Question

Given that $|z-1|=1$, where $z$ is a non zero point on the complex plane, then $\frac{z-2}{z}$ is equal to :

• A. $i\tan \left(\arg z\right)$
• B. $i\cot \left(\arg z\right)$
• C. $\cot \left(\arg z\right)$
• D. $\tan \left(\arg z\right)$

Answer 1

The correct option is A $i\tan \left(\arg z\right)$

Let $z=r(\cos \theta +i\sin \theta )$
Also, $|z-1|=1$
$\Rightarrow |r(\cos \theta +i\sin \theta )-1|=1$
$\Rightarrow \sqrt{(r\cos \theta -1{)}^2+r^2{\sin }^2\theta }=1$
$\Rightarrow (r\cos \theta -1{)}^2+r^2{\sin }^2\theta =1$
$\Rightarrow r^2-2r\cos \theta =0$
$\Rightarrow r=2\cos \theta$
and $z\overline{z}=|z{|}^2\Rightarrow \frac{1}{z}=\frac{\overline{z}}{|z{|}^2}$
Now $\frac{z-2}{z}=1-\frac{2}{z}=1-\frac{2\overline{z}}{|z{|}^2}$
$=1-\frac{2\cdot 2\cos \theta (\cos \theta -i\sin \theta )}{4{\cos }^2\theta }=1-1+i\tan \theta =i\tan \theta =i\tan \left(\arg \right(z\left)\right)$