Question

If $|z-1|=1$, then the value of $\tan \left(\frac{\text{arg}(z-1)}{2}\right)-\frac{2i}{z}$ is

• A. $1$
• B. $-1$
• C. $i$
• D. $-i$

Answer 1

The correct option is D $-i$

Given that $|z-1|=1$
Here $z-1=\cos \theta +i\sin \theta$
Where $\theta =\arg (z-1)$
$\Rightarrow z=1+\cos \theta +i\sin \theta \Rightarrow z=2{\cos }^2\frac{\theta }{2}+i2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\Rightarrow z=2\cos \frac{\theta }{2}(\cos \frac{\theta }{2}+i\sin \frac{\theta }{2})$
From above equation, we get
$|z|=2\cos \frac{\theta }{2},\arg z=\frac{\theta }{2}$
Hence
$\tan \left(\frac{\text{arg}(z-1)}{2}\right)-\frac{2i}{z}=\tan \frac{\theta }{2}-\frac{i}{\cos \frac{\theta }{2}(\cos \frac{\theta }{2}+i\sin \frac{\theta }{2})}=\tan \frac{\theta }{2}-i\frac{(\cos \frac{\theta }{2}-i\sin \frac{\theta }{2})}{\cos \frac{\theta }{2}}=-i$