SATHEE: Complex Numbers and Quadratic Equations

Complex Numbers and Quadratic Equations

Short Answer Type Questions

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

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Solution

Given expression $= (1 - i)^{n} (1 - {\frac{1}{i}}^{n})$

$\begin{aligned} = (1 - i)^{n} (i - 1)^{n} \cdot i^{- n} = (1 - i)^{n} (1 - i)^{n} (- 1)^{n} \cdot i^{- n} \\ = [(1 - i)^{2}]^{n} (- 1)^{n} \cdot i^{- n} = (1 + i^{2} - 2 i)^{n} (- 1)^{n} i^{- n} [∵ i^{2} = - 1] \\ = (1 - 1 - 2 i)^{n} (- 1)^{n} i^{- n} = (- 2)^{n} \cdot i^{n} (- 1)^{n} i^{- n} \\ = (- 1)^{2 n} \cdot 2^{n} = 2^{n} \end{aligned}$

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

Thinking Process

$Use i^{2} = - 1, i^{4} = (- 1)^{2} = 1, i^{3} = - i, and i^{5} = i to Solve it$

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Solution

Given that, $\sum_{n = 1}^{13} (i^{n} + i^{n + 1}), n \in N$

$\begin{matrix} = (i + i^{2} + i^{3} + i^{4} + i^{5} + i^{6} + i^{7} + i^{8} + i^{9} + i^{10} + i^{11} + i^{12} + i^{13}) \\ + (i^{2} + i^{3} + i^{4} + i^{5} + i^{6} + i^{7} + i^{8} + i^{9} + i^{10} + i^{11} + i^{12} + i^{13} + i^{14}) \\ = (i + 2 i^{2} + 2 i^{3} + 2 i^{4} + 2 i^{5} + 2 i^{6} + 2 i^{7} + 2 i^{8} + 2 i^{9} + 2 i^{10} + 2 i^{11} + 2 i^{12} + 2 i^{13} + i^{14}) \\ = i - 2 - 2 i + 2 + 2 i + 2 (i^{4}) i^{2} + 2 (i)^{4} i^{3} + 2 (i^{2})^{4} + 2 (i^{2})^{4} i + 2 (i^{2})^{5} \\ + 2 (i^{2})^{5} \cdot i + 2 (i^{2})^{6} + 2 (i^{2})^{6} \cdot i + (i^{2})^{7} \\ = i - 2 - 2 i + 2 + 2 i - 2 - 2 i + 2 + 2 i - 2 - 2 i + 2 + 2 i - 1 - 1 + i \end{matrix}$

Alternate Method

$\begin{aligned} \sum_{n = 1}^{13} (i^{n} + i^{n + 1}), n \in N = \sum_{n = 1}^{13} i^{n} (1 + i) \\ = (1 + i) [i + i^{2} + i^{3} + i^{4} + i^{5} + i^{6} + i^{7} + i^{8} + i^{9} + i^{10} + i^{11} + i^{12} + i^{13}] \\ = (1 + i) [i^{13}] [∵ i^{n} + i^{n + 1} + i^{n + 2} + i^{n + 3} = 0, where n \in N i.e., \sum_{n = 1}^{12} i^{n} = 0. \\ = (1 + i) i \\ [∵ (i^{4})^{3} . i = 1] \\ = (i^{2} + i) = i - 1 \end{aligned}$

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

Thinking Process

If two complex numbers $z_{1} = x_{1} + i y^{1}$ and $z_{2} = x_{2}$ +iy are equal

$i.e., z_{1} = z_{2} \Rightarrow x_{1} + i y_{1} = x_{2} + i y_{2}, then x_{1} = x_{2} and y_{1} = y_{2} .$

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Solution

Given that, ${(\frac{1 + i}{1 - i})}^{3} - {(\frac{1 - i}{1 + i})}^{3} = x + i y$

$∴ \begin{aligned} {(\frac{1 + i}{1 - i})}^{3} & = \frac{1 + i^{3} + 3 i (1 + i)}{1 - i^{3} - 3 i (1 - i)} = \frac{1 - i + 3 i + 3 i^{2}}{1 + i - 3 i + 3 i^{2}} \\ = \frac{2 i - 2}{- 2 i - 2} = \frac{i - 1}{- i - 1} = \frac{1 - i}{1 + i} \\ = \frac{(1 - i)}{(1 + i)} \frac{(1 - i)}{(1 - i)} = \frac{1 + i^{2} - 2 i}{1 + 1} = \frac{1 - 1 - 2 i}{2} \end{aligned}$

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

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

Using Eqs. (ii) and (iii) in Eq. (i), we get

$\begin{aligned} - i - i & = x + i y \\ \Rightarrow - 2 i & = x + i y \end{aligned}$

On comparing real and imaginary part of complex number, we get

$x = 0 and y = - 2$

So, $(x, y) = (0, - 2)$

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

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Solution

Given that,

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

$\begin{matrix} \Rightarrow & \frac{(1 + i^{2} + 2 i)}{2 - i} = x + i y \Rightarrow \frac{2 i}{2 - i} = x + i y \\ \Rightarrow & \frac{2 i (2 + i)}{(2 - i) (2 + i)} = x + i y \Rightarrow \frac{4 i + 2 i^{2}}{4 - i^{2}} = x + i y \end{matrix}$

$\Rightarrow \frac{4 i - 2}{4 + 1} = x + i y \Rightarrow \frac{- 2}{5} + \frac{4 i}{5} = x + i y$

On comparing both sides, we get

$x = - 2 / 5 \Rightarrow y = 4 / 5$

$\Rightarrow x + y = \frac{- 2}{5} + \frac{4}{5} = 2 / 5$

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

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Solution

Given that, ${(\frac{1 - i}{1 + i})}^{100} = a + i b$

$\begin{aligned} \Rightarrow {[\frac{(1 - i)}{(1 + i)} \cdot \frac{(1 - i)}{(1 - i)}]}^{100} = a + i b \Rightarrow {(\frac{1 + i^{2} - 2 i}{1 - i^{2}})}^{100} = a + i b \\ \Rightarrow {(\frac{- 2 i}{2})}^{100} = a + i b [∵ i^{2} = - 1] \\ \Rightarrow (i^{4})^{25} = a + i b \Rightarrow 1 = a + i b \\ Then, a = 1 and b = 0 [∵ i^{4} = 1] \\ ∴ (a, b) = (1, 0) \end{aligned}$

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

Thinking Process

To solve the above problem use the trigonometric formula $\cos θ = 2 \cos^{2} θ / 2$

$- 1 = 1 - 2 \sin^{2} θ / 2$ and $\sin θ = 2 \sin θ / 2 \cdot \cos θ / 2$ .

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Solution

Given that, $a = \cos θ + i \sin θ$

$\begin{aligned} ∴ \frac{1 + a}{1 - a} & = \frac{1 + \cos θ + i \sin θ}{1 - \cos θ - i \sin θ} \\ = \frac{1 + 2 \cos^{2} θ / 2 - 1 + 2 i \sin θ / 2 \cdot \cos θ / 2}{1 - 1 + 2 \sin^{2} θ / 2 - 2 i \sin θ / 2 \cdot \cos θ / 2} = \frac{2 \cos θ / 2 (\cos θ / 2 + i \sin θ / 2)}{2 \sin θ / 2 (\sin θ / 2 - i \cos θ / 2)} \\ = - \frac{2 \cos θ / 2 (\cos θ / 2 + i \sin θ / 2)}{2 i \sin θ / 2 (\cos θ / 2 + i \sin θ / 2)} = - \frac{1}{i} \cot θ / 2 \\ = \frac{+ i^{2}}{i} \cot θ / 2 = i \cot θ / 2 [∵ \frac{- 1}{i} = \frac{i^{2}}{i}] \end{aligned}$

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

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Solution

We have,

$(1 + i) z = (1 - i) \bar{z} \Rightarrow \frac{z}{\bar{z}} = \frac{(1 - i)}{(1 + i)}$

$\begin{matrix} \Rightarrow & \frac{z}{\bar{z}} = \frac{(1 - i)}{(1 + i)} \frac{(1 - i)}{(1 - i)} \Rightarrow \frac{z}{\bar{z}} = \frac{1 + i^{2} - 2 i}{1 - i^{2}} & [∵ i^{2} = - 1] \\ \Rightarrow & \frac{z}{\bar{z}} = \frac{1 - 1 - 2 i}{2} \Rightarrow \frac{z}{\bar{z}} = - i \end{matrix}$

$∴ z = - i \bar{z}$

Hence proved.

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

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Solution

Given that, $z = x + i y$

Then, $\bar{z} = x - i y$

Now, $z \bar{z} + 2 (z + \bar{z}) + b = 0$

$\Rightarrow (x + i y) (x - i y) + 2 (x + i y + x - i y) + b = 0$

$\Rightarrow x^{2} + y^{2} + 4 x + b = 0$ , which is the equation of a circle.

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.

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Solution

$\begin{aligned} Let z & = x + i y \\ Now, \frac{\bar{z} + 2}{\bar{z} - 1} & = \frac{x - i y + 2}{x - i y - 1} \\ = \frac{[(x + 2) - i y] [(x - 1) + i y]}{[(x - 1) - i y] [(x - 1) + i y]} \\ = \frac{(x - 1) (x + 2) - i y (x - 1) + i y (x + 2) + y^{2}}{(x - 1)^{2} + y^{2}} \\ = \frac{(x - 1) (x + 2) + y^{2} + i [(x + 2) y - (x - 1) y]}{(x - 1)^{2} + y^{2}} [∵ - i^{2} = 1] \end{aligned}$

Taking real part, $\frac{(x - 1) (x + 2) + y^{2}}{(x - 1)^{2} + y^{2}} = 4$

$\Rightarrow x^{2} - x + 2 x - 2 + y^{2} = 4 (x^{2} - 2 x + 1 + y^{2})$

$\Rightarrow 3 x^{2} + 3 y^{2} - 9 x + 6 = 0$ , which represents a circle.

Hence, $z$ lies on the circle.

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

Thinking Process

First use, $\arg (\frac{z_{1}}{z_{2}}) = \arg (z_{1}) - \arg (z_{2})$ . Also apply $\arg (z) = θ = \tan^{- 1} \frac{y}{x}$ , where $z = x + i y$

and then use the property $\tan^{- 1} x - \tan^{- 1} y = \tan^{- 1} (\frac{x - y}{1 + x y})$

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Solution

Let $z = x + i y$

Given that, $\arg (\frac{z - 1}{z + 1}) = π / 4$

$\begin{matrix} \Rightarrow & \arg (z - 1) - \arg (z + 1) = π / 4 \\ \Rightarrow & \arg (x + i y - 1) - \arg (x + i y + 1) = π / 4 \\ \Rightarrow & \arg (x - 1 + i y) - \arg (x + 1 + i y) = π / 4 \end{matrix}$

$\begin{matrix} \Rightarrow & \tan^{- 1} \frac{y}{x - 1} - \tan^{- 1} \frac{y}{x + 1} = π / 4 \\ \Rightarrow & \tan^{- 1} [\frac{\frac{y}{x - 1} - \frac{y}{x + 1}}{1 + (\frac{y}{x - 1}) (\frac{y}{x + 1})}] = π / 4 \\ \Rightarrow & \frac{y [\frac{x + 1 - x + 1}{x^{2} - 1}]}{\frac{x^{2} - 1 + y^{2}}{x^{2} - 1}} = \tan π / 4 \\ \Rightarrow & \frac{2 y}{x^{2} + y^{2} - 1} = 1 \\ \Rightarrow & x^{2} + y^{2} - 1 = 2 y \\ \Rightarrow & x^{2} + y^{2} - 2 y - 1 = 0, which represents a circle. \end{matrix}$

11. solve the equation $| z | = z + 1 + 2 i$ .

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Solution

The given equation is $| z | = z + 1 + 2 i$ …(i)

Let $z = x + i y$

From Eq. (i), $| x + i y | = x + i y + 1 + 2 i$

$\begin{matrix} \Rightarrow & \sqrt{x^{2} + y^{2}} = x + i y + 1 + 2 i & [∵ | z | + \sqrt{x^{2} = y^{2}}] \\ \Rightarrow & \sqrt{x^{2} + y^{2}} = (x + 1) + i (y + 2) \end{matrix}$

On squaring both sides, we get

$x^{2} + y^{2} = (x + 1)^{2} + i^{2} (y + 2)^{2} + 2 i (x + 1) (y + 2)$

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

On comparing real and imaginary parts,

$x^{2} + y^{2} = x^{2} + 2 x + 1 - y^{2} - 4 y - 4$

i.e., $2 y^{2} = 2 x - 4 y - 3$ …(ii)

and $2 (x + 1) (y + 2) = 0$

$(x + 1) = 0 or (y + 2) = 0$

$\Rightarrow x = - 1, or y = - 2$

For $x = - 1, 2 y^{2} = - 2 - 4 y - 3$

$2 y^{2} + 4 y + 5 = 0$ [using Eq. (ii)]

$\Rightarrow y = \frac{- 4 \pm \sqrt{16 - 2 \times 4 \times 5}}{4}$

$y = \frac{- 4 \pm \sqrt{- 24}}{4} \notin R$

For $y = - 2, 2 (- 2)^{2} = 2 x - 4 (- 2) - 3$ [using Eq. (ii)]

$\Rightarrow 8 = 2 x + 8 - 3$

$\Rightarrow 2 x = 3 \Rightarrow x = 3 / 2$

$∴ z = x + i y = 3 / 2 - 2 i$

Long Answer Type Questions

12. If $| z + 1 | = z + 2 (1 + i)$ , then find the value of $z$ .

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Solution

Given that,

$\begin{aligned} | z + 1 | & = z + 2 (1 + i) …(i) \\ z & = x + i y \end{aligned}$

Then,

$| x + i y + 1 | = x + i y + 2 (1 + i)$

$\begin{aligned} \Rightarrow | x + 1 + i y | = (x + 2) + i (y + 2) \\ \Rightarrow \sqrt{(x + 1)^{2} + y^{2}} = (x + 2) + i (y + 2) \end{aligned}$

On squaring both sides, we get

$\begin{aligned} (x + 1)^{2} + y^{2} & = (x + 2)^{2} + i^{2} (y + 2)^{2} + 2 i (x + 2) (y + 2) \\ \Rightarrow & x^{2} + 2 x + 1 + y^{2} & = x^{2} + 4 x + 4 - y^{2} - 4 y - 4 + 2 i (x + 2) (y + 2) \\ \Rightarrow & x^{2} + y^{2} + 2 x + 1 & = x^{2} - y^{2} + 4 x - 4 y + 2 i (x + 2) (y + 2) \end{aligned}$

On comparing real and imaginary parts, we get

$x^{2} + y^{2} + 2 x + 1 = x^{2} - y^{2} + 4 x - 4 y$ …(ii)

$\Rightarrow 2 y^{2} - 2 x + 4 y + 1 = 0 …(ii)$

$\begin{matrix} and & 2 (x + 2) (y + 2) = 0 \\ \Rightarrow & x + 2 = 0 or y + 2 & = 0 \\ x & = - 2 or y = - 2 …(iii) \end{matrix}$

For $x = - 2, 2 y^{2} + 4 + 4 y + 1 = 0$ [using Eq. (ii)]

$\Rightarrow 2 y^{2} + 4 y + 5 = 0$

$\Rightarrow 16 - 4 \times 2 \times 5 < 0$

$∴$ Discriminant, $D = b^{2} - 4 a c < 0$

$\Rightarrow 2 y^{2} + 4 y + 5$ has no real roots.

For $y = - 2, 2 (- 2)^{2} - 2 x + 4 (- 2) + 1 = 0$ [using Eq. (ii)]

$\Rightarrow 8 - 2 x - 8 + 1 = 0$

$\Rightarrow x = 1 / 2$

$∴ z = x + i y = \frac{1}{2} - 2 i$

13. If $\arg (z - 1) = \arg (z + 3 i)$ , then find $x - 1 : y$ , where $z = x + i y$ .

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Solution

Given that, $\arg (z - 1) = \arg (z + 3 i)$

$\begin{aligned} and let z & = x + i y \\ Now \arg (z - 1) & = \arg (z + 3 i) \\ \Rightarrow \arg (x + i y - 1) & = \arg (x + i y + 3 i) \\ \Rightarrow \arg (x - 1 + i y) & = \arg [x + i (y + 3)] \\ \Rightarrow \tan^{- 1} \frac{y}{x - 1} & = \tan^{- 1} \frac{y + 3}{x} \end{aligned}$

$\begin{aligned} \Rightarrow \frac{y}{x - 1} & = \frac{y + 3}{x} \Rightarrow x y = (x - 1) (y + 3) \\ \Rightarrow x y & = x y - y + 3 x - 3 \Rightarrow 3 x - 3 = y \end{aligned}$

$\begin{matrix} \Rightarrow & \frac{3 (x - 1)}{y} = 1 \Rightarrow \frac{x - 1}{y} = \frac{1}{3} \\ ∴ & (x - 1) : y = 1 : 3 \end{matrix}$

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

Thinking Process

If $z_{1} = x_{1} + i y_{1}$ and $z_{2} = x_{2} + i y_{2}$ are two complex numbers, then $| \frac{z_{1}}{z_{2}} | = | \frac{z_{1}}{z_{2}} |, (z_{2} \neq 0)$ , use this concept to solve the above problem. Also, we know that general equation of a circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , with centre $(- g, - f)$ and radius $= \sqrt{g^{2} + f^{2} - c}$ .

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Solution

Let $z = x + i y$

Given, equation is $| \frac{z - 2}{z - 3} | = 2 \Rightarrow | \frac{z - 2}{z - 3} | = 2$

$\Rightarrow | \frac{x + i y - 2}{x + i y - 3} | = 2$

$\Rightarrow | x - 2 + i y | = 2 ∣ x - 3 + i y$

$\Rightarrow \sqrt{(x - 2)^{2} + y^{2}} = 2 \sqrt{(x - 3)^{2} + y^{2}} [∵ | x + i y | = \sqrt{x^{2} + y^{2}}]$

On squaring both sides, we get

$\begin{aligned} x^{2} - 4 x + 4 + y^{2} & = 4 (x^{2} - 6 x + 9 + y^{2}) \\ \Rightarrow & 3 x^{2} + 3 y^{2} - 20 x + 32 & = 0 \\ \Rightarrow & x^{2} + y^{2} - \frac{20}{3} x + \frac{32}{3} & = 0 \end{aligned}$

On comparing the above equation with $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , we get

$\begin{aligned} \Rightarrow 2 g = \frac{- 20}{3} \Rightarrow g = \frac{- 10}{3} \\ and 2 f = 0 \Rightarrow f = 0 and c = \frac{32}{3} \\ ∴ Centre = (- g, - f) = (10 / 3, 0) \\ Also, radius (r) = \sqrt{(10 / 3)^{2} + 0 - 32 / 3} [∵ r = \sqrt{g^{2} + f^{2} - c}] \\ = \frac{1}{3} \sqrt{(100 - 96)} = 2 / 3 \end{aligned}$

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

Thinking Process

If $z = x$ +iy is a purely imaginary number, then its real part must be equal to zero i.e., $x = 0$ ,

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Solution

Let

$\begin{aligned} z & = x + i y \\ \frac{z - 1}{z + 1} & = \frac{x + i y - 1}{x + i y + 1}, z \neq - 1 \\ = \frac{x - 1 + i y}{x + 1 + i y} = \frac{(x - 1 + i y) (x + 1 - i y)}{(x + 1 + i y) (x + 1 - i y)} \end{aligned}$

$\begin{aligned} = \frac{(x^{2} - 1) + i y (x + 1) - i y (x - 1) - i^{2} y^{2}}{(x + 1)^{2} - (i y)^{2}} \\ \Rightarrow \frac{z - 1}{z + 1} & = \frac{(x^{2} - 1) + y^{2} + i [y (x + 1) - y (x - 1)]}{(x + 1)^{2} + y^{2}} \end{aligned}$

Given that, $\frac{z - 1}{z + 1}$ is a purely imaginary numbers.

Then, $\frac{(x^{2} - 1) + y^{2}}{(x + 1)^{2} + y^{2}} = 0$

$\begin{matrix} \Rightarrow & x^{2} - 1 + y^{2} = 0 \Rightarrow x^{2} + y^{2} = 1 \\ \Rightarrow & \sqrt{x^{2} + y^{2}} = \sqrt{1} \Rightarrow | z | = 1 & [∵ | z | = \sqrt{x^{2} + y^{2}}] \end{matrix}$

16. $z_{1}$ and $z_{2}$ are two complex numbers such that $| z_{1} | = | z_{2} |$ and $\arg (z_{1}) + \arg (z_{2}) = π$ , then show that $z_{1} = - {\bar{z}}_{2}$ .

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Solution

Let $z_{1} = r_{1} (\cos θ_{1} + i \sin θ_{1})$ and $z_{2} = r_{2} (\cos θ_{2} + i \sin θ_{2})$ are two complex numbers.

$\begin{aligned} Given that, | z_{1} | & = | z_{2} | \\ and \arg (z_{1}) + \arg (z_{2}) & = π \\ If | z_{1} | & = | z_{2} | \\ \Rightarrow r_{1} & = r_{2} \\ and if \arg (z_{1}) + \arg (z_{2}) & = π \\ \Rightarrow θ_{1} + θ_{2} & = π \\ \Rightarrow θ_{1} & = π - θ_{2} \end{aligned}$

$Now z_{1} = r_{1} (\cos θ_{1} + i \sin θ_{1})$

$\begin{matrix} \Rightarrow z_{1} = r_{2} (\cos (π - θ_{2}) + i \sin (π - θ_{2})] & [∵ r_{1} = r_{2} and θ_{1} = (π - θ_{2})] \\ \Rightarrow z_{1} = r_{2} (- \cos θ_{2} + i \sin θ_{2}) \\ \Rightarrow z_{1} = - r_{2} (\cos θ_{2} - i \sin θ_{2}) \\ \Rightarrow z_{1} = - [r_{2} (\cos θ_{2} - i \sin θ_{2})] & [∵ {\bar{z}}_{2} = r_{2} (\cos θ_{2} - i \sin θ_{2})] \\ \Rightarrow z_{1} = - {\bar{z}}_{2} \end{matrix}$

17. If $| z_{1} | = 1 (z_{1} \neq - 1)$ and $z_{2} = \frac{z_{1} - 1}{z_{1} + 1}$ , then show that the real part of $z_{2}$ is zero.

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Solution

Let

$\begin{aligned} z_{1} & = x + i y \\ \Rightarrow | z_{1} | & = \sqrt{x^{2} + y^{2}} = 1 [because | z_{1} | = 1, given] …(i) \\ Now, z_{2} & = \frac{z_{1} - 1}{z_{1} + 1} = \frac{x + i y - 1}{x + i y + 1} \\ = \frac{x - 1 + i y}{x + 1 + i y} = \frac{(x - 1 + i y) (x + 1 - i y)}{(x + 1 + i y) (x + 1 - i y)} \\ = \frac{x^{2} - 1 + i y (x + 1) - i y (x - 1) - i^{2} y^{2}}{(x + 1)^{2} - i^{2} y^{2}} \\ = \frac{x^{2} - 1 + i x y + i y - i x y + i y + y^{2}}{(x + 1)^{2} + y^{2}} \end{aligned}$

$\begin{aligned} = \frac{x^{2} + y^{2} - 1 + 2 i y}{(x + 1)^{2} + y^{2}} = \frac{1 - 1 + 2 i y}{(x + 1)^{2} + y^{2}} [∵ x^{2} + y^{2} = 1] \\ = 0 + \frac{2 y i}{(x + 1)^{2} + y^{2}} \end{aligned}$

Hence, the real part of $z_{2}$ is zero.

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

Thinking Process

First let, $z = r (\cos θ + i \sin θ)$ , then conjugate of $z$ i.e., $\bar{z} = r (\cos θ - i \sin θ)$ . Use the property $\arg (\frac{z_{1}}{z_{2}}) = \arg (z_{1}) - \arg (z_{2})$

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Solution

Let $z_{1} = r_{1} (\cos θ_{1} + i \sin θ_{1})$ ,

Then, $z_{2} = {\bar{z}}_{1} = r_{1} (\cos θ_{1} - i \sin θ_{1}) = r_{1} [\cos (- θ_{1}) + \sin (- θ_{1})]$

Also, let $z_{3} = r_{2} (\cos θ_{2} + i \sin θ_{2})$ ,

Then, $z_{4} = {\bar{z}}_{3} = r_{2} (\cos θ_{2} - i \sin θ_{2})$

$\begin{aligned} \arg \frac{z_{1}}{z_{4}} + \arg \frac{z_{2}}{z_{3}} & = \arg (z_{1}) - \arg (z_{4}) + \arg (z_{2}) - \arg (z_{3}) \\ = θ_{1} - (- θ_{2}) + (- θ_{1}) - θ_{2} [∵ \arg (z) = θ] \\ = θ_{1} + θ_{2} - θ_{1} - θ_{2} = 0 \end{aligned}$

19. If $| z_{1} | = | z_{2} | = \dots = | z_{n} | = 1$ , then show that

$| z_{1} + z_{2} + z_{3} + \dots + z_{n} | = | \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} + \dots + \frac{1}{z_{n}} | .$

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Solution

Given that,

$\Rightarrow | z_{1} | = | z_{2} | = \dots = | z_{n} | = 1$

$\Rightarrow | z_{1} |^{2} = | z_{2} |^{2} = \dots = | z_{n} |^{2} = 1$

$\Rightarrow z_{1} {\bar{z}}_{1} = z_{2} {\bar{z}}_{2} = z_{3} {\bar{z}}_{3} = \dots = z_{n} {\bar{z}}_{n} = 1$

$\Rightarrow z_{1} = \frac{1}{{\bar{z}}_{1}}, z_{2} = \frac{1}{{\bar{z}}_{2}} = \dots = z_{n} = \frac{1}{\overset{―}{z_{n}}}$

Now, $| z_{1} + z_{2} + z_{3} + z_{4} + \dots + z_{n} |$

$\begin{aligned} = | \frac{z_{1} {\bar{z}}_{1}}{{\bar{z}}_{1}} + \frac{z_{2} {\bar{z}}_{2}}{{\bar{z}}_{2}} + \frac{z_{3} {\bar{z}}_{3}}{{\bar{z}}_{3}} + \dots + \frac{z_{n} {\bar{z}}_{n}}{{\bar{z}}_{n}} | ∵ z_{1} \bar{z} = 1, where z_{1} = \frac{1}{z}, z_{1} = \frac{\bar{z}}{\bar{z} - \bar{z}}, z_{1} = \bar{z} \\ = | \frac{| z_{1} |^{2}}{{\bar{z}}_{1}} + \frac{| z_{2} |^{2}}{{\bar{z}}_{2}} + \frac{| z_{3} |^{2}}{z_{3}} + \dots + \frac{| z_{n} |^{2}}{{\bar{z}}_{n}} | \\ = | \frac{1}{{\bar{z}}_{1}} + \frac{1}{{\bar{z}}_{2}} + \frac{1}{{\bar{z}}_{3}} + \dots + \frac{1}{{\bar{z}}_{n}} | = \sqrt{\frac{1}{{\bar{z}}_{1}} + \frac{1}{{\bar{z}}_{2}} + \frac{1}{{\bar{z}}_{3}}} \end{aligned}$

$= | \frac{1}{z_{1}} + \frac{1}{z_{2}} + \dots + \frac{1}{z_{n}} |$

Hence, proved.

20. If the complex numbers $z_{1}$ and $z_{2}$ , $\arg (z_{1}) - \arg (z_{2}) = 0$ , then show that $| z_{1} - z_{2} | = | z_{1} | - | z_{2} |$ .

Show Answer

Solution

Let $z_{1} = r_{1} (\cos θ_{1} + i \sin θ_{1})$

and $z_{2} = r_{2} (\cos θ_{2} + i \sin θ_{2})$

$\Rightarrow \arg (z_{1}) = θ_{1} and \arg (z_{2}) = θ_{2}$

Given that, $\arg (z_{1}) - \arg (z_{2}) = 0$

$θ_{1} - θ_{2} = 0 \Rightarrow θ_{1} = θ_{2}$

$z_{2} = r_{2} (\cos θ_{1} + i \sin θ_{1})$

$z_{1} - z_{2} = (r_{1} \cos θ_{1} - r_{2} \cos θ_{1}) + i (r_{1} \sin θ_{1} - r_{2} \sin θ_{1})$

$\begin{aligned} | z_{1} - z_{2} | & = \sqrt{(r_{1} \cos θ_{1} - r_{2} \cos θ_{1})^{2} + (r_{1} \sin θ_{1} - r_{2} \sin θ_{1})^{2}} \\ = \sqrt{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} \cos^{2} θ_{1} - 2 r_{1} r_{2} \sin^{2} θ_{1}} \\ = \sqrt{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} (\sin^{2} θ_{1} + \cos^{2} θ_{1})} \\ = \sqrt{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2}} = \sqrt{(r_{1} - r_{2})^{2}} \end{aligned}$

$\begin{aligned} \Rightarrow | z_{1} - z_{2} | & = r_{1} - r_{2} \\ = | z_{1} | - | z_{2} | \end{aligned}$

21. solve the system of equations $R e (z^{2}) = 0, | z | = 2$ .

Show Answer

Solution

Given that, $R e (z^{2}) = 0, | z | = 2$

Let $z = x + i y$

$\begin{matrix} ∵ & \sqrt{x^{2} + y^{2}} = 2 \\ \Rightarrow & x^{2} + y^{2} = 4 …(i) \end{matrix}$

and $R e (z) = x$

Also, $z = x + i y$

$\Rightarrow z^{2} = x^{2} + 2 i x y - y^{2}$

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

$\Rightarrow R e (z^{2}) = x^{2} - y^{2}$ $[∵ R e (z^{2}) = 0]$

$\Rightarrow x^{2} - y^{2} = 0$ …(ii)

From Eqs. (i) and (ii),

$x^{2} + x^{2} = 4$

$\Rightarrow 2 x^{2} = 4 \Rightarrow x^{2} = 2$

$\Rightarrow x = \pm \sqrt{2}$

$∴ y = \pm \sqrt{2}$

$∵ z = x + i y$

$\Rightarrow z = \sqrt{2} \pm i \sqrt{2}, - \sqrt{2} \pm i \sqrt{2}$

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

Show Answer

Solution

Given equation is $z + \sqrt{2} | (z + 1) | + i = 0$

Let $z = x + i y$

$\Rightarrow x + i y + \sqrt{2} | x + i y + 1 | + i = 0$

$\Rightarrow x + i (1 + y) + \sqrt{2} \sqrt{(x + 1)^{2} + y^{2}} = 0$

$\Rightarrow x + i (1 + y) + \sqrt{2} \sqrt{(x^{2} + 2 x + 1 + y^{2})} = 0$

$\Rightarrow x + \sqrt{2} \sqrt{x^{2} + 2 x + 1 + y^{2}} = 0$

$\Rightarrow x^{2} = 2 (x^{2} + 2 x + 1 + y^{2}$

$\Rightarrow x^{2} + 4 x + 2 y^{2} + 2 = 0$ …(ii)

$1 + y = 0$

$\Rightarrow y = - 1$

For $y = - 1, x^{2} + 4 x + 2 + 2 = 0$ [using Eq. (ii)]

$\Rightarrow x^{2} + 4 x + 4 = 0 \Rightarrow (x + 2)^{2} = 0$

$\Rightarrow x + 2 = 0 \Rightarrow x = - 2$

$∴ z = x + i y = - 2 - i$

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

Show Answer

Solution

Given that, $z = \frac{1 - i}{\cos \frac{π}{3} + i \sin \frac{π}{3}} = \frac{- \sqrt{2} [\frac{- 1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}]}{\cos \frac{π}{3} + i \sin \frac{π}{3}}$

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

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

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

$= - \sqrt{2} [\cos \frac{5 π}{12} + i \sin \frac{5 π}{12}]$

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

Show Answer

Solution

Let $z = r_{1} (\cos θ_{1} + i \sin θ_{1})$ and $w = r_{2} (\cos θ_{2} + i \sin θ_{2})$

Also, $| z w | = | z | | w | = r_{1} r_{2} = 1$

$∴ r_{1} r_{2} = 1$

Further, $\arg (z) = θ_{1}$ and $\arg (w) = θ_{2}$

$\begin{aligned} But \arg (z) - \arg (w) & = \frac{π}{2} \\ \Rightarrow θ_{1} - θ_{2} & = \frac{π}{2} \\ \Rightarrow \arg (\frac{z}{w}) & = \frac{π}{2} \\ Now, to prove \bar{z} w & = - i \\ LHS & = \bar{z} w \\ = r_{1} (\cos θ_{1} - i \sin θ_{1}) r_{2} (\cos θ_{2} + i \sin θ_{2}) \\ = r_{1} r_{2} [\cos (θ_{2} - θ_{1}) + i \sin (θ_{2} - θ_{1})] \\ = r_{1} r_{2} [\cos (- π / 2) + i \sin (- π / 2)] \\ = 1 [0 - i] \\ = - i = R H S \end{aligned}$

Hence proved.

Fillers

25. Fill in the blanks of the following.

(i) For any two complex numbers $z_{1}, z_{2}$ 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 $i + i^{2} + i^{3} +$ …… upto 1000 terms is ……

(v) Multiplicative inverse of $1 + i$ is ……

(vi) If $z_{1}$ and $z_{2}$ are complex numbers such that $z_{1} + z_{2}$ is a real number, then $z_{1} =$ ……

(vii) $\arg (z) + \arg \bar{z}$ where, $(\bar{z} \neq 0)$ is ……

(viii) If $| z + 4 | \leq 3$ , 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 $z =$ ……

Show Answer

Solution

(i) $| a z_{1} - b z_{2} |^{2} + | b z_{1} + a z_{2} |^{2}$

$\begin{aligned} = | a z_{1} |^{2} + | b z_{2} |^{2} - 2 R e (a z_{1} \cdot b {\bar{z}}_{2}) + | b z_{1} |^{2} + | a z_{2} |^{2} + 2 R e (a z_{1} \cdot b {\bar{z}}_{2}) \\ = (a^{2} + b^{2}) | z_{1} |^{2} + (a^{2} + b^{2}) | z_{2} |^{2} \\ = (a^{2} + b^{2}) (| z_{1} |^{2} + | z_{2} |^{2}) \end{aligned}$

(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) $i + i^{2} + i^{3} + \dots$ upto 1000 terms $= i + i^{2} + i^{3} + i^{4} + \dots i^{1000} = 0$

$[∵ i^{n} + i^{n + 1} + i^{n + 2} + i^{n + 3} = 0, where n \in N i . e ., \sum_{n = 1}^{1000} i^{n} = 0]$

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

(vi) Let $z_{1} = x_{1} + i y_{1}$ and $z_{2} = x_{2} + i y_{2}$

$z_{1} + z_{2} = (x_{1} + x_{2}) + i (y_{1} + y_{2})$ , which is real.

If $z_{1} + z_{2}$ is real, then $y_{1} + y_{2} = 0$

$\Rightarrow y_{1} = - y_{2}$

$∵ z_{2} = x_{2} - i y_{1}$

$\Rightarrow z_{2} = {\bar{z}}_{1}$ [when $x_{1} = x_{2}$ ]

(vii) $\arg (z) + \arg (\bar{z}), (\bar{z} \neq 0)$

$\Rightarrow θ + (- θ) = 0$

(viii) Given that, $| z + 4 | \leq 3$

For the greatest value of $| z + 1 |$ .

$\begin{aligned} \Rightarrow | z + 1 | & = | z + 4 - 3 | \leq | z + 4 | + | - 3 | \\ = | z + 4 - 3 | \leq 3 + 3 \\ = | z + 4 - 3 | \leq 6 \end{aligned}$

So, greatest value of $| z + 1 | = 6$

For, now, least value of $| z + 1 |$ , we know that the least value of the modulus of a complex number is zero. So, the least value of $| z + 1 |$ is zero.

(ix) Given that,

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

$\begin{matrix} \Rightarrow \frac{| x + i y - 2 |}{| x + i y + 2 |} = \frac{π}{6} \Rightarrow \frac{| x - 2 + i y |}{| x + 2 + i y |} = \frac{π}{6} \\ \Rightarrow 6 | x - 2 + i y | = π | x + 2 + i y | \\ \Rightarrow 6 \sqrt{(x - 2)^{2} + y^{2}} = π \sqrt{(x + 2)^{2} + y^{2}} \\ \Rightarrow 36 [x^{2} + 4 - 4 x + y^{2}] = π^{2} [x^{2} + 4 x + 4 + y^{2}] \\ \Rightarrow (36 - π^{2}) x^{2} + (36 - π^{2}) y^{2} - (144 + 4 π^{2}) x + 144 + 4 π^{2} = 0, which is a circle. \end{matrix}$

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

$\begin{matrix} Let z = x + i y = r (\cos θ + i \sin θ) \\ \Rightarrow | z | = r = 4 and \arg (z) = θ \\ ∵ \tan θ = \frac{5 π}{6} \\ \Rightarrow z = 4 (\cos \frac{π}{6} + i \sin \frac{π}{6}) = 4 [\cos (π - π / 6) + i \sin (π - π / 6)] \\ = 4 [- \cos \frac{π}{6} + i \sin \frac{π}{6}] = 4 [\frac{- \sqrt{3}}{2} + i \frac{1}{2}] = - 2 \sqrt{3} + 2 i \end{matrix}$

True/False

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 $a r r o w$ 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 the points $(1, 0)$ and $(0, 1)$ .

(v) If $z$ is a complex number such that $z \neq 0$ and $R e (z) = 0$ , then, $I m (z^{2}) = 0$ .

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

(vii) Let $z_{1}$ and $z_{2}$ be two complex numbers such that $| z_{1} + z_{2} | = | z_{1} | + | z_{2} |$ , then $\arg (z_{1} - z_{2}) = 0$ .

(viii) 2 is not a complex number.

Show Answer

Solution

(i) False

We can compare two complex numbers when they are purely real. Otherwise comparison of complex number is not possible.

(ii) False

$(x, y) [\begin{matrix} - 1 \\ 0 \end{matrix}] = [\begin{matrix} 0 \\ - y \end{matrix}], which is false.$

(iii) True

$\begin{aligned} Let \begin{aligned} z & = x + i y \\ | z | + | z - 1 | & = \sqrt{x^{2} + y^{2}} + \sqrt{(x - 1)^{2} + y^{2}} \end{aligned} \end{aligned}$

If $x = 0, y = 0$ , then the value of $| z | + | z - 1 | = 1$ .

(iv) True

$\begin{aligned} Let z & = x + i y \\ | z - 1 | & = | z - i | \\ Then, | x - 1 + i y | & = | x - i (1 - y) | \\ (x - 1)^{2} + y^{2} & = x^{2} + (1 - y)^{2} \\ x^{2} - 2 x + 1 + y^{2} & = x^{2} + 1 + y^{2} - 2 y \\ - 2 x + 1 & = 1 - 2 y \\ - 2 x + 2 y & = 0 \\ x - y & = 0 …(i) \end{aligned}$

Equation of a line through the points $(1, 0)$ and $(0, 1)$ ,

$\begin{aligned} y - 0 = \frac{1 - 0}{0 - 1} (x - 1) \\ \Rightarrow y = - (x - 1) \Rightarrow x + y = 1 …(ii) \end{aligned}$

which is perpendicular to the line $x - y = 0$ .

(v) False

Let $z = x + i y, z \neq 0$ and $R e (z) = 0$

i.e., $x = 0$

$∴ z = i y$

$I m (z^{2}) = i^{2} y^{2} = - y^{2}$ which is real.

(vi) True

Given inequality, $| z - 4 | < | z - 2 |$

Let $z = x + i y$

$∴ \begin{aligned} | x - 4 + i y | & < | x - 2 + i y | \\ \Rightarrow \sqrt{(x - 4)^{2} + y^{2}} & < \sqrt{(x - 2)^{2} + y^{2}} \end{aligned}$

$\Rightarrow (x - 4)^{2} + y^{2} < (x - 2)^{2} + y^{2}$

$\Rightarrow x^{2} - 8 x + 16 + y^{2} < x^{2} - 4 x + 4 + y^{2}$

$\Rightarrow - 8 x + 16 < - 4 x + 4$

$\Rightarrow - 8 x < - 4 x - 12$

$\Rightarrow - 4 x < - 12$

$\Rightarrow 4 x > 12$

$\Rightarrow x > 3$

(vii) False

$\begin{aligned} Let & z_{1} = x_{1} + i y_{1} and z_{2} = x_{2} + i y_{2} \\ Given that, & | z_{1} + z_{2} | = | z_{1} | + | z_{2} | \end{aligned}$

$\begin{aligned} | x_{1} + i y_{1} + x_{2} + i y_{2} | & = | x_{1} + i y_{1} | + | x_{2} + i y_{2} | \\ \Rightarrow \sqrt{(x_{1} + x_{2})^{2} + (y_{1} + y_{2})^{2}} & = \sqrt{(x_{1}^{2} + y_{1}^{2})} + \sqrt{(x_{2}^{2} + y_{2}^{2})} \end{aligned}$

On squaring both sides, we get

$\begin{aligned} (x_{1} + x_{2})^{2} + (y_{1} + y_{2})^{2} = (x_{1}^{2} + y_{1}^{2}) + (x_{2}^{2} + y_{2}^{2}) + 2 \sqrt{(x_{1}^{2} + y_{1}^{2}) (x_{2}^{2} + y_{2}^{2})} \\ \Rightarrow & x_{1}^{2} + x_{2}^{2} + 2 x_{1} x_{2} + y_{1}^{2} + y_{2}^{2} + 2 y_{1} y_{2} = x_{1}^{2} + y_{1}^{2} + x_{2}^{2} + y_{2}^{2} + 2 \sqrt{(x_{1}^{2} + y_{1}^{2}) (x_{2}^{2} + y_{2}^{2})} \\ \Rightarrow & 2 x_{1} x_{2} + 2 y_{1} y_{2} = 2 \sqrt{(x_{1}^{2} + y_{1}^{2}) (x_{2}^{2} + y_{2}^{2})} \\ \Rightarrow & x_{1} x_{2} + y_{1} y_{2} = \sqrt{(x_{1}^{2} + y_{1}^{2}) (x_{2}^{2} + y_{2}^{2})} \end{aligned}$

On squaring both sides, we get

$x_{1}^{2} x_{2}^{2} + y_{1}^{2} y_{2}^{2} + 2 x_{1} x_{2} y_{1} y_{2} = x_{1}^{2} x_{2}^{2} + y_{1}^{2} x_{2}^{2} + x_{1}^{2} y_{2}^{2} + y_{1}^{2} y_{2}^{2}$

$\Rightarrow (x_{1} y_{2} - x_{2} y_{1})^{2} = 0$

$\Rightarrow x_{1} y_{2} = x_{2} y_{1}$

$\Rightarrow \frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}}$

$\Rightarrow (\frac{y_{1}}{x_{1}}) - (\frac{y_{2}}{x_{2}}) = 0$

$\Rightarrow \arg (z_{1}) - \arg (z_{2}) = 0$

(viii) True

We know that, 2 is a real number.

Since, 2 is not a complex number.

27. Match the statements of Column A and Column B.

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

Show Answer

Solution

$\begin{aligned} (i) Given, z & = i + \sqrt{3} = r (\cos θ + \sin θ) \\ ∵ r \cos θ & = \sqrt{3}, r \sin θ = 1 \\ \Rightarrow r^{2} & = 1 + 3 = 4 \Rightarrow r = 2 [∵ r > 0] \\ \Rightarrow \tan α & = | \frac{r \sin θ}{r \cos θ} | = \frac{1}{\sqrt{3}} \\ \Rightarrow \tan α & = \frac{1}{\sqrt{3}} \Rightarrow α = \frac{π}{6} \\ ∵ x & > 0, y > 0 \\ and \arg (z) & = θ = \frac{π}{6} \end{aligned}$

So the polar form of $z$ is $2 (\cos \frac{π}{6} + i \sin \frac{π}{6})$ .

(ii) Given that, $z = - 1 + \sqrt{- 3} = - 1 + i \sqrt{3}$

$∴ \tan α = | \frac{\sqrt{3}}{- 1} | = \sqrt{3}$

$\Rightarrow \tan α = \tan \frac{π}{3} \Rightarrow α = \frac{π}{3}$

$\begin{matrix} ∵ & x < 0, y > 0 \\ \Rightarrow & θ = π - α = π - \frac{π}{3} = \frac{2 π}{3} \end{matrix}$

(iii) Given that, $| z + 2 | = | z - 2 |$

$\begin{matrix} \Rightarrow | x + 2 + i y | = | x - 2 + i y | \\ \Rightarrow (x + 2)^{2} + y^{2} = (x - 2)^{2} + y^{2} \\ \Rightarrow x^{2} + 4 x + 4 = x^{2} - 4 x + 4 \Rightarrow 8 x = 0 \\ ∴ x = 0 \end{matrix}$

It is a straight line which is a perpendicular bisector of segment joining the points $(- 2, 0)$ and $(2, 0)$

(iv) Given that, $| z + 2 i | = | z - 2 i |$

$\begin{matrix} \Rightarrow & | x + i (y + 2) | & = | x + i (y - 2) | \\ \Rightarrow & x^{2} + (y + 2)^{2} & = x^{2} + (y - 2)^{2} \\ \Rightarrow & 4 y & = 0 \Rightarrow y = 0 \end{matrix}$

It is a straight line, which is a perpendicular bisector of segment joining $(0, - 2)$ and $(0, 2)$ .

(v) Given that, $| z + 4 i | \geq 3 = | x + i y + 4 i | \geq 3$

$\begin{aligned} \Rightarrow & = | x + i (y + 4) | \geq 3 \\ \Rightarrow & = \sqrt{x^{2} + (y + 4)^{2}} \geq 3 \\ \Rightarrow & = x^{2} + (y + 4)^{2} \geq 9 \\ \Rightarrow & = x^{2} + y^{2} + 8 y + 16 \geq 9 \\ = x^{2} + y^{2} + 8 y + 7 \geq 0 \end{aligned}$

Which represent a circle. On or outside having centre $(0, - 4)$ and radius 3 .

(vi) Given that, $| z + 4 | \leq 3$

$\begin{matrix} \Rightarrow & | x + i y + 4 | \leq 3 \\ \Rightarrow & | x + 4 + i y | \leq 3 \\ \Rightarrow & \sqrt{(x + 4)^{2} + y^{2}} \leq 3 \\ \Rightarrow & (x + 4)^{2} + y^{2} \leq 9 \\ \Rightarrow & x^{2} + 8 x + 16 + y^{2} \leq 9 \\ \Rightarrow & x^{2} + 8 x + y^{2} + 7 \leq 0 \end{matrix}$

It represent the region which is on or inside the circle having the centre $(- 4, 0)$ and radius 3.

(vii) Given that,

$\begin{aligned} z & = \frac{1 + 2 i}{1 - i} = \frac{(1 + 2 i) (1 + i)}{(1 - i) (1 + i)} \\ = \frac{1 + 2 i + i + 2 i^{2}}{1 - i^{2}} = \frac{1 - 2 + 3 i}{1 + 1} = \frac{- 1 + 3 i}{2} \end{aligned}$

$∴ \bar{z} = \frac{- 1}{2} - \frac{3 i}{2}$

Hence, $(\frac{- 1}{2}, \frac{- 3}{2})$ lies in third quadrant.

(viii) Given that, $z = 1 - i$

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

Hence, $(\frac{1}{2}, \frac{1}{2})$ lies in first quadrant. Hence, the correct matches are

(a) $\to$ (v),

(b) $\to$ (iii),

(c) $\to$ (i),

(d) $\to$ (iv),

(e) $\to$ (ii),

(f) $\to$ (vi),

(g) $\to$ (viii),

(h) $\to$ (vii)

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

Show Answer

Solution

$\begin{aligned} Given that, z = \frac{2 - i}{(1 - 2 i)^{2}} = \frac{2 - i}{1 + 4 i^{2} - 4 i} \\ = \frac{2 - i}{1 - 4 - 4 i} = \frac{2 - i}{- 3 - 4 i} \\ = \frac{(2 - i)}{- (3 + 4 i)} = - [\frac{(2 - i) (3 - 4 i)}{(3 + 4 i) (3 - 4 i)}] \\ = - (\frac{6 - 8 i - 3 i + 4 i^{2}}{9 + 16}) = - \frac{(- 11 i + 2)}{25} \\ = \frac{- 1}{25} (2 - 11 i) \Rightarrow z = \frac{1}{25} (- 2 + 11 i) \\ ∴ \bar{z} = \frac{1}{25} (- 2 - 11 i) = \frac{- 2}{25} - \frac{11}{25} i \end{aligned}$

29. If $| z_{1} | = | z_{2} |$ , is it necessary that $z_{1} = z_{2}$ .

Show Answer

Solution

$\begin{aligned} Given that, | z_{1} | & = | z_{2} | \\ Let z_{1} & = x_{1} + i y_{1} and z_{2} = x_{2} + i y_{2} \\ \Rightarrow | x_{1} + i y_{1} | & = | x_{2} + i y_{2} | \end{aligned}$

$\begin{aligned} \Rightarrow x_{1}^{2} + y_{1}^{2} & = x_{2}^{2} + y_{2}^{2} \\ \Rightarrow x_{1}^{2} & = x_{2}^{2}, y_{1}^{2} = y_{2}^{2} \\ \Rightarrow x_{1} & = \pm x_{2}, y_{1} = \pm y_{2} \\ \Rightarrow z_{1} & = x_{1} + i y_{1} or z_{1} = \pm x_{2} \pm i y_{2} \end{aligned}$

Hence, it is not neccessary that $z_{1} = z_{2}$ .

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

Show Answer

Solution

Given that, $\frac{(a^{2} + 1)^{2}}{2 a - i} = x + i y \Rightarrow \frac{(a^{2} + 1)^{2}}{(2 a - i)} = x + i y$

$\begin{matrix} \Rightarrow & \frac{(a^{2} + 1)^{2} (2 a + i)}{(2 a - i) (2 a + i)} & = x + i y \\ \Rightarrow & \frac{(a^{2} + 1)^{2} (2 a + i)}{4 a^{2} + 1} & = x + i y \\ \Rightarrow x & = \frac{2 a (a^{2} + 1)^{2}}{4 a^{2} + 1} and y = \frac{(a^{2} + 1)^{2}}{4 a^{2} + 1} \\ ∴ & x^{2} + y^{2} & = 4 a^{2} [\frac{(a^{2} + 1)^{2}}{4 a^{2} + 1}] + [\frac{(a^{2} + 1)^{2}}{4 a^{2} + 1}] \\ = \frac{(4 a^{2} + 1) (a^{2} + 1)^{4}}{(4 a^{2} + 1)^{2}} = \frac{(a^{2} + 1)^{4}}{(4 a^{2} + 1)} \end{matrix}$

31. Find the value of $z$ , if $| z | = 4$ and $\arg (z) = \frac{5 π}{6}$ .

Show Answer

Solution

Let $z = r (\cos θ + i \sin θ)$

$\begin{matrix} \begin{aligned} Also, | z | & = r = 4 and θ = \frac{5 π}{6} [∵ \arg (z) = θ] \\ ∴ & z = 4 [\cos \frac{5 π}{6} + i \sin \frac{5 π}{6}] [∵ z = r (\cos θ + i \sin θ)] \\ = 4 [\cos (π - \frac{π}{6}) + i \sin (π - \frac{π}{6})] \\ = 4 [- \cos \frac{π}{6} + i \sin \frac{π}{6}] \\ = 4 [- \frac{\sqrt{3}}{2} + i \frac{1}{2}] = - 2 \sqrt{3} + 2 i \end{aligned} \end{matrix}$

32. Find the value of $| (1 + i) \frac{(2 + i)}{(3 + i)} |$ .

Thinking Process

First, convert the given expression in the formed $a + i b$ , then use $| a + i b | = \sqrt{a^{2} + b^{2}}$ .

Show Answer

Solution

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

$\begin{aligned} = | \frac{1 + 3 i}{3 + i} | = | \frac{(1 + 3 i) (3 - i)}{(3 + i) (3 - i)} | \\ = | \frac{3 + 9 i - i - 3 i^{2}}{9 - i^{2}} | = | \frac{3 + 8 i + 3}{9 + 1} | = | \frac{6 + 8 i}{10} | \\ = \sqrt{\frac{6^{2}}{100} + \frac{8^{2}}{100}} = \sqrt{\frac{36 + 64}{100}} = \sqrt{\frac{100}{100}} = 1 \end{aligned}$

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

Thinking Process

Let $z = a + i b$ , then the polar form of $z$ is $r (\cos θ + i \sin θ)$ , where $r = | z | = \sqrt{a^{2} + b^{2}}$ and $\tan θ = \frac{b}{a} \cdot$ Here, $θ$ is argument or amplitude of $z$ i.e., $\arg (z) = θ$ . The principal argument is a unique value of $θ$ such that $- π \leq θ \leq π$ .

Show Answer

Solution

Given that,

$\begin{aligned} z = (1 + i \sqrt{3})^{2} \\ z = 1 - 3 + 2 i \sqrt{3} \Rightarrow z = - 2 + i 2 \sqrt{3} \end{aligned}$

$\begin{matrix} \Rightarrow & \tan α = | \frac{2 \sqrt{3}}{- 2} | = | - \sqrt{3} | = \sqrt{3} [∵ \tan α = | \frac{I m (z)}{R e (z)} |] \\ \Rightarrow & \tan α = \tan \frac{π}{3} \Rightarrow α = \frac{π}{3} \\ ∵ & R e (z) < 0 and I m (z) > 0 \\ \Rightarrow & \arg (z) = π - \frac{π}{3} \Rightarrow= \frac{2 π}{3} \end{matrix}$

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

Thinking Process

$\begin{aligned} If z_{1} = x_{1} tiy and z_{2} = x_{2} + i y_{2}, then | z_{1} | = \sqrt{x_{1}^{2} + y_{1}^{2}} and | z_{2} | = \sqrt{x_{2}^{2} + y_{2}^{2}} . \\ Also, use the modulus property i.e., | \frac{z_{1}}{z_{2}} | = \frac{| z_{1} |}{| z_{2} |} \end{aligned}$

Show Answer

Solution

Let $z = x + i y$

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

$\Rightarrow | \frac{z - 5 i}{z + 5 i} | = \frac{| x + i (y - 5) |}{| x + i (y + 5) |} [∵ | \frac{z - 5 i}{z + 5 i} | = 1]$

$\Rightarrow | \frac{z - 5 i}{z + 5 i} | = \frac{\sqrt{x^{2} + (y - 5)^{2}}}{\sqrt{x^{2} + (y + 5)^{2}}}$

On squaring both sides, we get

$x^{2} + (y - 5)^{2} = x^{2} + (y + 5)^{2}$

$\begin{matrix} \Rightarrow & - 10 y & = + 10 y \\ \Rightarrow & 20 y & = 0 \\ ∴ & y & = 0 \end{matrix}$

So, $z$ lies on real axis.

Objective Type Questions

35. $\sin x + i \cos 2 x$ and $\cos x - i \sin 2 x$ are conjugate to each other for

(a) $x = n π$

(b) $x = n + \frac{1}{2} \frac{π}{2}$

(d) No value of $x$

Show Answer

Solution

(d) Let $z = \sin x + i \cos 2 x$

and $\bar{z} = \sin x - i \cos 2 x$

Given that, $\bar{z} = \cos x - i \sin 2 x$

$∴ \sin x - i \cos 2 x = \cos x - i \sin 2 x$

$\Rightarrow \sin x = \cos x$ and $\cos 2 x = \sin 2 x$

$\Rightarrow \tan x = 1$ and $\tan 2 x = 1$

$\Rightarrow \tan x = \tan \frac{π}{4}$ and $\tan 2 x = \tan \frac{π}{4}$

$\Rightarrow x = n π + \frac{π}{4}$ and $2 x = n π + \frac{π}{4}$

$\Rightarrow 2 x - x = 0 \Rightarrow x = 0$

36. 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) $(2 n + 1) \frac{π}{2}$

(d) None of these

where, $n \in N$

Thinking Process

First, convert the given expansion into a +ibform and then check whether the complex number a +ib is purely real.

Show Answer

Solution

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

It is given that $z$ is a purely real.

$\begin{aligned} ∴ \frac{- 3 \sin α}{1 + 4 \sin^{2} α} & = 0 \\ \Rightarrow - 3 \sin α & = 0 \Rightarrow \sin α = 0 \\ α & = n π \end{aligned}$

37. If $z = x$ +iy lies in the third quadrant, then $\frac{\bar{z}}{z}$ also lies in the third quadrant, if

(a) $x > y > 0$

(b) $x < y < 0$

(d) $y > x > 0$

Show Answer

Solution

(b) Given that, $z = x + i y$ lies in third quadrant.

$x < 0 and y < 0 .$

Now, $\frac{\bar{z}}{z} = \frac{x - i y}{x + i y} = \frac{(x - i y) (x - i y)}{(x + i y) (x - i y)} = \frac{x^{2} - y^{2} - 2 i x y}{x^{2} + y^{2}}$

$\frac{\bar{z}}{z} = \frac{x^{2} - y^{2}}{x^{2} + y^{2}} - \frac{2 i x y}{x^{2} + y^{2}}$

Since, $\frac{\bar{z}}{z}$ also lies in third quadrant.

$\begin{matrix} ∴ \frac{x^{2} - y^{2}}{x^{2} + y^{2}} < 0 and \frac{- 2 x y}{x^{2} + y^{2}} < 0 \\ x^{2} - y^{2} < 0 and - 2 x y < 0 \\ \Rightarrow x^{2} < y^{2} and x y > 0 \\ So, x < y < 0 \end{matrix}$

38. The value of $(z + 3) (\bar{z} + 3)$ is equivalent to

(a) $| z + 3 |^{2}$

(b) $| z - 3 |$

(d) None of these

Show Answer

Solution

(a) Given that, $(z + 3) (\bar{z} + 3)$

Let $z = x + i y$

$\Rightarrow (z + 3) (\bar{z} + 3) = (x + i y + 3) (x + 3 - i y)$

$= (x + 3)^{2} - (i y)^{2} = (x + 3)^{2} + y^{2}$

$= | x + 3 + i y |^{2} = | z + 3 |^{2}$

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

(a) $x = 2 n + 1$

(b) $x = 4 n$

(d) $x = 4 n + 1$

where, $n \in N$

Show Answer

Solution

(b) Given that, ${(\frac{1 + i}{1 - i})}^{x} = 1$

$\begin{aligned} \Rightarrow & {[\frac{(1 + i) (1 + i)}{(1 - i) (1 + i)}]}^{x} & = 1 \Rightarrow {[\frac{1 + 2 i + i^{2}}{1 - i^{2}}]}^{x} = 1 \\ \Rightarrow & {[\frac{2 i}{1 + 1}]}^{x} & = 1 \Rightarrow {[\frac{2 i}{2}]}^{x} = 1 \\ \Rightarrow & i^{x} & = 1 \Rightarrow i^{x} = (i^{4 n}) \\ \Rightarrow & x & = 4 n \end{aligned}$

40. $A$ real value of $x$ satisfies the equation $(\frac{3 - 4 i x}{3 + 4 i x}) = α - i β (α, β \in R)$ , if $α^{2} + β^{2}$ is equal to

(a) 1

(b) -1

(d) -2

Show Answer

Solution

(a) Given equation, $(\frac{3 - 4 i x}{3 + 4 i x}) = α - i β (α, β \in R)$

$\begin{matrix} \Rightarrow & [\frac{3 - 4 i x}{3 + 4 i x}] = α - i β \\ Now, & (α - i β) = \frac{(3 - 4 i x) (3 - 4 i x)}{(3 + 4 i x) (3 - 4 i x)} = \frac{9 + 16 i^{2} x^{2} - 24 i x}{9 - 16 i^{2} x^{2}} \\ \Rightarrow & α - i β = \frac{9 - 16 x^{2} - 24 i x}{9 + 16 x^{2}} \\ \Rightarrow & α - i β = \frac{9 - 16 x^{2}}{9 + 16 x^{2}} - \frac{i 24 x}{9 + 16 x^{2}} …(i) \\ ∴ & α + i β = \frac{9 - 16 x^{2}}{9 + 16 x^{2}} + \frac{i 24 x}{9 + 16 x^{2}} …(ii) \end{matrix}$

$\begin{aligned} So, (α - i β) (α + i β) & = (\frac{9 - 16 x^{2}}{9 + 16 x^{2}}) - {(\frac{i 24 x}{9 + 16 x^{2}})}^{2} \\ ∴ α^{2} + β^{2} & = \frac{81 + 256 x^{4} - 288 x^{2} + 576 x^{2}}{(9 + 16 x^{2})^{2}} \\ = \frac{81 + 256 x^{4} + 288 x^{2}}{(9 + 16 x^{2})^{2}} \\ = \frac{(9 + 16 x^{2})^{2}}{(9 + 16 x^{2})^{2}} = 1 \end{aligned}$

41. 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) $\arg (z_{1} z_{2}) = \arg (z_{1}) \cdot \arg (z_{2})$

(d) $| z_{1} + z_{2} | \geq | z_{1} | - | z_{2} |$

Show Answer

Solution

(a) Let $z_{1} = r_{1} (\cos θ_{1} + i \sin θ_{1})$

$\Rightarrow | z_{1} | = r_{1}$

and $z_{2} = r_{2} (\cos θ_{2} + i \sin θ_{2})$

$\Rightarrow | z_{2} | = r_{2}$

Now, $z_{1} z_{2} = r_{1} r_{2} [\cos θ_{1} \cos θ_{2} + i \sin θ_{1} \cos θ_{2} + i \cos θ_{1} \sin θ_{2} + i^{2} \sin θ_{1} \sin θ_{2}]$ $= r_{1} r_{2} [\cos (θ_{1} + θ_{2}) + i \sin (θ_{1} + θ_{2})]$

$\Rightarrow | z_{1} z_{2} | = r_{1} r_{2}$

$∴ | z_{1} z_{2} | = | z_{1} | | z_{2} |$ [using Eqs. (i) and (ii)]

42. 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) $1 + 2 i$

(b) $- 1 - 2 i$

(d) $- 1 + 2 i$

Thinking Process

Here, $z < α$ is a complex number, where modulus is $r$ and argument $(θ + α)$ . If $P (z)$ rotates in clockwise sense through an angle $α$ , then its new position will be $z (θ - i α)$ .

Show Answer

Solution

(b) Given that, $z = 2 - i$

It is rotated about origin through an angle $\frac{π}{2}$ in the clockwise direction

$∴$ New position $= z e^{- i π / 2} = (2 - i) e^{- i π / 2}$

$\begin{aligned} = (2 - i) [\cos (\frac{- π}{2}) + i \sin (\frac{- π}{2})] = (2 - i) [0 - i] \\ = - 2 i - 1 = - 1 - 2 i \end{aligned}$

43. If $x, y \in R$ , then $x + i y$ is a non-real complex number, if

(a) $x = 0$

(b) $y = 0$

(d) $y \neq 0$

Show Answer

Solution

(d) Given that, $x, y \in R$

Then, $x + i y$ is non-real complex number if and only if $y \neq 0$ .

44. If $a + i b = c + i d$ , then

(a) $a^{2} + c^{2} = 0$

(b) $b^{2} + c^{2} = 0$

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

Thinking Process

If two complex numbers $z_{1} = x_{1} + i y_{1}$ and $z_{2} = x_{2} + i y_{2}$ are equal, then

$| z_{1} | = | z_{2} | \Rightarrow \sqrt{x_{1}^{2} + y_{1}^{2}} = \sqrt{x_{2}^{2} + y_{2}^{2}}$

Show Answer

Solution

(d) Given that,

$\begin{matrix} a + i b = c + i d \\ \Rightarrow | a + i b | = | c + i d | \\ \Rightarrow \sqrt{a^{2} + b^{2}} = \sqrt{c^{2} + d^{2}} \end{matrix}$

On squaring both sides, we get

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

45. The complex number $z$ which satisfies the condition $| \frac{i + z}{i - z} | = 1$ lies on

(a) circle $x^{2} + y^{2} = 1$

(b) the $X$ -axis

(d) the line $x + y = 1$

Show Answer

Solution

(b) Given that, $| \frac{i + z}{i - z} | = 1$

Let $z = x + i y$

$\begin{aligned} ∴ | \frac{x + i (y + 1)}{- x - i (y - 1)} | = 1 \Rightarrow \frac{x^{2} + (y + 1)^{2}}{x^{2} + (y - 1)^{2}} = 1 \\ \Rightarrow x^{2} + (y + 1)^{2} = x^{2} + (y - 1)^{2} \\ \Rightarrow 4 y = 0 \Rightarrow y = 0 \end{aligned}$

So, $z$ lies on $X$ -axis (real axis).

46. If $z$ is a complex number, then

(a) $| z^{2} | > | z |$

(b) $| z^{2} | = | z |^{2}$

(d) $| z^{2} | \geq | z |^{2}$

Show Answer

Solution

(b) If $z$ is a complex number, then $z = x + i y$

$\begin{aligned} | z | = | x + i y | and | z |^{2} = | x + i y |^{2} \\ \Rightarrow | z |^{2} = x^{2} + y^{2} …(i) \\ and z^{2} = (x + i y)^{2} = x^{2} + i^{2} y^{2} + i 2 x y \\ z^{2} = x^{2} - y^{2} + i 2 x y \\ \Rightarrow | z^{2} | = \sqrt{(x^{2} - y^{2})^{2} + (2 x y)^{2}} \\ \Rightarrow | z^{2} | = \sqrt{x^{4} + y^{4} - 2 x^{2} y^{2} + 4 x^{2} y^{2}} \\ \Rightarrow | z^{2} | = \sqrt{x^{4} + y^{4} + 2 x^{2} y^{2}} = \sqrt{(x^{2} + y^{2})^{2}} \\ \Rightarrow | z^{2} | = x^{2} + y^{2} …(ii) \end{aligned}$

From Eqs. (i) and (ii),

$| z |^{2} = | z^{2} |$

47. $| z_{1} + z_{2} | = | z_{1} | + | z_{2} |$ is possible, if

(a) $z_{2} = {\bar{z}}_{1}$

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

(d) $| z_{1} | = | z_{2} |$

Show Answer

Solution

$\Rightarrow | r_{1} (\cos θ_{1} + i \sin θ_{1}) + r_{2} (\cos θ_{2} + i \sin θ_{2}) | = | r_{1} (\cos θ_{1} + i \sin θ_{1}) | + | r_{2} (\cos θ_{2} + i \sin θ_{2}) |$

$\Rightarrow | (r_{1} \cos θ_{1} + r_{2} \cos θ) + i (r_{1} \sin θ_{1} + r_{2} \sin θ_{2}) | = r_{1} + r_{2}$

$\Rightarrow \sqrt{r_{1}^{2} \cos^{2} θ_{1} + r_{2}^{2} \cos^{2} θ_{2} + 2 r_{1} r_{2} \cos θ_{1} \cos θ_{2} + r_{1}^{2} \sin^{2} θ_{1} + r_{2}^{2} \sin^{2} θ_{2}}$

$\sqrt{r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} [\cos (θ_{1} - θ_{2})}] = r_{1} + r_{2}$

$\Rightarrow \sqrt{+ 2 r_{1} r_{2} \sin θ_{1} \sin θ_{2}} = r_{1} + r_{2}$

On squaring both sides, we get

$\begin{aligned} r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} \cos (θ_{1} - θ_{2}) = r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} \\ \Rightarrow & 2 r_{1} r_{2} [1 - \cos (θ_{1} - θ_{2})] = 0 \\ \Rightarrow & 1 - \cos (θ_{1} - θ_{2}) = 0 \\ \Rightarrow & \cos (θ_{1} - θ_{2}) = 1 \\ \Rightarrow & \cos (θ_{1} - θ_{2}) = \cos 0^{\circ} \\ \Rightarrow & θ_{1} - θ_{2} = 0^{\circ} \\ \Rightarrow & θ_{1} = θ_{2} \\ ∴ & \arg (z_{1}) = \arg (z_{2}) \end{aligned}$

48. 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) $n π + (- 1)^{n} \frac{π}{4}$

(d) None of these

Show Answer

Solution

(c) Given expression $= \frac{1 + i \cos θ}{1 - 2 i \cos θ} = \frac{(1 + i \cos θ) (1 + 2 i \cos θ)}{(1 - 2 i \cos θ) (1 + 2 i \cos θ)}$

$\begin{aligned} = \frac{1 + i \cos θ + 2 i \cos θ + 2 i^{2} \cos^{2} θ}{1 - 4 i^{2} \cos^{2} θ} \\ = \frac{1 + 3 i \cos θ - 2 \cos^{2} θ}{1 + 4 \cos^{2} θ} \end{aligned}$

For real value of $θ, \frac{3 \cos θ}{1 + 4 \cos^{2} θ} = 0$

$\begin{matrix} \Rightarrow & 3 \cos θ & = 0 \\ \Rightarrow & \cos θ & = \cos \frac{π}{2} \\ \Rightarrow & θ & = 2 n π \pm \frac{π}{2} \end{matrix}$

49. The value of $\arg (x)$ , when $x < 0$ is

(a) 0

(b) $\frac{π}{2}$

(d) None of these

Show Answer

Solution

$\begin{aligned} z & = x + 0 i and x < 0 \\ | z | & = \sqrt{(- 1)^{2} + (0^{2})} = 1 \end{aligned}$

Since, the point $(x, 0)$ represent $z = x + 0 i$ lies on the negative side of real axis.

$∴$ Principal $\arg (z) = π$

50. If $f (z) = \frac{7 - z}{1 - z^{2}}$ , where $z = 1 + 2 i$ , then $| f (z) |$ is equal to

(a) $\frac{| z |}{2}$

(b) $| z |$

(d) None of these

Show Answer

Solution

(a)

$\begin{aligned} Let z & = 1 + 2 i \\ \Rightarrow | z | & = \sqrt{1 + 4} = \sqrt{5} \\ Now, f (z) & = \frac{7 - z}{1 - z^{2}} = \frac{7 - 1 - 2 i}{1 - (1 + 2 i)^{2}} \\ = \frac{6 - 2 i}{1 - 1 - 4 i^{2} - 4 i} = \frac{6 - 2 i}{4 - 4 i} \\ = \frac{(3 - i) (2 + 2 i)}{(2 - 2 i) (2 + 2 i)} \\ = \frac{6 - 2 i + 6 i - 2 i^{2}}{4 - 4 i^{2}} = \frac{6 + 4 i + 2}{4 + 4} \\ = \frac{8 + 4 i}{8} = 1 + \frac{1}{2} i \\ f (z) & = 1 + \frac{1}{2} i \end{aligned}$

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

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