$$ \begin{align} \int_{0}^{2\pi}\cot(x+ib)\mathrm{d}x &=\int_{0}^{2\pi}\frac{\cos(x+ib)}{\sin(x+ib)}\mathrm{d}x\\ &=\int_{0}^{2\pi}\frac{i(e^{ix-b}+e^{-ix+b})}{(e^{ix-b}-e^{-ix+b})}\mathrm{d}x\\ &=\int_{\left | z \right |=1}\frac{(ze^{-b}+\frac1ze^{b})}{(z^2e^{-b}-e^{b})}\mathrm{d}z\\ &=\int_{\left | z \right |=1}\frac{(z^2e^{-b}+e^{b})}{(z^3e^{-b}-ze^{b})}\mathrm{d}z\\ &=2\pi iRes_{z=0} \frac{(z^2e^{-b}+e^{b})}{(z^3e^{-b}-ze^{b})} \\ &=-2\pi i\\ \end{align} $$
but I know from my book that
