Cauchy Theorem and Integral Formula

Question

I know that for $0<\eta<1 , \frac{(e^z-1)}{z} $ is holomorphic on {$z \in \mathbb{C}:|z-2|<1+\eta$ }

So I was wondering whether this makes

$\int_{|z-2|=1}\frac{(e^z-1)^2}{z} dz$ and $\int_{|z|=1}\frac{(e^z-1)^2}{z^n}dz$ = 0 by Cauchy's Theorem and Cauchy's integral formula respectively ?

Answer 1

The first integral is zero because of Cauchy's theorem. For the second one we have

$$f^{(n)}(0)=\frac{n!}{2 \pi i}\int_{|z|=1}\frac{f(z)}{z^{n+1}}$$ where $f(z)=(e^z-1)^2.$ Since $f''(0)=2$ we have that

$$2=\frac{2!}{2 \pi i}\int_{|z|=1}\frac{f(z)}{z^{3}}.$$ That is

$$\int_{|z|=1}\frac{f(z)}{z^{3}}=2\pi i\ne 0.$$

Answer 2

first integral is equal to 0 as the function is holomorphic on |z-2|=1. the second one you need to look at more examples sheet 4 on the module webpage. its very similar to 4ii. basically, as the denominator is Z^n and 0 exists in the region of integration, we need to differentiate the numerator n-1 times, evaluate at z=o, and then multiply by (2(pi)(i))/(n-1)!. the answer comes to (pi)(i)((2^n)-4)/((n-1)!)

sorry i dont know latex