Question

The mirror image of the curve $Arg\left(\frac{z+i}{z-1}\right)=\frac{pi}{4}$across the real line $x-y=0$

Answer 1

Find the mirror image of the curve across the real line.

Given,

Mirror image of the curve $Arg\left(\frac{z+i}{z-1}\right)=\frac{pi}{4}$

The real line $x-y=0$

We know,

We know, $z=x+iy$
$\frac{z+i}{z-1}=\frac{x+iy+1}{x+iy-1}$
$\Rightarrow \frac{[x+i(y+1\left)\right]\left[\right(x-1)-iy]}{\left[\right(x-1)+iy]\left[\right(x-1)-iy]}=\frac{x(x-1)+y(y+1)+i\left[\right(x-1\left)\right(y+1)-xy]}{(x-1{)}^2+y^2}$
Let
$X=\frac{x(x-1)+y(y+1)}{(x-1{)}^2+y^2}$
and
$$\begin{gathered} \Rightarrow Y=\frac{(x-1)(y+1)-xy}{(x-1{)}^2+y^2} \\ \Rightarrow \frac{z+i}{z-1}=X+iY \end{gathered}$$
Now, given
$Arg\left(\frac{z+i}{z-1}\right)=\frac{\pi }{4}$

That is,
$$\begin{gathered} {\tan }^{-1}\left(\frac{Y}{X}\right)=\frac{\pi }{4} \\ \Rightarrow Y=X \end{gathered}$$
Then, $(x-1)(y+1)-xy=x(x-1)+y(y+1)$
$\Rightarrow (x-1{)}^2+(y+1{)}^2=1$
This is an equation for a circle on the Argand Plane with a radius of 1 and a center of $(1,-i)$ in the 4th quadrant. Its inverse across the real line $x-y=0$ or $x=y$ which is a line passing through the origin with a slope of $45°$, is a circle in the second quadrant of the Argand Plane with a center at $(-1,\mathrm{i})$.
The polar opposite will be
$Arg\left(\frac{z-i}{z+1}\right)=\frac{\pi }{4}$

Hence, mirror image is $Arg\left(\frac{z-i}{z+1}\right)=\frac{\pi }{4}$