Proof.

First, assume that is not an essential singularity of , which we will prove later. Then must be a pole by trichotomy, as a removable singularity implies boundedness (Proposition 4.3.1). Therefore, is a polynomial of degree , where .

Since , it is true that is entire. Since

it follows that has no zeros in . By the Fundamental Theorem of Algebra (Theorem 3.3.1), if , then has a complex zero, which is a contradiction. Hence, must be linear, and all functions in are in the form of , where and are constants. In other words, any holomorphic automorphism on is a composition of a rotation, a dilation, and a translation.

We will now prove the primary assumption; the singularity at cannot be an essential singularity of . Let be arbitrary. Then by the Casorati–Weierstrass Theorem (Theorem 4.2.3), and , such that . Equivalently, , such that . Since is continuous on by holomorphy, by Theorem 1.2.13, it is bounded, which is a contradiction.