4.13

4.13 Show that \(\int_{\gamma}z^n dz=0\) for an closed piecewise smooth \(\gamma\) and an integer \(n \neq -1\). (If \(n\) is negative, assume that \(\gamma\) does not ass through the origin, since otherwise the integral is not defined.)
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Proof: Assume \(\gamma\) is a closed, smooth, piecewise path and \(n \in \mathbb{Z}\) where \(n \neq -1\). Also, assume when \(z < 0\), then the path of \(\gamma\) does not include the origin.
Consider \(\frac{1}{n+1}*z^{n+1}\) [because \(z \neq -1\) is defined for all other values of \(z\)]. THen by Proposition 2.15: \[\begin{aligned} (\frac{1}{n+1}*z^{n+1})'&=\frac{1}{n+1}*(z^{n+1})'\\ &=\frac{1}{n+1}*(n+1)z^n)\\ &=z^n\end{aligned}\] Therefore, \(\frac{1}{n+1}*z^{n+1}\) is the antiderivative of \(z^n\). Then by Corollary 4.16, \(\int_{\gamma}z^n dz=0\).