find the answer to this integral $\oint_{|z-i|=1/2}\frac{z^2}{z-i}+\ln\frac{z+1}{z-1}\,dz$

Question

find $$\oint_{|z-i|=1/2}\frac{z^2}{z-i}+\ln\frac{z+1}{z-1}\,dz$$ my approach: I know that I should find the points in which this function is not analytic and the points should be in the zone of the circle and the final answer is $\frac{2i\pi}{n!}f^{n}(z_0)$
but my problem is that for $\ln\frac{z+1}{z-1}$ I don't know which points does it give any help would be appreciated

Answer 1

The evaluation of a contour integral is not always of the form $2\pi i\frac{f^{(n)}(a)}{n!}$ - this is just the standard result for Cauchy formula-type integrals. You should also recall that if you have a closed contour and a function holomorphic everywhere on the interior of that contour (and continuous on the boundary) then it integrates to $0$.

Let $z$ be such that $|z-i|\le1/2$ and $z=x+iy$ for real $x,y$. Compute: $$\begin{align}\frac{z+1}{z-1}&=\frac{(x+1)+iy}{(x-1)+iy}\\&=\frac{[(x+1)+iy][(x-1)-iy]}{(x-1)^2+y^2}\\&=\frac{(x^2-1+y^2)-2iy}{(x-1)^2+y^2}\end{align}$$The imaginary part is $0$ if and only if $y=0$, and $y$ is never zero for $z$ in this region hence the ratio is never purely real. The standard branch cuts of the logarithm occur on the positive and negative real axes - a point is on the cut only if it is purely real, and we’ve just determined that $(z+1)(z-1)^{-1}$ is never purely real in the region under consideration. So, there are no issues with the logarithm - $\ln\frac{z+1}{z-1}$ is analytic on the interior and boundary of this contour if we use standard branches. Therefore it integrates to $0$ (Cauchy integral theorem).

As for the $z^2/(z-i)$ term, this is un the standard Cauchy integral formula $f(z)/(z-a)$ form. As the contour is closed the integral of this term is $2\pi i\cdot f(a)=-2\pi i$ where $f(z)=z^2$ and $a=i$.

Your final answer is $-2\pi i+0=-2\pi i$.