# Complex integral question

• Applejacks
In summary, the conversation discusses calculating an integral involving a counterclockwise orientation for the unit circle and the residue theorem. The question involves finding the integral of \frac{z+i}{z^3+2z^2} dz, with |z|=1 as the lower limit. The conversation also delves into the use of the residue theorem to find the poles located inside the contour and how to handle multiple singularities within the contour. Finally, the conversation concludes with the understanding that the integral is equal to 2\pi i\sum~\mathrm{residues}.

## 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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|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?

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

Awesome. I finally understand this topic! Thanks Vela and Lanedance.