This post proves that if is a function that is holomorphic (analytic) and injective, then for all in . The condition of having nonzero derivative is equivalent to the condition of conformal (preserves angles). Hence, this result can be stated as “A holomoprhic and injective function is conformal.”

(Proof modified from Stein-Shakarchi Complex Analysis)

We prove by contradiction. Suppose to the contrary for some . Using Taylor series,

Since , for all near , with , and where is analytic.

For sufficiently small , we write where .

Since on a small circle centered at , and has at least two zeroes inside that circle, Rouche’s theorem implies that has at least two zeroes there.

Since the zeroes of a non-constant holomorphic function are isolated, for all but sufficiently close to .

Let , be the two roots of . Note that since , , . If , then for some analytic function . This means which is a contradiction.

Thus , which implies that is not injective.

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