How to prove the graph of a certain complex function is a circle To be specific, the function is
$$\frac{2e^{z}-3}{3e^z-2}$$
where $z = 1+bi$ , $0\le b \le 2\pi$. 
I plot the graph, it is a circle whose center is not at the origin, but I don't know the center's coordinate. If I know the center, I can prove it by showing the distance to the center is fixed. But in this case, the center is unclear, so how to show that it is a circle? 
Thanks!
 A: $w = e^{bi}$ for $0 \le b \le 2\pi$ traverses the unit circle.  You then transform that unit circle by $f(w) = \frac{2ew-3}{3ew-2}$, which is a Möbius transformation.  One of the basic properties of these is that they take circles to circles (or straight lines, in the case where a point of the first circle is a root of the denominator).  In this case $3ew-2=0$ for $w = 2/(3e)$, which is not on the unit circle, so it is indeed a circle.  Now, which circle?
The first thing to notice is that, because complex conjugation takes the unit circle to itself, and $f(\overline{w}) = \overline{f(w)}$, the complex conjugation takes the image circle to itself.  That says the centre must be on the real line.  Now two points of the circle on the real line are $f(1) = (2e-3)/(3e-2)$ and $f(-1) = (-2e-3)/(-3e-2) = (2e+3)/(3e+2)$.  Thus the centre must be the midpoint of these, namely $$ \frac{6(e^2-1)}{9 e^2-4}$$
A: Let $g(z)=e^z$ and $h(z)=\dfrac{2z-3}{3z-2}$, so $$f(z)=\frac{2e^{z}-3}{3e^z-2}=h(g(z))$$
The function $g(z)$ maps $z=1+ib$ to the circle $e\cos b+ie\sin b$ with center in origin and radius $e$. 
The function $h(z)$ is a Mobius function and maps circle to circle. It maps the circle $e\cos b+ie\sin b$ with center $\Big(\dfrac{6e^2-6}{9e^2-4},0\Big)$ and radius $\dfrac{5e}{9e^2-4}$. 
