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How can I calculate $$\int_{\gamma}zdz~\text{with }\gamma:[0,1]\rightarrow\mathbb{C},t\mapsto te^{2\pi i t}$$ by using Cauchy's integral formula? The line $\gamma$ isn't even closed. Has anyone a hint?

Thank you.

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HINT:

Since $z$ is an entire function, the value of the integral along any path from $0$ to $1$ independent of the path.

SPOILER ALERT: Scroll over the highlighted area to reveal the solution.

Cauchy's Integral Theorem states that if $f(z)$ is analytic in and on a closed rectifiable contour $C$, then $$\oint_C f(z)\,dz=0$$Now, this is equivalent to stating that the integral from any point $z_1$ to any other point $z_2$ is path independent. Therefore, for the problem at hand we have from Cauchy's Integral Theorem $$\begin{align}\oint_C z\,dz=\int_\gamma z\,dz+\int_1^0 x\,dx=0\tag 1 \end{align}$$where we have formed $C$ by adding the straight line path along the real axis that connects the end points of $\gamma$. Then from $(1)$ it is easy to see that $$\int_\gamma z\,dz=\int_0^1 x\,dx=\frac12$$and we are done!

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  • $\begingroup$ Thank you! This is exactly what I did. But since the exercise is 'by using Cauchy's integral formula' I am not sure if this is the right solution. Is there a way to make $\gamma$ closed and use Cachy directly? $\endgroup$ – user337060 Jul 13 '16 at 15:17
  • $\begingroup$ You're welcome. My pleasure. And , yes we can and have applied Cauchy's Integral Theorem and not Cauchy's Integral Formula. To apply the former, we closed the contour by simply connecting the endpoints $0$ and $1$. Then, $$\oint_C z\,dz=0 \implies \int_\gamma z\,dz+\int_1^0 x\,dx=0 \implies \int_\gamma z\,dz=\int_0^1 x\,dx=\frac12$$ $\endgroup$ – Mark Viola Jul 13 '16 at 15:21
  • $\begingroup$ What do you mean by 'simply connecting the endpoints $0$ and $1$? What curve? $\endgroup$ – user337060 Jul 13 '16 at 15:24
  • $\begingroup$ It doesn't matter which curve we take that connects the endpoints. So, we may take the straight line path along the real axis. $\endgroup$ – Mark Viola Jul 13 '16 at 15:29
  • $\begingroup$ Why it doesn't matter which curve we take to connecting the endpoints? :/ $\endgroup$ – user337060 Jul 13 '16 at 16:10
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I would not know how to use Cauchy in this problem, but you could notice $f(z)=z$ has a primitive function on $\mathbb{C}$, namely $z^2/2$, so the integral is just the endpoints of the path $\gamma$ plugged into the primitive, by the Fundamental theorem of Calculus. So $$\int_{\gamma} z dz = \frac{1}{2}.$$

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