I'm studying for an exam in Complex Analysis and I'm stuck with this simple problem:
$\int_{\gamma}(z-1)\cos(e^z) dz$
where
$\gamma(t)=t^2-1+i\sin(\pi t)$, t $\in$ [-1,1]
I don't have a exact image of the domain which $\gamma$ bounds but I'm pretty sure it is at least star-shaped. If I'm not completely wrong it should look like a stretched circle around $-\frac{1}{2}$. However, I don't know how to show that this domain is convex or star-shaped in a simple way. Another approach would be to show that $\gamma$ is homologous to 0, which is easy to see if my picture of the path is correct but how do I show this? I don't think $f(z)=(z-1)\cos(e^z)$ has a primitive so calculating that won't help. Calculating it the traditional way by integrating $f(\gamma(t))\gamma(t)'$ is the approach they clearly don't want you to take.. if I didn't miss something.
This is a exam question so there should be a somewhat easy way to show that the integral is 0 as $f(z)$ is holomorphic everywhere and $\gamma$ closed, continuous (and the domain inside star-shaped, $\gamma$ homologous to 0). Maybe I also miss a simple argument here... probably even.
There's some misunderstanding. It's not important if the region bounded by a curve is star-shaped or not, but the relation between the domain of the function and the curve (does the curve surrounds some 'holes'). Here the domain is the whole $\Bbb C$ (star-shaped domain) and $f$ is holomorphic, so we can make use of Cauchy's integral theorem, which gives that the integral is zero.