$ \int_0^i\!e^z\,\mathrm{d}z $ I have a function that I need to integrate. I can't figure out how to do so though.
$$ \int_0^i\!e^z\,\mathrm{d}z $$
My intuition is telling me that the answer is $ e^i - 1 = \cos(1) + i\sin(1) - 1 $, but I can't find anything in my book that suggests this is true. The closest thing I have is $ \int_C f(z)\,\mathrm{d}z = F(z(b)) - F(z(a)) $, where $ C = z(t), a \leq t \leq b $ and $ f(z) = F'(z) $. This proposition is only true for a line integral, so I'm not sure if/how it applies to the above. A step in the right direction would be very helpful, or an explanation of the general procedure for doing this type of integration. Thanks for the help.

Clearly $\frac{d e^z}{dz} =e^z$. Choose $z(t) = it$, and $C$ the line that connects $0$ and $i$. Then $z(0)= 0,z(1) = i$. Thus your integral becomes $$\int_0^i e^z dz = e^z|_{z(0)}^{z(1)} = e^i-e^0 = e^i-1$$
