# Complex integral question

## Homework Statement

Assuming a counterclockwise orientation for the unit circle, calculate
∫ $\frac{z+i}{z^3+2z^2}$ dz
|z|=1

## Homework Equations

f'(a)=$\frac{n!}{2i\pi}$=∫$\frac{f(z)}{{z-a}^(n+1)}$
???

## The Attempt at a Solution

I don't understand these types of questions. What does the |z| have to do with the integral? It's written on the bottom limit of the integral in case that wasn't clear.

f(z)=$\frac{z+i}{z+2}$
2$\pi$f'(0)=$\frac{\pi}{2}$+$\pi$i

Edit: Nevermind. I found out how the formula works. I still don't understand what the |z| does. What would happen if it was |z|=10 for example?

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lanedance
Homework Helper
|z|=1 is represents the closed curve (unit circle) around which the contour integration should be performed.

do you know the residue theorem, could be pretty useful here

I somewhat understand the residue theorem and how singularities work. I'm not sure how that explains what the circle does. As long as the singularity is located in the circle, then I can do the integral. If I had something like |z-2|=1, then a problem would occur since the singularity is outside the circle. How do I adjust it then?

lanedance
Homework Helper
you don't the residue theorem relates the integral around a closed curve to the poles located inside the curve - so you need to find which poles are located with the unit circle

So if I had something like the expression below, where C is the square with vertices ($\pm2+2i$,$\pm2-2i$)

∫ $\frac{cos(z)}{z(z^2+8)}$ dz
C

Would I ignore the poles at +/-isqrt(8) and proceed to use the CIF?

2$\pi$i*f(0)=2$\pi$i*$\frac{cos(0)}{(0^2+8)}$=$\frac{i\pi}{4}$

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vela
Staff Emeritus
Homework Helper
Yup, you only care about the poles inside the contour.

To answer your question in post 3: If there are no singularities inside the contour, the integral is equal to 0.

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To answer your question in post 3: If there are no singularities inside the contour, the integral is equal to 0.
omg I feel so stupid for not seeing something this simple...

Thanks for the help so far. I have one last question. What happens when you have multiple singularities within the contour. For example:

∫ $\frac{sin(z)cos(z)}{((z-\frac{\pi}{4}}$)(z+$\frac{\pi}{4}$) dz
|z|=$\pi$

Edit: blah I can't fix the fractions in the integral. The z+pi/4 is on the denominator

Do we treat each one individually and then add them up?

2$\pi$i*f($\frac{\pi}{4}$)=2$\pi$i*$\frac{1}{2*(\pi/2}$)=2i

2$\pi$i*f($\frac{-\pi}{4}$)=2$\pi$i*$\frac{-1}{2*(-\pi/2)}$=2i

2i+2i=4i

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vela
Staff Emeritus
Yes, that's what you do. The integral is equal to $2\pi i\sum~\mathrm{residues}$.