Compute the path integral $\int_{0}^{i}{\sin(z)}dz$

Question

I'm pretty stumped on this question. This is what I've got so far.

$\int_{0}^{i}{\sin(z)}dz = -{\cos(i)} + {\cos(0)} = 1 - {\cos(i)}$.

I am using the following theorem: Let $D\subseteq \mathbb C$ be a domain and let $f:D \rightarrow \mathbb C$ be a continuous function. A primitive of $f$ on $D$ is an analytic function $F:D \rightarrow \mathbb C$ such that $F'=f$ on $D$.

THM: If $f$ is continuous on $D$ and if $f$ has a primitive $F$ on $D$, then any curve $\gamma : [a,b] \rightarrow D$ we have that $\int_{\gamma} f(z)dz = F(\gamma(b)) - F(\gamma(a))$.

Answer 1

Your answer is correct, but it can be simplified further. Use the fact that $\cos z = \frac12(\exp(iz)+\exp(-iz))$, which can be proven using Euler's identity. I'll leave the rest to you.