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Let D be a simply connected region in $\mathbb{C}$ and let C be a simple closed curve contained in D. Let $f(z)$ be analytic in D. Suppose that $z_0$ is a point enclosed by C. Then $f(z_0)=\frac{1}{2\pi i} \int_{C} \frac{f(z)}{z-z_0}$.

I am writing for a more in depth explanation as to why if $z_0=2$ and C is the circle centered at the origin with radius $5$, that

$\int_{C} \frac{e^{z^2}}{z-2}=2 \pi i e^4$

Where is $e^{z^2}$ coming from and how do I evaluate such an integral using the theorem given above?

Any help is greatly appreciated

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  • $\begingroup$ You should look at the proof of the theorem to have deeper understanding. $\endgroup$ – Mhenni Benghorbal Dec 4 '13 at 3:06
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$e^{z^2}$ is coming from the problem.

The question is asking for

$$\int_C \frac{e^{z^2}}{z-2} dz$$

Now, looking at this, it looks exactly like CIF. So, lets set $f(z)=e^{z^2}$ and $z_0=2$.

As $f(z)$ is analytic inside the circle of radius $5$ and $2$ is also inside the circle of radius $5$, CIF tells us that

$$f(z_0) = \frac{1}{2\pi i} \int_C \frac{e^{z^2}}{z-z_0} dz \,.$$

Replacing $z_0=2$ and $f(z)=e^{z^2}$ you get

$$e^{2^2}= \frac{1}{2\pi i} \int_C \frac{e^{z^2}}{z-2} dz \,.$$

Now multiply both sides by $2 \pi i$.

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  • $\begingroup$ Awesome. Thank you, very clear now. $\endgroup$ – user7090 Dec 4 '13 at 3:06

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