The value of analytic function $f(z)$ is defined to be the value of $f(1/t)$ at $t=0$ as an element of $\Bbb C\cup \{\infty\}$. We can consider a meromorphic function as a function from $\Bbb C\cup \{\infty\}$ to $\Bbb C\cup \{\infty\}$
Suppose that the Laurent expansion at the original of such a meromorphic function is of the form $$\frac{b_N}{z^N}+\dots\frac{b_1}{z}+a_0+a_1z+\dots$$ where $b_n\not=0$ and the series converges for $0\lt \vert z\vert \lt \infty$.
Also, assume that $f:\Bbb C\cup \{\infty\}\to \Bbb C\cup \{\infty\}$ is one to one.
Prove or disprove that $f$ is Möbius transformation.
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Are you familiar with the fact that every meromorhic function on $\mathbb{C}_\infty$ is rational? To see this, suppose that $f$ is meromorphic on $\mathbb{C}_\infty$. Suppose that $f$ is not identically zero. Then since $\mathbb{C}_\infty$ is compact, $f$ must have a finite number of zeros and poles. Consider the rational function $$r(z):=\prod_{i}(z-\lambda_i)^{e_i},$$ where the $\lambda_i$'s are the zeros and poles of $f$ and the $e_i$'s are the corresponding orders. Then $g=f/r$ is a meromorphic function on $\mathbb{C}_\infty$ with no zeros and poles, and therefore it must be constant.
Now, if you assume that $f:\mathbb{C}_\infty \to \mathbb{C}_\infty$ is one-to-one, then it is easy to check that $f$ must be a Möbius transformation. Indeed, $f$ has degree one.