Question

For $|z-1|=1$, find tan $\left[arg\frac{\left(\right(z-1)}{(2-2\frac{i}{z}}\right)\left)\right].$

• A. $i$
• B. $1$
• C. $-i$
• D. $-1$

Answer 1

The correct option is D $-1$

Let $z=x+iy$
Hence, $|z-1|=1$
$(x-1{)}^2+y^2=1$
Hence by using parametric equation of circle.
$x-1=cos\theta$
$x=1+cos\theta$ and
$y=sin\theta$
Then, $z=(1+cos\theta )+isin\theta$
Let $\theta =\frac{\pi }{2}$
Then, $z=1+i$ ...(i)
Hence, $z-1=i$ and
$2-2i\frac{1}{z}$
$=2(1-\frac{i}{1+i})$
$=2\left(\frac{1+i-i}{1+i}\right)$
$=\frac{2}{1+i}$
$=1-i$...(ii)
Hence, $\frac{z-1}{2-\frac{2i}{z}}$
$=\frac{i}{1-i}$
$=\frac{i(1+i)}{2}$
$=\frac{i-1}{2}$
$=\frac{1}{\sqrt{2}}\left(\frac{-1+i}{\sqrt{2}}\right)$
$=\frac{1}{\sqrt{2}}.e^{i\frac{3\pi }{4}}$
Hence, $tan\left(\frac{3\pi }{4}\right)$
$=-1$.