Compute the integral $I=\int_\gamma f(z)\, dz$, where $f(z)=\frac{e^{z^2}}{z^2-6z}$ and $\gamma=\{z:|z-2|=3\}$.

Question

Compute the integral $I=\int_\gamma f(z)\,dz,\ $ where $f(z)=\frac{e^{z^2}}{z^2-6z}$ and $\gamma=\{z:|z-2|=3\}$.

I thought of using $$f(z_0)=\frac{1}{2\pi i}\int\frac{f(z_0)}{z-z_0}\,dz$$

Rewriting $$\int_\gamma\frac{e^{z^2}}{z^2-6z}=\int_\gamma\frac{e^{z^2}}{z(z-6)}=\int_\gamma\frac{\frac{e^{z^2}}{z}}{(z-6)} $$

The circumference $\gamma$ determines the domain in which we want the function $\frac{e^{z^2}}{z}$ to operate, which contains $(z-6)$.

So the applying the Cauchy formula:

$$\int f(z)\,dz=\pi i\frac{e^{6^2}}{3} $$

However the solution states the value of integral is $\frac{-\pi i}{3}\cdot$

Question:

What am I doing wrong?

Thanks in advance!

Answer 1

The point $z=0$ is contained in the contour $\{z:|z-2|\le3\}$ and thus the integral is equivalent to $$2\pi i\,{\rm{Res}}_{z=0} zf(z)=2\pi i\cdot\frac{e^{0^2}}{0-6}=-\frac{\pi i}3.$$ The point $z=6$ is outside this contour so it is not a singularity that concerns us in this question.

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The correct step in your working should be $$\int_\gamma\frac{e^{z^2}}{z^2-6z}\,dx=\int_\gamma\frac{\frac{e^{z^2}}{z-6}}{z}\,dz=2\pi i\left(\frac{e^{z^2}}{z-6}\right)\bigg|_{z=0}$$ from which the same result follows.