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Mathematics Question on Complex Numbers and Quadratic Equations

A complex number zz is the said to be unimodular if ∣z∣=1.|z|=1 . Suppose z1z_{1} and z2z_{2} are complex number such that z1−2z22−z1z2\frac{z_{1}-2 z_{2}}{2-z_{1} z_{2}} is unimodular and z2z _{2} is not unimodular. Then the point z1z_{1} lies on a :

A

Straight line parallel to x-axis

B

Straight line parallel to y-axis

C

Circle of radius 2

D

Circle of radius 2\sqrt{2}

Answer

Circle of radius 2

Explanation

Solution

∣z1−2z22−z1zˉ2∣=1\left|\frac{z_{1}-2 z_{2}}{2-z_{1} \bar{z}_{2}}\right|=1
(z1−2z2)(zˉ1−2zˉ2)=(2−z1zˉ2)(2−zˉ1z2)\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)
∣z1∣2−2z1zˉ2−2z2zˉ1+4∣z2∣2\left|z_{1}\right|^{2}-2 z_{1} \bar{z}_{2}-2 z_{2} \bar{z}_{1}+4\left|z_{2}\right|^{2}
=4−2zˉ1z2−2z1zˉ2+∣z1∣2∣z2∣2=4-2 \bar{z}_{1} z_{2}-2 z_{1} \bar{z}_{2} +\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}
∣z1∣2∣z2∣2−∣z1∣2−4∣z2∣2+4=0\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}-\left|z_{1}\right|^{2}-4\left|z_{2}\right|^{2}+4=0
(∣z1∣2−4)(∣z2∣2−1)=0\left(\left|z_{1}\right|^{2}-4\right)\left(\left|z_{2}\right|^{2}-1\right)=0
⇒∣z1∣=2\Rightarrow\left|z_{1}\right|=2