Find out the image of $\gamma$ under $f(z)=\frac{z+1}{z-1}$

Question

Find out the image of $\gamma = \{z : \mathbb{Re}(z) = \mathbb{Im}(z)\}$ under $f(z) = \frac{z+1}{z-1}$
My attempt: I want to use the symmetry as $f(1)=\infty$ and $f(-1)=0 \Rightarrow \{|z|=1\} \rightarrow \mathbb{Re}(w) = 0$. $\gamma \rightarrow$ circle of some $r$.
I don't understand the way to find the center of the circle and that radius $r$

Answer 1

Hint...start by writing $$w=\frac{z+1}{z-1}$$ and rearrange so $z$ is the subject.

Now put $w=u+iv$ so that you get a simplified expression for $z$ in terms of $u$ and $v$.

Setting the real and imaginary parts equal will give a Cartesian equation in terms of $u$ and $v$.

Answer 2

I don't see how $f(\pm 1)$ can help you, since $\pm 1\notin\gamma$. Nor is the result a circle, as we'll see.

Let me offer another approach. The elements of $\gamma$ are those complex numbers of the form $z=x+xi$, so $u+iv:=f(z)=\frac{2x^2-1-2xi}{2x^2-2x+1}$. The parameterisation $u=\frac{2x^2-1}{2x^2-2x+1},\,v=\frac{-2x}{2x^2-2x+1}$ implies $au^2+buv+cv^2=1$ iff $(2x^2-2x+1)^2=a(2x^2-1)^2-2bx(2x^2-1)+4cx^2$. A little algebra gives $a=1,\,b=2,\,c=3$. Since $b^2-4ac=-8<0$, the shape is a non-circular ellipse. You can see the result here.