I'm stuck on this particular integration. Here is my problem:
$$\int_{\mathcal{C}} Log (z)\,\mathrm{d}z,$$
where Log (z) is the branch cut of the complex logarithm and C(t) = $\ e^{it}$ with $t$ from -$\pi$ to $\pi$.
Because of the Branch cut of Log (z), I know that it is impossible to use the Couchy-Gorsat Theorem for this. If not, how can I solve this problem?
Best Answer
@Peter is correct. Make the substitution $z=e^{it}$; then $dz=ie^{it}dt$ and we can write $$\int_{\mathcal{C}}\log z dz=\int_{-\pi}^\pi \log(e^{it})(ie^{it}dt)=\int_{-\pi}^\pi -te^{it}dt$$ By parts, we let $u=-t;\ dv=e^{it}ht$ and the integral becomes $$-t\frac{e^{it}}{i}\bigg|_{-\pi}^\pi-\int_{-\pi}^\pi\frac{e^{it}}{i}(-dt)$$ Simplifying and evaluating, $$=\left(-\pi\frac{-1}{i}-\pi\frac{-1}{i}\right)-(-1-(-1))=\boxed{-2\pi i}$$