# Analytic bijection preserves order of poles

If $f:U\rightarrow\mathbb{C}$ has a pole at $z$ of order $n$ and $\phi:V\rightarrow U$ is an analytic bijection with $\phi(w)=z$, show that $w$ is a pole of $f\circ\phi:V\rightarrow\mathbb{C}$ of order $n$.

I have come up with a proof by contradiction. Just want to know whether there is simpler/direct proof because it is a , somehow, really obvious.

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## 1 Answer

Without loss of generality, you may assume that $w = z = 0$. If we expand $\phi$ is a power series around $0$, we have $$\phi(\eta) = \lambda \eta + O(\eta^2),$$ where $\lambda\neq 0$, because $\phi$ is a bijective analytic map. Since $f$ has a pole of order $n$ at $0$, the function $g(\zeta) = \zeta^nf(\zeta)$ is holomorphic on $U$ and satisfies $g(0)\neq 0$. We then compute $$(g\circ\phi)(\eta) = (\phi(\eta))^n(f\circ \phi)(\eta) = [\lambda^n\eta^n + O(\eta^{n+1})](f\circ\phi)(\eta).$$ We can write $\lambda^n\eta^n + O(\eta^{n+1})$ as $\lambda^n\eta^n h(\eta)$, where $h$ is holomorphic and $h(0) = 1$. We conclude that $$(g\circ \phi)(\eta) = \lambda^n\eta^nh(\eta)(f\circ \phi)(\eta).$$ The left hand side of the expression evaluated at $0$ is $g(\phi(0)) = g(0)\neq0$, and thus the right hand side of the expression cannot vanish or be infinite at $0$. Since $\lambda\neq 0$ and $h(0) = 1$, it follows that $\eta^n(f\circ \phi)(\eta)$ cannot vanish or be infinite at $0$. This exactly says that $f\circ \phi$ has a pole of order $n$ at $0$.

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