Use Residue Theorems or Laurent Series to evaluate integral

[monnapomona]

Homework Statement

Evaluate the integral using any method:
∫_C (z¹⁰) / (z - (1/2))(z¹⁰ + 2), where C : |z| = 1​

Homework Equations

∫_C f(z) dz = 2πi*(Σ^k_i=1 Res_{p_i} f(z)

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: (Σ^∞_n=0 (1/2)ⁿ / zⁿ⁺¹)*(Σ^∞_n=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**

 

[vela]

What are the poles and where are they located?

 

[monnapomona]

What are the poles and where are they located?

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

 

[vela]

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

 

[monnapomona]

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

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?

 

[vela]

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?

 

[monnapomona]

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?

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

 

[vela]

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 $n \in \mathbb{Z}$.

 

[monnapomona]

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 $n \in \mathbb{Z}$.

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.

 

[vela]

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.

 

[Ray Vickson]

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 $n \in \mathbb{Z}$.

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