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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!

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    It is a four dimensional graph. So it can't be plotted on paper. :) – 2017-01-31
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    no, it's from complex plane to complex plane – 2017-01-31
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    Complex plane is 2-dimensional. Also sorry for my mistake but the domain is a straight line. – 2017-01-31

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$w = e^{bi}$ 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 $$ \frac{6(e^2-1)}{9 e^2-4}$$

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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 $\dfrac{5e}{9e^2-4}$.