Use the residues theorem to calulate

$${\int}_{|z-i|=3}\frac{{e}^{{z}^{2}}-1}{{z}^{3}-i{z}^{2}}\text{}\mathrm{d}z$$

$${\int}_{|z-i|=3}\frac{{e}^{{z}^{2}}-1}{{z}^{3}-i{z}^{2}}\text{}\mathrm{d}z$$

snaketao0g
2022-10-23
Answered

Use the residues theorem to calulate

$${\int}_{|z-i|=3}\frac{{e}^{{z}^{2}}-1}{{z}^{3}-i{z}^{2}}\text{}\mathrm{d}z$$

$${\int}_{|z-i|=3}\frac{{e}^{{z}^{2}}-1}{{z}^{3}-i{z}^{2}}\text{}\mathrm{d}z$$

You can still ask an expert for help

occuffick24

Answered 2022-10-24
Author has **13** answers

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

$$\begin{array}{rcl}{\int}_{|z-i|=3}\frac{{e}^{{z}^{2}}-1}{{z}^{3}-i{z}^{2}}dz& =& 2\pi i(\text{Res}(f,i)+\text{Res}(f,0)\\ & =& 2\pi i\left(\frac{{e}^{{z}^{2}}-1}{{z}^{2}}{|{}_{z=i}+\frac{d}{dz}\frac{{e}^{{z}^{2}}-1}{z-i}|}_{z=0}\right)\\ & =& 2\pi i(\frac{e-1}{e}+1)\\ & =& \frac{2\pi (2e-1)i}{e}.\end{array}$$

$$\begin{array}{rcl}{\int}_{|z-i|=3}\frac{{e}^{{z}^{2}}-1}{{z}^{3}-i{z}^{2}}dz& =& 2\pi i(\text{Res}(f,i)+\text{Res}(f,0)\\ & =& 2\pi i\left(\frac{{e}^{{z}^{2}}-1}{{z}^{2}}{|{}_{z=i}+\frac{d}{dz}\frac{{e}^{{z}^{2}}-1}{z-i}|}_{z=0}\right)\\ & =& 2\pi i(\frac{e-1}{e}+1)\\ & =& \frac{2\pi (2e-1)i}{e}.\end{array}$$

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