# Use the residues theorem to calulate int_(|z-i|=3)(e^(z^2)-1)/(z^3-iz^2)dz

Use the residues theorem to calulate
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occuffick24
No, you miss the residue at z=0 since there are two poles z=0, z=i inside |z-i|=3. The correct answer is
$\begin{array}{rcl}{\int }_{|z-i|=3}\frac{{e}^{{z}^{2}}-1}{{z}^{3}-i{z}^{2}}dz& =& 2\pi i\left(\text{Res}\left(f,i\right)+\text{Res}\left(f,0\right)\\ & =& 2\pi i\left(\frac{{e}^{{z}^{2}}-1}{{z}^{2}}{|{}_{z=i}+\frac{d}{dz}\frac{{e}^{{z}^{2}}-1}{z-i}|}_{z=0}\right)\\ & =& 2\pi i\left(\frac{e-1}{e}+1\right)\\ & =& \frac{2\pi \left(2e-1\right)i}{e}.\end{array}$