Complex Numbers 2 Question 6
6. A complex number $z$ is said to be unimodular, if $|z| \neq 1$. If $z _1$ and $z _2$ are complex numbers such that $\frac{z _1-2 z _2}{2-z _1 \bar{z} _2}$ is unimodular and $z _2$ is not unimodular.
Then, the point $z _1$ lies on a
(2015 Main)
(a) straight line parallel to $X$-axis
(b) straight line parallel to $Y$-axis
(c) circle of radius 2
(d) circle of radius $\sqrt{2}$
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Answer:
Correct Answer: 6. (c)
Solution:
- PLAN If $z$ is unimodular, then $|z|=1$. Also, use property of modulus i.e. $z \bar{z}=\left.z\right|^{2}$
Given, $z _2$ is not unimodular i.e. $\left|z _2\right| \neq 1$
and $\frac{z _1-2 z _2}{2-z _1 \bar{z} _2}$ is unimodular.
$$ \begin{aligned} & \Rightarrow \quad\left|\frac{z _1-2 z _2}{2-z _1 \bar{z} _2}\right|=1 \Rightarrow\left|z _1-2 z _2\right|^{2}=2-\left.z _1 \bar{z} _2\right|^{2} \\ & \Rightarrow \quad\left(z _1-2 z _2\right)\left(\bar{z} _1-2 \bar{z} _2\right)=\left(2-z _1 \bar{z} _2\right)\left(2-\bar{z} _1 z _2\right) \quad \quad\left[z \bar{z}=|z|^{2}\right] \\ & \Rightarrow\left|z _1\right|^{2}+4\left|z _2\right|^{2}-2 \bar{z} _1 z _2-2 z _1 \bar{z} _2 \\ & \quad=4+\left|z _1\right|^{2}\left|z _2\right|^{2}-2 \bar{z} _1 z _2-2 z _1 \bar{z} _2 \Rightarrow\left(\left|z _2\right|^{2}-1\right)\left(\left|z _1\right|^{2}-4\right)=0 \\ & \because \quad \quad\left|z _2\right| \neq 1 \\ & \therefore \quad \quad \quad z _1 \mid=2 \\ & \text { Let } \quad z _1=x+i y \Rightarrow x^{2}+y^{2}=(2)^{2} \end{aligned} $$
$\therefore$ Point $z _1$ lies on a circle of radius 2 .
