Use Residue Theorems or Laurent Series to evaluate integral

• monnapomona
In summary: They are ##\sqrt[10]{2}\left(\cos\left(\frac{\pi}{5} + \frac{2\pi n}{10}\right) + i \sin\left(\frac{\pi}{5} + \frac{2\pi n}{10}\right)\right)## for ##n=0, 1, 2, ..., 9##. So, the poles are at ##z = \frac{1}{2}, \sqrt[10]{2}\left(\cos\left(\frac{\pi}{5} + \frac{2\pi n}{10}\right) + i \sin\left(\frac{\pi}{5} + \frac{2\pi n
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*(Σki=1 Resp_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: (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**

What are the poles and where are they located?

vela said:
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)?

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

vela said:
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?

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?

vela said:
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) ?

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}##.

vela said:
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.

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.

vela said:
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!

1. How do I use residue theorems to evaluate an integral?

To use residue theorems, first find the singularities of the function within the contour of integration. Then, calculate the residues at each singularity and use the residue theorem to evaluate the integral.

2. Can residue theorems be used for any type of integral?

Residue theorems can be used for integrals that are in the form of a contour integral and have singularities within the contour of integration. This includes integrals with poles and essential singularities.

3. What is the difference between a pole and an essential singularity?

A pole is a singularity where the function becomes infinite, while an essential singularity is a singularity where the function has no limit as it approaches the singularity. Poles have finite residues, while essential singularities have infinite residues.

4. How do I use Laurent series to evaluate an integral?

To use Laurent series, first find the annulus of convergence for the function. Then, expand the function into its Laurent series and use the coefficients to evaluate the integral.

5. Are there any limitations to using residue theorems or Laurent series to evaluate integrals?

Residue theorems and Laurent series can only be used to evaluate integrals that are in the form of a contour integral and have singularities within the contour of integration. Additionally, the function must have a finite number of singularities within the contour of integration.

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