We can do this by letting $z = e^{it}$, where $t$ goes from $0$ to $2\pi$. Then, $dz = ie^{it} dt$.
Now, we can substitute this into the integral:
$$\oint_C e^{iz} dz = \int_0^{2\pi} e^{ie^{it}}(ie^{it}) dt$$
Now, we can use the Cauchy integral formula, which
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