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complex.pdf page42
Let $U$ be a star-like domain. Then if $f:U→\Bbb C$ is holomorphic and $γ:[a,b]→U$ is a closed path in $U$ we have$$\int_{\gamma} f(z)\mathrm{\ d}z=0$$Proof. The proof proceeds similarly to the proof of Theorem 5.22: by Theorem 5.19 it suffices to show that $f$ has a primitive in $U$. To show this, let $z_0 \in U$ be a point for which the line segment from $z_0$ to every $z \in U$ lies in $U$. Let $\gamma_z=z_0+t\left(z-z_0\right)$ be a parametrization of this curve, and define
\[
F(z)=\int_{\gamma_z} f(\zeta)\mathrm{\ d}\zeta
\]
We claim that $F$ is a primitive for $f$ on $U$. Indeed pick $\epsilon>0$ such that $B(z, \epsilon) \subseteq U$. Then if $w \in B(z, \epsilon)$ note that the triangle $T$ with vertices $z_0, z, w$ lies entirely in $U$ by the assumption that $U$ is star-like with respect to $z_0$. It follows from Theorem 7.2 that $\int_T f(\zeta)\mathrm{\ d}\zeta=0$, and hence if $\eta(t)=w+t(z-w)$ is the straight-line path going from $w$ to $z$ (so that $T$ is the concatenation of $\gamma_w, \eta$ and $\gamma_z^{-}$) we have
\begin{aligned}
\left|\frac{F(z)-F(w)}{z-w}-f(z)\right| &=\left|\int_\eta \frac{f(\zeta)}{z-w}\mathrm{\ d}\zeta-f(z)\right| \\
&=\left|\int_0^1 f(w+t(z-w))\mathrm{\ d}t-f(z)\right| \\
&=\left|\int_0^1(f(w+t(z-w))-f(z)\mathrm{\ d}t\right|\\
& \leq \sup _{t \in[0,1]}|f(w+t(z-w))-f(z)|,
\end{aligned}
which, since $f$ is continuous at $w$, tends to zero as $w \rightarrow z$ so that $F'(z)=f(z)$ as required. |
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