Using a branch cut to integrate a contour

Question

I came across this homework question and I have no clue how to approach it.

\begin{align}&\text{Let}\quad {\rm f}\left(z\right)= \frac{z^2 + 3}{\left(z^{2} - 1\right)\left(z - 1\right)} =\frac{1}{z + 1} - \frac{2}{\left(z - 1\right)^{2}} - \frac{1}{z}, \\[3mm]&\text{and}\ \Gamma\ \text{be the left half of the ellipse}\ x^{2} + 4y^{2} = 4\ \text{from}\ {\rm i}\ \text{to}\ -{\rm i}. \\[3mm]&\text{ By using the branch } \log_0\left(z\right) = \ln\left(\left\vert z\right\vert\right) + {\rm i}{\rm arg}\left(z\right), \text{ for }\ 0 \leq {\rm arg}\left(z\right) \leq 2\pi \\[3mm]&\text{ of the logarithm, or otherwise, evaluate } \int_\Gamma f(z)dz. \end{align}

I can't apply Cauchy's Integral Formula as the contour is clearly not closed and I'm unsure how the branch cut comes into it. Could someone please gently nudge me in the right direction ?.

Answer 1

For example, $\mathrm{Log}(z)$ is not the antiderivative of $1/z$. In particular, $\mathrm{Log}(z)$ is not differentiable at $z=-2$.

However, you can choose a different branch of the logarithm that is differentiable everywhere along your contour, and its derivative would be $1/z$.

Answer 2

The integrand has a simple antiderivative: $$ F(z) = \log(1+z) + \frac{2}{z-1} - \log z $$ satisfies $$ F'(z) = \frac{1}{1+z}-\frac{2}{(z-1)^2}-\frac{1}{z}=f(z). $$ Choosing the branch cuts in $F(z)$ as the text suggests, the situation should be as in this figure:

Now, the path never crosses the branch cuts of the logarithm, so as long as we hold to our choice $0<\theta<2\pi$, we should be safe. By the complex version of the fundamental theorem of calculus: $$ \int_{\Gamma}f(z) dz = F(-i)-F(i). $$ Explicitly: $$ F(-i) = \log\left(\sqrt{2} e^{i7\pi/4}\right)+ \frac{\sqrt 2}{e^{i5\pi/4}}-\log\left(e^{i3\pi/2}\right) $$ $$ F(i) = \log\left(\sqrt{2} e^{i\pi/4}\right)+ \frac{\sqrt 2}{e^{i3\pi/4}}-\log\left(e^{i\pi/2}\right) $$ so that $$ \int_\Gamma f(z) dz= i\frac{3}{2}\pi + 2i -i\pi=i\left(\frac{\pi}{2}+2\right). $$