Use Residue Theorems or Laurent Series to evaluate integral

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1. Jun 3, 2015

monnapomona

1. The problem statement, all variables and given/known data
Evaluate the integral using any method:
C (z10) / (z - (1/2))(z10 + 2), where C : |z| = 1​

2. Relevant equations
C f(z) dz = 2πi*(Σki=1 Resp_i f(z)

3. The attempt at a solution
Rewrote the function as (1/(z-(1/2)))*(1/(1+(2/z^10))). Not sure if Laurent series expansion is the best choice for this problem but I ended up getting: (Σn=0 (1/2)n / zn+1)*(Σn=0 (-1)n 2n/(z10)n)

I get stuck at this point but i tried working out the series and get: 1 - 2/z12 + 1/z23+ ...

So would the residue just be 1 and the ∫C f(z) dz = 2πi?

**sorry in advance for my formatting**

2. Jun 4, 2015

vela

Staff Emeritus
What are the poles and where are they located?

3. Jun 4, 2015

monnapomona

Not sure if this is correct but would one pole be +1/2 (Used this function (1/(z-(1/2))) for my reasoning)?

4. Jun 4, 2015

vela

Staff Emeritus
You just need to find where the denominator vanishes, so $z=1/2$ is definitely one pole. Where are the rest?

5. Jun 4, 2015

monnapomona

That's where I'm kind of stuck. For the other function in the denominator, (1/(1+2/z^10)), it doesn't go to 0 even if z = 0?

6. Jun 4, 2015

vela

Staff Emeritus
Oh, look at the integrand before you messed with it. In other words, when does the denominator of $\frac{z^{10}}{(z-1/2)(z^{10}+2)}$ vanish?

7. Jun 4, 2015

monnapomona

Would the poles be z = 1/2, (-2)^(1/10) ?

8. Jun 4, 2015

vela

Staff Emeritus
Expand on what you mean by (-2)^(1/10). Remember, you're working with complex numbers, so taking fractional powers is a little more involved. Consider solving the equation $z^{10} = -2e^{2\pi n i}$ where $n \in \mathbb{Z}$.

9. Jun 4, 2015

monnapomona

Went to office hours today, it turns out for this question, it wasn't limited to just the residue theorems and Laurent series expansion methods. I tried using Cauchy integral formula, where z0 = 1/2 and f(z) = (z^10 / (z^10 + 2)) and got 2πi / 2049 as the final answer.

10. Jun 4, 2015

vela

Staff Emeritus
Regardless of which method you end up using, you still need to determine where the poles are. Just make sure you know how to do that.

11. Jun 4, 2015

Ray Vickson

The number $-2$ has ten 10th roots!