Complex Integral over a Unit Circle

[Applejacks]

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.

From the answers, it writes:
f(z)=\frac{z+i}{z+2}
2\pif'(0)=\frac{\pi}{2}+\piiEdit: 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?

 

[lanedance]

|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

 

[Applejacks]

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]

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

start with finding the poles of the function

 

[Applejacks]

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

 

[vela]

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.

 

[Applejacks]

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|=\piEdit: 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\pii*f(\frac{\pi}{4})=2\pii*\frac{1}{2*(\pi/2})=2i

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

2i+2i=4i

 

[vela]

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

 

[Applejacks]

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