# Compute an integral using Cauchy's integral formula

I am revising complex analysis and have come across this example which I want to solve.

Example: Compute the integral $$\int_\gamma \frac{e^{2z} + \sin z}{z - \pi} \,\text{d}z$$ where $\gamma$ is the circle $|z-2|=2$ traversed counter-clockwise.

I think that it needs to be solved using Cauchy's integral formula, however I am struggling to apply the formula to get a solution. I would appreciate any hints and tips on how to get to the answer.

• In which way are you struggling to apply it? – quid Jan 3 '16 at 11:24
• Can you show your work so far? – piepi Jan 3 '16 at 11:28
• The solution i was given says that applying the conditions of Cauchy's integral formula we find that this integral is equal to 2pi*ie^(2pi), i can't work out how to get to this answer – Robert Thompson Jan 3 '16 at 12:00

$$\int_\gamma \frac{e^{2z} + \sin z}{z - \pi} \,\text{d}z$$

Here is the formula $f(z)=\frac{1}{2\pi i}\displaystyle\int_\gamma \frac{f(z)}{z-z_0}$

Let us denote $f(z)=e^{2z}+\sin z$ so $f(\pi)\cdot 2 \pi i=\displaystyle\int_\gamma \frac{e^{2z} + \sin z}{z - \pi} \,\text{d}z$

In our case $z_0=\pi$ now $f(\pi)=e^{2\pi}+\sin(\pi)=e^{2\pi}$

So the final answer is $e^{2\pi}\cdot 2\pi i$

As you said you apply Cauchy's Integral Formula with $a = \pi$ and $f(z)= e^{2z} + \sin z$ to get that $2\pi i f(\pi)$ is equal to your integral. (Check the conditions are fulfilled though.)

As $\sin \pi = 0$ this yields exactly what your answer says.