The mirror image of the curve arg $\left(\frac{z+i}{z-1}\right)=\frac{\pi }{4}. i=\sqrt{-1}$ in the line x-y=0, is

$\because$ The image of z in the line $y=x$is $i\overline{z}$

$\therefore$ The image of the given curve is

$\begin{array}{l}\arg \left(\frac{i\overline{z}+i}{i\overline{z}-1}\right)=\frac{\pi }{4}\\ \Rightarrow \arg \left(\frac{\overline{z}+1}{\overline{z}+i}\right)=\frac{\pi }{4}\\ \Rightarrow -\arg \left(\frac{z+1}{z-i}\right)=\frac{\pi }{4}[\because \arg (\overline{z})=-\arg (z\left)\right]\\ \Rightarrow \arg \left(\frac{z-i}{z+1}\right)=\frac{\pi }{4}\left[\because \arg \left(\frac{z_1}{z_2}\right)=-\arg \left(\frac{z_2}{z_1}\right)\right]\end{array}$