Evaluate $\int_0^i e^z dz$

Question

Evaluate $\int_0^i e^z dz$.

Let $z(t) = it$, where $ 0\leq t \leq 1$.

Then, $\int_0^i e^z \, dz = \int_0^1 ie^{it} dt = e^i - e^0 = e^i - 1 = \cos{1} + i\sin{1}$.

This looks fairly simple, but I just want to make sure that I am doing this correctly.

Answer 1

You can also use the fact that the anti-derivative of $e^z$ is $e^z$ and that $e^z$ is holomorphic everywhere to get that $$ \begin{align} \int_0^ie^z\,\mathrm{d}z &=\left.e^z\vphantom{\int}\,\right]_0^i\\ &=e^i-1 \end{align} $$ independent of the path taken between $0$ and $i$.

However, parametrizing the straight line from $0$ to $i$, as you've done, is fine (once you bring back the lost $-1$).