Question

If $|z-i|=1$ and $arg\left(z\right)=\theta ,\theta \in (0,\frac{\pi }{2}),$ the the value of $\cot \theta -\frac{2}{z}$ is equal to

• A. $0$
• B. $i$
• C. $-i$
• D. $1$

Answer 1

The correct option is B $i$

We have, $z-i=e^{i\alpha }\Rightarrow z=i+e^{i\alpha }=\cos \alpha +i(1+\sin \alpha )$

$\therefore \theta =arg\left(z\right)={\tan }^{-1}\left(\frac{1+\sin \alpha }{\cos \alpha }\right)\Rightarrow \tan \theta =\frac{1+\sin \alpha }{\cos \alpha }$

$\therefore \cot \theta -\frac{2}{z}=\frac{\cos \alpha }{1+\sin \alpha }-\frac{2}{\cos \alpha +i(1+\sin \alpha )}$

$=\frac{\cos \alpha }{1+\sin \alpha }-\frac{2[\cos \alpha -i(1+\sin \alpha )]}{{\cos }^2\alpha +{(1+\sin \alpha )}^2}$

$=\frac{\cos \alpha }{1+\sin \alpha }-\frac{2[\cos \alpha -i(1+\sin \alpha )]}{2(1+\sin \alpha )}=i$