Complex Numbers 1 Question 2

2. All the points in the set $S=\frac{\alpha+i}{\alpha-i}: \alpha \in \mathbf{R} \quad(i=\sqrt{-1})$ lie on a

(2019 Main, 9 April I)

(a) circle whose radius is $\sqrt{2}$.

(b) straight line whose slope is -1 .

(c) circle whose radius is 1.

(d) straight line whose slope is 1 .

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Answer:

Correct Answer: 2. (c)

Solution:

  1. Let $x+i y=\frac{\alpha+i}{\alpha-i}$

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

On comparing real and imaginary parts, we get

$$ x=\frac{\alpha^{2}-1}{\alpha^{2}+1} \text { and } y=\frac{2 \alpha}{\alpha^{2}+1} $$

$$ \begin{aligned} & \text { Now, } x^{2}+y^{2}={\frac{\alpha^{2}-1}{\alpha^{2}+1}}^{2}+{\frac{2 \alpha}{\alpha^{2}+1}}^{2} \\ & =\frac{\alpha^{4}+1-2 \alpha^{2}+4 \alpha^{2}}{\left(\alpha^{2}+1\right)^{2}}=\frac{\left(\alpha^{2}+1\right)^{2}}{\left(\alpha^{2}+1\right)^{2}}=1 \\ & \Rightarrow \quad x^{2}+y^{2}=1 \end{aligned} $$

Which is an equation of circle with centre $(0,0)$ and radius 1 unit.

So, $S=\frac{\alpha+i}{\alpha-i} ; \alpha \in \mathbf{R}$ lies on a circle with radius 1 .



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