The equation $\arg \left(\frac{\mathrm{z}-1}{\mathrm{z}+1}\right)=\frac{\mathrm{\pi }}{4}$ represents a circle with

We have, $\arg \left(\frac{\mathrm{z}-1}{\mathrm{z}+1}\right)=\frac{\mathrm{\pi }}{4}$

$\begin{array}{l}\Rightarrow \arg (\mathrm{z}-1)-\arg (\mathrm{z}+1)=\frac{\mathrm{\pi }}{4}\\ \text{ Let }\mathrm{z}=\mathrm{x}+\mathrm{iy}\\ \arg \left[\right(\mathrm{x}-1)+\mathrm{iy}]-\arg \left[\right(\mathrm{x}+1)+\mathrm{iy}]=\frac{\mathrm{\pi }}{4}\\ \Rightarrow {\tan }^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}-1}\right)-{\tan }^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}+1}\right)=\frac{\mathrm{\pi }}{4}\\ \Rightarrow \left(\frac{\frac{\mathrm{y}}{\mathrm{x}-1}-\frac{\mathrm{y}}{\mathrm{x}+1}}{1+\frac{\mathrm{y}}{\mathrm{x}-1}\cdot \frac{\mathrm{y}}{\mathrm{x}+1}}\right)=\tan \frac{\mathrm{\pi }}{4}\\ \Rightarrow \frac{\mathrm{y}(\mathrm{x}+1)-\mathrm{y}(\mathrm{x}-1)}{\left({\mathrm{x}}^2-1\right)+{\mathrm{y}}^2}=1\\ \Rightarrow 2\mathrm{y}={\mathrm{x}}^2+{\mathrm{y}}^2-1\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{y}-1=0\\ \Rightarrow {\mathrm{x}}^2+(\mathrm{y}-1{)}^2=2\\ \Rightarrow {\mathrm{x}}^2+(\mathrm{y}-1{)}^2=(\sqrt{2}{)}^2\end{array}$

Which is a circle with Centre (0, 1) and Radius = $\sqrt{2}$ units