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I am trying to prove that all analytic bijective map from $\mathbb{C}$ to $ \mathbb{C}$ has the form of $f(z)=az+b$ where $a$ is nonzero.

I tried to approach this problem by using taylor series and Cauchy integral formula, but how do I use the property of bijection?

user335468
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  • This is being marked as a duplicate, but the the answers in the link provided are way overkill. All we need here is Casorati-Weierstrass. I'll provide a sketch in the next comment. – zhw. Aug 07 '17 at 21:58
  • Basic idea: $f$ can't have an essential singularity at $\infty.$ If it did, then by Casorati-Weierstrass, $f$ would send ${|z|>1}$ to a dense subset of $\mathbb C.$ But by injectivity and the open mapping theorem, $f(D(0,1))$ is an open set disjoint from $f({|z|>1}),$ contradiction. Thus $f$ does not have an essential singularity at $\infty,$ which implies $f$ is a holomorphic polynomial. But the only holomorphic injective polynomials have the form $az+b, a\ne 0,$ and we're done. – zhw. Aug 07 '17 at 21:58

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