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Let $\gamma$ be a continuously differentiable closed curve in the complex plane with parameter interval $[a,b]$, and assume that $\gamma (t) \ne 0$ for all $ t \in [a,b]$. Define the index of $\gamma$ to be $$\text{Ind} (\gamma) = \frac{1}{2\pi i}\int_{a}^{b} \frac{\gamma'(t)}{\gamma (t)} dt$$

Prove that this is always an integer.

In the hint it's mentioned there is a function $\phi$ on $[a,b]$ with $\phi ' =\frac{\gamma'}{\gamma}$ and $\phi(a)=0$.

I can understand that the primitive of $\frac{\gamma'}{\gamma}$ exists since it is continuous and integrable but cannot understand why $\phi(a)=0$ which plays a vital role in the proof.

Martin Sleziak
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Sruti Sarkar
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1 Answers1

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Consider the map \begin{align} \varphi(t)=\exp\Big(\int^t_a \frac{\gamma'(s)}{\gamma(s)}\,ds\Big), \qquad t\in[a,b]. \end{align} We will show that $\phi(b)=1$. The fundamental theorem of calculus implies that \begin{align} \varphi'(t)=\frac{\varphi(t)\gamma'(t)}{\gamma(t)}, \end{align} Then \begin{align} \frac{d}{dt}\Big( \frac{\varphi}{\gamma}\Big)=\frac{(\gamma-z)\varphi'-\varphi\gamma'}{(\gamma)^2}=\frac{\varphi\gamma'-\varphi\gamma'}{(\gamma)^2}=0 \end{align} Thus, $\frac{\phi}{\gamma}$ is constant in $[a,b]$. In particular, \begin{align} \frac{\varphi(b)}{\gamma(b)}=\frac{\varphi(a)}{\gamma(a)}= \frac{1}{\gamma(b)} \end{align} since $\varphi(a)=1$ and $\gamma(b)=\gamma(a)$. Consequently, $\varphi(b)=1$. From this it must be that $$\int^b_a\frac{\gamma'}{\gamma}=2\pi in$$ for some integer $n$ (recall that $e^z=1$ iff $z\in 2\pi i\mathbb{Z}$).

Mittens
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