Evaluate $\int_C{(z^3-e^z)dz}$ where C is the arc of the circle centered at $0$ from $(3,0)$ to $(0,-3)$.
I tried using $\gamma(t)=3\cos(t)+3i\sin(t)$ at first, but it gets confusing.
In the other exercises we saw in class, the assistant always uses short cuts such as "the integral is $0$ if $f$ is holomorphic", or "the integral of $z^n$ is $2\pi i$ if $n=-1$", etc. But I don't quite know how to approach this one. Should I separate the integral in two parts and evaluate $z^3$ and $e^z$ separately? Also the function is whole, so I can't use the residue theorem.
See here.
The line is the arc of the circle $\rho=3$ with $t\in[-\pi/2,0]$. $$z^3-e^z;\;z=3e^{it};\;dz=3ie^{it}dt$$ $$\int_{C}(z^3-e^z)\,dz=\int_{-\pi/2}^0\left(27 e^{3 i t}-e^{3 e^{i t}}\right) \left(3 i e^{i t}\right)\,dt=\left[\frac{81}{4} e^{4 i t}-e^{3 e^{i t}}\right]_{-\pi/2}^0=e^{-3 i}-e^3$$