt{(x_{1}^{2} + y_{1}^{2}) (x_{2}^{2} + y_{2}^{2})}$
$\Rightarrow 2 x_{1} x_{2} + 2 y_{1} y_{2} = 2 \sqrt{(x_{1}^{2} + y_{1}^{2}) (x_{2}^{2} + y_{2}^{2})} \Rightarrow x_{1} x_{2} + y_{1} y_{2} = \sqrt{(x_{1}^{2} + y_{1}^{2}) (x_{2}^{2} + y_{2}^{2})}$
$\Rightarrow x_{1}^{2} x_{2}^{2} + y_{1}^{2} y_{2}^{2} + 2 x_{1} x_{2} y_{1} y_{2} = x_{1}^{2} x_{2}^{2} + y_{1}^{2} y_{2}^{2} + x_{1}^{2} y_{2}^{2} + y_{1}^{2} y_{2}^{2} \Rightarrow {(x_{1} y_{2} - x_{2} y_{1})}^{2} = 0 \Rightarrow x_{1} y_{2} = x_{2} y_{1} \Rightarrow \frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}} \Rightarrow (\frac{y_{1}}{x_{1}}) - (\frac{y_{2}}{x_{2}}) = 0 \Rightarrow \arg (z_{1}) - \arg (z_{2}) = 0$

(viii) True,
2 is a real number and not a complex number.

Page No 94:

Question 27:

Match the statements of Column A and Column B.

Column A	Column B
(a) The polar form of $i + \sqrt{3}$ is	(i) Perpendicular bisector of segment joining (–2, 0) and (2, 0)
(b) The amplitude of $- 1 + \sqrt{3}$ is	(ii) On or outside the circle having centre at (0, – 4) and radius 3.
(c) If \|z + 2\| = \|z − 2\|, then locus of z is	(iii) $\frac{2 π}{3}$
(d) If \|z + 2i\| = \|z − 2i\|, then locus of z is	(iv) Perpendicular bisector of segment joining (0, – 2) and (0, 2).
(e) Region represented by \|z + 4i\| ≥ 3 is	(v) $2 (\cos \frac{π}{6} + i \sin \frac{π}{6})$
(f) Region represented by \|z + 4\| ≤ 3 is	(vi) On or inside the circle having centre (– 4, 0) and radius 3 units.
(g) Conjugate of $\frac{1 + 2 i}{1 - i}$ lies in	(vii) First quadrant
(h) Reciprocal of 1 – i lies in	(viii) Third quadrant

Answer:

Column A	Column B
(a) The polar form of $i + \sqrt{3}$ is	(iii) $\frac{2 π}{3}$
(b) The amplitude of $- 1 + \sqrt{3}$ is	(v) $2 (\cos \frac{π}{6} + i \sin \frac{π}{6})$
(c) If \|z + 2\| = \|z − 2\|, then locus of z is	(ii) On or outside the circle having centre at (0, – 4) and radius 3.
(d) If \|z + 2i\| = \|z − 2i\|, then locus of z is	(iv) Perpendicular bisector of segment joining (0, – 2) and (0, 2).
(e) Region represented by \|z + 4i\| ≥ 3 is	(i) Perpendicular bisector of segment joining (–2, 0) and (2, 0)
(f) Region represented by \|z + 4\| ≤ 3 is	(vi) On or inside the circle having centre (– 4, 0) and radius 3 units.
(g) Conjugate of $\frac{1 + 2 i}{1 - i}$ lies in	(viii) First quadrant
(h) Reciprocal of 1 – i lies in	(vii) Third quadrant

(i) z = i +

\sqrt{3}

|z| = |i +

\sqrt{3}

\sqrt{1^{2} + {(\sqrt{3})}^{2}}

also, z lies in first quadrant.

\Rightarrow \arg (z) = \tan^{- 1} \frac{1}{\sqrt{3}} = \frac{π}{6}

So, polar from of ≠ is 2

(\cos \frac{π}{6} + i \sin \frac{π}{6})

(ii) z =

- 1 + \sqrt{- 3} = - 1 + i \sqrt{3}

Here, z lies in the second quadrant.

\Rightarrow \arg (z) = π - \tan^{- 1} |\frac{\sqrt{3}}{- 1}|

= π - \tan^{- 1} \sqrt{3} = π = \frac{π}{3} = \frac{2 π}{3}

(iii) |z + 2| = |z – 2|
⇒ |x + 2 + iy| = |x – 2 + iy|
⇒ (x + 2)² + y² = (x – 2)² + y²
⇒ x² + 4x + 4 = x² – 4x + 4
⇒ 8x = 0
⇒ x = 0
It is a straight line which is a perpendicular bisector of segment joining the points (–2, 0) and (2, 0)

(iv) |z + 2i| = |z – 2i|
putting z = x + iy, we get
|x + i(y + 2)|² = |x + i(y – 2)|²
⇒ x² + (y + 2)² = x² + (y – z)²
⇒ 4y = 0
⇒ y = 0
It is a straight line which is a perpendicular bisector of segment joining (0, –2) and (0, 2).

(v) |z + 4i) ≥ 3
⇒ |x + iy + 4i| ≥ 3
⇒ |x + i (y + 3)| ≥ 3

⇒

\sqrt{x^{2} + {(y + 4)}^{2}}

≥ 3
⇒ x² + y² + 8y + 16 ≥ 9
⇒ x² + y² + 8y + 7 ≥ 0
This represents the region on or outside the circle having centre (0, –4) and radius 3.

(vi) |z + 4| ≤ 3
⇒ |x + iy + 4| ≤ 3
⇒ |x + 4 + iy| ≤ 3

⇒

\sqrt{{(x + 4)}^{2} + y^{2}} \leq 3

⇒ (x + 4)² + y² ≤ 9
⇒ x² + 8x + 16 + y² ≤ 9
⇒ x² + 8x + y² + 7 ≤ 0
This represents the region on or inside the circle having center (–4, 0) and radius 3.

(vii) z

= \frac{1 + 2 i}{1 - i}

= \frac{(1 + 2 i) (1 + i)}{(1 + i) (1 + 2 i)} = \frac{1 + 2 i + i + 2 i^{2}}{1 - i^{2}} = \frac{1 - 2 + 3 i}{1 + 1} = \frac{- 1}{2} + \frac{3 i}{2}

Hence,

\bar{z}

lies in the third quadrant.

(viii) z = 1 – i

\Rightarrow \frac{1}{2} = \frac{1}{1 - i}

= \frac{1 + i}{(1 - i) (1 + i)} = \frac{1 + i}{1 - i^{2}} = \frac{1}{2} (1 + i)

This, the reciprocal of z lies is the first quadrant.

Page No 94:

Question 28:

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

Answer:

$z = \frac{2 - i}{{(1 - 2 i)}^{2}} = \frac{2 - i}{1 + 4 i^{2} - 4 i} = \frac{2 - i}{1 - 4 - 4 i} = \frac{2 - i}{- 3 - 4 i}$

$= \frac{(2 - i)}{(- 3 + 4 i)} = - [\frac{(2 - i) (3 - 4 i)}{(3 + 4 i) (3 - 4 i)}]$

$= \frac{6 - 8 i - 3 i + 4 i^{2}}{9 + 16} = - \frac{(- 11 i + 2)}{25} = \frac{- 1}{25} (2 - 11 i) = \frac{1}{25} (- 2 + 11 i)$

$\bar{z} = \frac{1}{25} (- 2 - 11 i) = \frac{- 2}{25} - \frac{11}{25} i$

Page No 94:

Question 29:

If |z₁| = |z₂|, is it necessary that z₁ = z₂?

Answer:

If |z₁| = |z₂| then z₁ and z₂ are at the same distance from origin. But, if arg(z₁) ≠ arg(z₂), then z₁ and z₂ are different.
So, if |z₁| = |z₂|, it is not necessary that z₁ = z₂.

Page No 94:

Question 30:

If $\frac{{(a^{2} + 1)}^{2}}{2 a - i} = x + i y$ , what is the value of x² + y²?

Answer:

$|\frac{{(a^{2} + 1)}^{2}}{2 a - i}| = |x + i y|$
$\Rightarrow \frac{|{(a^{2} + 1)}^{2}|}{|2 a - i|} = |x + i y| \Rightarrow \frac{{(a^{2} + 1)}^{2}}{\sqrt{{(2 a)}^{2} + {(- 1)}^{2}}} = \sqrt{x^{2} + y^{2}} \Rightarrow x^{2} + y^{2} = \frac{{(a^{2} + 1)}^{4}}{4 a^{2} + 1}$

Page No 95:

Question 31:

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

Answer:

Let z = |z| (cosθ + i sinθ), where θ = arg|–z|

Now, |z| = 4 and arg(z) = $\frac{5 π}{6}$

$\Rightarrow z = 4 [\cos \frac{5 π}{6} + i \sin \frac{5 π}{6}] (z lies in II quadrant)$

$= 4 [\frac{- \sqrt{3}}{2} = \frac{i}{2}] = - 2 \sqrt{3} + 2 i$

Page No 95:

Question 32:

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

Answer:

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

$= |\frac{(z + i + 2 i + i^{2})}{(3 + i)}| = |\frac{2 + 3 i - 1}{3 + i}| = |\frac{1 + 3 i}{3 + i}|$

$= |\frac{(1 + 3 i) (3 - i)}{(3 + i) (3 - i)}| = |\frac{3 + 9 i - i - 3 i^{2}}{9 - i^{2}}|$

$= |\frac{3 + 8 i + 3}{9 + 1}| = |\frac{6 + 8 i}{10}| = |\frac{6^{2}}{100} + \frac{8^{2}}{100}| = \sqrt{\frac{100}{100}} = 1$

Page No 95:

Question 33:

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

Answer:

Let z = a + ib
Then, the polar from of z is r(cosθ + i sinθ) where r = |z| = $\sqrt{a^{2} + b^{2}}$ and tanθ = $\frac{b}{a}$ , and arg(z) = θ.
The principal argument is a unique value of θ such that $- π \leq θ \leq π$ .
Now,
$z = {(1 + i \sqrt{3})}^{2} = 1 - 3 + 2 i \sqrt{3} = - 2 + i 2 \sqrt{3}$

$\Rightarrow \tan α = |\frac{2 \sqrt{3}}{- 2}| (\tan α = |\frac{Im (z)}{Re (z)}|) = |- \sqrt{3}| = \sqrt{3}$

$\Rightarrow \tan α = \tan \frac{π}{3} \Rightarrow α = \frac{π}{3}$
$∵ Re (z) < 0 and Im (z) > 0 \Rightarrow \arg (z) = π - \frac{π}{3} = \frac{2 π}{3}$

Page No 95:

Question 34:

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

Answer:

If z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂ then |z₁| = $\sqrt{x_{1}^{2} + y_{1}^{2}} and |z_{2}| = \sqrt{x_{2}^{2} + y_{2}^{2}}$
Let z = x + iy

$|\frac{z - 5 i}{z + 5 i}| = |\frac{x + i y - 5 i}{x + i y + 5 i}|$

$= |\frac{x + i (y - 5)}{x + i (y + 5)}| = \frac{\sqrt{x^{2} + {(y - 5)}^{2}}}{\sqrt{x^{2} + {(y - 5)}^{2}}}$
$\Rightarrow x^{2} + {(y + 5)}^{2} = x^{2} + {(y - 5)}^{2} [∵ |\frac{z - 5 i}{z + 5 i}| = 1]$
⇒ –10y = 10y
⇒ 20y = 0
⇒ y =0
So, z lies on real axis.

Page No 95:

Question 35:

Choose the correct answer from the given four options:
sin x + i cos 2x and cos x – i sin 2x are conjugate to each other for:
(A) x = nπ
(B) $x = (n + \frac{1}{2}) \frac{π}{2}$
(C) x = 0
(D) No value of x

Answer:

Let z = sinx + i cos2x

and $\bar{z}$ = sinx – i cos2x
Given,
$\bar{z}$ = cosx – i sin2x
⇒ sinx – i cos2x = cosx – i sin2x
⇒ sinx = cosx and cos2x = sin2x
⇒ tanx = 1 and tan2x = 1

⇒ tanx tan $\frac{π}{4}$ and tan2x = tan $\frac{π}{4}$

⇒ x = nπ + $\frac{π}{4}$ and 2x = nπ + $\frac{π}{4}$

⇒ 2x – x = 0
⇒ x = 0
Hence, the correct answer is option C.

Page No 95:

Question 36:

Choose the correct answer from the given four options:
The real value of α for which the expression $\frac{1 - i \sin α}{1 + 2 i \sin α}$ is purely real is :
(A) $(n + 1) \frac{π}{2}$

(B) $(2 n + 1) \frac{π}{2}$

(C) nπ

(D) None of these, where n ∈ N

Answer:

$z = \frac{1 - i \sin α}{1 + 2 i \sin α}$
$= \frac{(1 - i \sin α) (1 - 2 i \sin α)}{(1 + 2 i \sin α) (1 - 2 i \sin α)} = \frac{1 - i \sin α - 2 i \sin α + 2 i^{2} \sin^{2} α}{1 - 4 i^{2} \sin^{2} α} = \frac{1 - 3 i \sin α - 2 \sin^{2} α}{1 + 4 \sin^{2} α} = \frac{1 - 2 \sin^{2} α}{1 + 4 \sin^{2} α} - \frac{3 i \sin α}{1 + 4 \sin^{2} α}$

It is given that z is purely real.

$∴ \frac{- 3 \sin α}{1 + 4 \sin^{2} α} = 0$

⇒ –3sinα = 0
⇒ sinα = 0
⇒ α = nπ
Hence, the correct answer is option C.

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Question 37:

Choose the correct answer from the given four options:
If z = x + iy lies in the third quadrant, then $\frac{z}{z}$ also lies in the third quadrant if
(A) x > y > 0
(B) x < y < 0
(C) y < x < 0
(D) y > x > 0

Answer:

$\begin{array}{rcl} \frac{\bar{z}}{2} & = & \frac{x - i y}{x + i y} \\ = & \frac{(x - i y) (x - i y)}{(x + i y) (x - i y)} \\ = & \frac{x^{2} - y^{2} - 2 i x y}{x^{2} + y^{2}} \\ = & \frac{x^{2} - y^{2}}{x^{2} + y^{2}} - \frac{2 i x y}{x^{2} + y^{2}} \end{array}$

Since, $\frac{\bar{z}}{2}$ also lies in the third quadrant:

$∴ \frac{x^{2} - y^{2}}{x^{2} + y^{2}} < 0$ and $\frac{- 2 x y}{x^{2} + y^{2}} < 0$

x² – y² < 0 and –2xy < 0

x² < y² and xy > 0

So, x < y < 0.

Hence, the correct answer is option B.

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Question 38:

Choose the correct answer from the given four options:
The value of $(z + 3) (z + 3)$ is equivalent to
(A) |2 + 3|²
(B) |z −3|
(C) z² + 3
(D) None of these

Answer:

Let z = x + iy

$\begin{array}{rcl} \Rightarrow & (z + 3) (\bar{z} + 3) = (x + i y + 3) (x + 3 - i y) \\ = & {(x + 3)}^{2} - {(i y)}^{2} \\ = & {(x + 3)}^{2} + y^{2} \\ = & {|x + 3 + i y|}^{2} \\ = & {|z + 3|}^{2} \end{array}$

Hence, the correct answer is option A.

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Question 39:

Choose the correct answer from the given four options:
If ${(\frac{1 + i}{1 - i})}^{x} = 1$ , then
(A) x = 2n +1
(B) x = 4n
(C) x = 2n
(D) x = 4n + 1, where n ∈ N

Answer:

${(\frac{1 + i}{1 - i})}^{x} = 1 \Rightarrow {[\frac{(1 + i) (1 + i)}{(1 - i) (1 + i)}]}^{x} = 1 \Rightarrow {[\frac{1 + 2 i + i^{2}}{1 - i^{2}}]}^{x} = 1 \Rightarrow {[\frac{2 i}{1 + 1}]}^{x} = 1 \Rightarrow i^{x} = 1 \Rightarrow i^{x} = (i^{4 n}) [∵ i^{4 n} = 1, n \in N] \Rightarrow x = 4 n$

Hence, the correct answer is option B.

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Question 40:

Choose the correct answer from the given four options:
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

Answer:

$\begin{array}{rcl} α - i^{β} & = & \frac{(3 - 4 i x) (3 - 4 i x)}{(3 + 4 i x) (3 - 4 i x)} \\ = & \frac{9 + 16 i^{2} x^{2} - 24 i x}{9 - 16 i^{2} x^{2}} \\ = & \frac{9 - 16 x^{2} - 24 i x}{9 + 16 x^{2}} \\ = & \frac{9 - 16 x^{2}}{9 + 16 x^{2}} - \frac{i 24 x}{9 + 16 x^{2}} \end{array}$

$\Rightarrow α + i β = \frac{9 - 16 x^{2}}{9 + 16 x^{2}} + \frac{i 24 x}{9 + 16 x^{2}} So, (α - i β) (α + i β) = {(\frac{9 - 16 x^{2}}{9 + 16 x^{2}})}^{2} - {(\frac{i 24 x}{9 + 16 x^{2}})}^{2}$

$\begin{array}{rcl} \Rightarrow & α^{2} + β^{2} = \frac{81 + 256 x^{4} - 288 x^{2} + 576 x^{2}}{{(9 + 16 x^{2})}^{2}} \\ = & \frac{81 + 256 x^{4} + 288 x^{2}}{{(9 + 16 x^{2})}^{2}} \\ = & \frac{81 + 256 x^{4} + 288 x^{2}}{{(9 + 16 x^{2})}^{2}} \\ = & \frac{{(9 + 16 x^{2})}^{2}}{{(9 + 16 x^{2})}^{2}} \\ = & 1 \end{array}$

Hence, the correct answer is option A.

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Question 41:

Choose the correct answer from the given four options:
Which of the following is correct for any two complex numbers z₁ and z₂?
(A) |z₁z₂| = |z₁| |z₂|
(B) arg (z₁z₂) = arg (z₁). arg (z₂)
(C) |z₁ + z₂| = |z₁| + |z₂|
(D) |z₁ + z₂| ≥ |z₁| – |z₂|

Answer:

Let z₁ = r₁ (cosθ + isinθ₁) and z₂ = r₂ (cosθ₂ + isinθ₂) be two complex numbers.
⇒ |z₁| = r₁ and |z₂| = r₂
Then,
z₁z₂ = r₁r₂ [cosθ₁ + cosθ₂ + isinθ₁ cosθ₂ + icosθ₁ sinθ₂ + i²sinθ₁sinθ₂ ]
= r₁r₂ [cos(θ₁ + θ₂) + isin(θ₁ + θ₂)]
⇒ |z₁z₂| = r₁r₂
⇒ |z₁z₂| = |z₁| |z₂|

Hence, the correct answer is option A.

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Question 42:

Choose the correct answer from the given four options:
The point represented by the complex number 2 – i is rotated about origin through an angle $\frac{π}{2}$ in the clockwise direction, the new position of point is:
(A) 1 + 2i
(B) –1 – 2i
(C) 2 + i
(D) –1 + 2i

Answer:

z < iα is a complex number, where modulus is r and argument (θ + α). If point P(2) rotates in clockwise sense through an angle α, its new position will be 2(θ – iα)
Given that z = 2 – i.
It is rotated about origin through an angle $\frac{π}{2}$ in the clockwise direction.
⇒ New position $= z e^{- \frac{i π}{2}}$

$\Rightarrow (2 - i) e^{- \frac{i π}{2}} = (2 - i) [\cos (\frac{- π}{2}) + i \sin (\frac{- π}{2})] \begin{matrix} \Rightarrow (2 - i) [0 - i] & = - 2 i - 1 \\ = - 1 - 2 i \end{matrix}$

Hence, the correct answer is option B.

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Question 43:

Choose the correct answer from the given four options:
Let x, y ∈ R, then x + iy is a non real complex number if:
(A) x = 0
(B) y = 0
(C) x ≠ 0
(D) y ≠ 0

Answer:

Given x, y∈R
x + iy is non-real complex number if and only if g ≠ 0

Hence, the correct answer is option D.

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Question 44:

Choose the correct answer from the given four options:
If a + ib = c + id, then
(A) a² + c² = 0
(B) b² + c² = 0
(C) b² + d² = 0
(D) a² + b² = c² + d²

Answer:

a + ib = c + id

⇒ |a + ib| = |c + id| [if z₁ = z₂, |z₁| = |z₂|]

$\Rightarrow \sqrt{a^{2} + b^{2}} = \sqrt{c^{2} + d^{2}}$

⇒ a² + b² = c² + d²

Hence, the correct answer is option D.

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Question 45:

Choose the correct answer from the given four options:
The complex number z which satisfies the condition $|\frac{i + z}{i - z}| = 1$ lies on
(A) circle x² + y² = 1
(B) the x-axis
(C) the y-axis
(D) the line x + y = 1.

Answer:

Let z = x + iy

Then,

$|\frac{i + z}{i - z}| = 1 \Rightarrow |\frac{i + x + i y}{i - (x - i y)}| = 1 \Rightarrow |\frac{x + i (y + 1)}{- x - i (y - 1)}| = 1 \Rightarrow \frac{x^{2} + {(y + 1)}^{2}}{x^{2} + {(y - 1)}^{2}} = 1$

⇒ x² + (y + 1)² = x² + (y – 1)²

⇒ x² + y² + 1 + 2y = x² + y² + 1 – 2y

⇒ 4y = 0

⇒ y = 0

so, z lies on the x-axis.

Hence, the correct answer is option B.

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Question 46:

Choose the correct answer from the given four options:
If z is a complex number, then
(A) |z²| > |z|²
(B) |z²| = |z|²
(C) |z²| < |z|²
(D) |z²| ≥ |z|²

Answer:

Let z be a complex number.
Then,
z = x + iy
⇒ |z| = |x + iy|
⇒ |z|² = |x + iy|²
⇒ |z|² = x + y²
and z² = x² – y² + i2xy

$\Rightarrow |z^{2}| = \sqrt{{(x^{2} - y^{2})}^{2} + {(2 x y)}^{2}} = \sqrt{x^{4} + y^{4} - 2 x^{2} y^{2} + 4 x^{2} y^{2}} = \sqrt{x^{4} + y^{4} + 2 x^{2} y^{2}} = \sqrt{{(x^{2} + y^{2})}^{2}} = x^{2} + y^{2}$
⇒ |z|² = |z|²

Hence, the correct answer is option B.

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Question 47:

Choose the correct answer from the given four options:
|z₁ + z₂| = |z₁| + |z₂| is possible if
(A) $z_{2} = z_{1}$

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

(C) arg (z₁) = arg (z₂)

(D) |z₁| = |z₂|

Answer:

$|z_{1} + z_{2}| = |z_{1}| + |z_{2}| \Rightarrow |r_{1} ({cosθ}_{1} + i \sin θ_{1}) + r_{2} ({cosθ}_{2} + i \sin θ_{2})| = |r_{1} ({cosθ}_{1} + i \sin θ_{1})| + |r_{2} ({cosθ}_{2} + i \sin θ_{2})| \Rightarrow |(r_{1} {cosθ}_{1} + r_{2} {cosθ}_{2}) + i (r_{1} \sin θ_{1} + r_{2} \sin θ_{2})| = r_{1} + r_{2} \Rightarrow \sqrt{r_{1}^{2} \cos^{2} θ_{1} + r_{2}^{2} \cos^{2} θ_{2} + 2 r_{1} r_{2} θ_{1} {cosθ}_{2} + r_{1}^{2} \sin^{2} θ_{1} + r_{2}^{2} \sin^{2} θ_{2} + 2 r_{1} r_{2} \sin θ_{1} \sin θ_{2}} = r_{1} + r_{2} \Rightarrow \sqrt{r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} [\cos (θ_{1} - θ_{2})]} = r_{1} + r_{2} \Rightarrow r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} \cos (θ_{1} - θ_{2}) = r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} \Rightarrow 2 r_{1} r_{2} [1 - \cos (θ_{1} - θ_{2})] = 0 \Rightarrow \cos (θ_{1} - θ_{2}) = 1 \Rightarrow \cos (θ_{1} - θ_{2}) = \cos 0 ° \Rightarrow θ_{1} - θ_{2} = 0 \Rightarrow θ_{1} = θ_{2} \Rightarrow \arg (z_{1}) = \arg (z_{2})$

Hence, the correct answer is option C.

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Question 48:

Choose the correct answer from the given four options:
The real value of θ for which the expression $\frac{1 + i \cos θ}{1 - 2 i \cos θ}$ is a real number is:
(A) $n π + \frac{π}{4}$

(B) $n π + {(- 1)}^{n} \frac{π}{4}$

(C) $2 n π \pm \frac{π}{2}$

(D) none of these.

Answer:

$\begin{array}{rcl} \frac{1 + i cosθ}{1 - 2 i cosθ} & = & \frac{(1 + i cosθ) (1 + 2 i cosθ)}{(1 - 2 i cosθ) (1 + 2 i cosθ)} \\ = & \frac{1 + i \cos θ + 2 i \cos θ + 2 i^{2} \cos^{2} θ}{1 - 4 i^{2} \cos^{2} θ} \\ = & \frac{1 + 3 i \cos θ - 2 i \cos^{2} θ}{1 + 4 \cos^{2} θ} \end{array}$

For real value of θ,

$\frac{3 cosθ}{1 + 4 \cos^{2} θ} = 0 \Rightarrow 3 cosθ = 0 \Rightarrow cosθ = \cos \frac{π}{2} \Rightarrow θ = 2 n π \pm \frac{π}{2}$

Hence, the correct answer is option C.

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Question 49:

Choose the correct answer from the given four options:
The value of arg (x) when x < 0 is:
(A) 0
(B) $\frac{π}{2}$
(C) π
(D) none of these

Answer:

Let z = x + 0i and x < 0

$\begin{array}{rcl} \Rightarrow & |z| = \sqrt{{(- 1)}^{2} + {(0)}^{2}} \\ = & 1 \end{array}$

Since, the point (x, 0) represents

z = x + 0i and lies on the negative side of real x-axis,

∴ Principal arg(z) = π.

Hence, the correct answer is option C.

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Question 50:

Choose the correct answer from the given four options:
If $f (z) = \frac{7 - z}{1 - z^{2}}$ , where z = 1 + 2i, then |f (z)| is
(A) $\frac{|z|}{2}$

(B) |z|

(C) 2|z|

(D) none of these.

Answer:

z = 1 + 2i

$\begin{array}{rcl} \Rightarrow & |z| = \sqrt{1 + 4} \\ = & \sqrt{5} \end{array}$
Now,

$\begin{array}{rcl} f (z) & = & \frac{7 - z}{1 - 2^{2}} \\ = & \frac{7 - 1 - 2 i}{1 - {(1 + 2 i)}^{2}} \\ = & \frac{6 - 2 i}{1 - 1 - 4 i^{2} - 4 i} \\ = & \frac{6 - 2 i}{4 - 4 i} \\ = & \frac{3 - i}{2 - 2 i} \\ = & \frac{(3 - i) (2 + 2 i)}{(2 - 2 i) (2 + 2 i)} \\ = & \frac{6 - 2 i + 6 i - 2 i^{2}}{4 - 4 i^{2}} \\ = & \frac{6 + 4 i + 2}{4 + 4} \\ = & \frac{8 + 4 i}{8} \\ = & 1 + \frac{i}{2} \end{array}$

$|f (z)| = \sqrt{1 + \frac{1}{4}} = \frac{\sqrt{5}}{2} = \frac{|z|}{2}$

Hence, the correct answer is option A.

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