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In the course notes for my class, we have an example of calculating a complex integral about a path. The specific example uses the Winding number but I am having some difficulty understanding.

For a path $\alpha : [t_1, t_2]\to U \subseteq \mathbf C$ and a point $a\in\mathbf C$ which does not lie on the path $\alpha$, we define the winding number $\eta(\alpha, a)$ of $\alpha$ about $a$ as follows. We write $\alpha(t) = a + r(t)e^{i\theta(t)}$ where $r(t) = |\alpha(t) - a|$ and $\theta(t)$ is chosen continuously with $0\le\theta(t_1)\le2\pi$ (it can be shown that the map $\theta(t)$ exists and is uniquely determined), and then we set $$\eta(\alpha, a) = \dfrac{\theta(t_2) - \theta(t_1)}{2\pi}.$$ If $\alpha$ is a loop then we have $\alpha(t_1) = \alpha(t_2)$ and so $e^{i\theta(t_1)} = e^{i\theta(t_2)}$ and hence $\theta(t_2) - \theta(t_1)$ will be a multiple of $2\pi$. Thus for a loop $\alpha$, we have $\eta(\alpha, a)\in\mathbf Z$.

Specifically, how does one go about choosing a function $\theta(t)$

For example, given the integral $\int_\alpha \frac{1}{z}$ where $\alpha(t) = (1+ \frac 38t)e^{i\pi t}$. Here $t_1 = 0, t_2 = \frac83$ The answer is $\ln\frac21 + 2\pi i \frac 43$. I am not sure how this is evaluated.

Edit: In the same question, there is another integral $\int_\alpha \frac{1}{z+1}$ along the same path, how does $\theta(t)$ and $r(t)$ change?

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  • $\begingroup$ What are $t_1$ and $t_2$ in your example? $\endgroup$ – Lord Shark the Unknown Jul 30 '18 at 2:04
  • $\begingroup$ Oh whoops, t1 is 0 and t2 is 8/3. I've edited it into the question, thanks. $\endgroup$ – CyclicGroupOfOrder2 Jul 30 '18 at 2:06
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On your path, $\theta(t)=\pi t$ is an allowable value. But to evaluate $\int_\alpha\frac{dz}{z}$ you need to take a branch $\ell(t)$ for $\ln\alpha(t)$ on your curve. In general that would be $\ln|\alpha(t)|+i\theta(t)$, and here would be $\ln(1+3t/8)+i\pi t$. The value of the integral is then $\ell(t_2)-\ell(t_1)$, which should come out to your given value.

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  • $\begingroup$ Thanks, could you also take a look at my edit. Much appreciated $\endgroup$ – CyclicGroupOfOrder2 Jul 30 '18 at 2:26

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