Locus of $\mathcal{z},\ if\arg{\left( \frac{\mathcal{z -}1}{\mathcal{z} + 1} \right) = \frac{\pi}{2}}$, is

Question

Dr. Abubacker · Accepted Answer

(b)
We have, $\arg(\mathcal{z -}1) - \arg\begin{array}{r}
\  \
\left( \mathcal{z} + 1 ight) = \frac{\pi}{2}
\end{array}$
[image:5_image3.emf]
It is clear from the figure that it a semi circle

Q. Locus of $\mathcal{z},\ if\arg{\left( \frac{\mathcal{z -}1}{\mathcal{z} + 1} \right) = \frac{\pi}{2}}$, is

Explanation

Reviewed by Dr. Abubacker

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