Complex Numbers Ques 5
- 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: 5.(c)
Solution:
- 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 .
BETA
