$\int_{\gamma}^{}\frac{1}{z}dz$, $\gamma$ is the ellipse $x^2+4y^2=1$ traversed once with the positive orientation

Question

I am unable to find the integral $\int_{\gamma}^{}\frac{1}{z}dz$, $\gamma$ is the ellipse $x^2+4y^2=1$ traversed once with the positive orientation. This maybe possible to be done using Cauchy-Goursat but I cannot seem to find anyway whatsoever to solve this. Hope someone could help me out. Thanks

Answer 1

If you take the circle $\;S^1:=\left\{z\in\Bbb C\;;\;|z|=1\iff z= e^{it}\;,\;\;t\in[0,2\pi]\right\}\;$ , then

$$z=e^{it}\iff dz=ie^{it}dt\implies\oint\limits_{S^1}\frac{dz}{z}=\int\limits_0^{2\pi} i\,dt=2\pi i$$

Since the function in the integral is analytic everywhere except at the origin, it doesn't matter whether we integrate on the given ellipse or in the above circle.

Answer 2

This integral can be viewed as a limiting case of the integrals $$I_\epsilon=\int_{\gamma_\epsilon} \frac{1}{z} dz $$ where $\gamma_\epsilon$ is the ellipse after removing an $\epsilon$-neighborhood of the point $(-1,0)$. For such contours you may use the principal branch of the logarithm as a primitive function and obtain

$$I_\epsilon=\log B_\epsilon-\log A_\epsilon \sim \log1+i \pi-(\log 1-i \pi)=2 \pi i. $$ where the $\sim$ means as $\epsilon \to 0$.