Complex Analysis Practice Problems

[karan4496]

Homework Statement

a) Find the radius of convergence of the following complex series and the complex point, where the center of the disk of convergence is located:

\sum_{n=1}^{inf} 4^n (z-i-5)^{2n}
b) Find the Laurent series of the following function, f(z), about the singularity, z = 2, and find the residue of f(z)

f(z) = \frac{1}{z(z-2)^3}c) Evaluate the following integral:

\int_{0}^{inf} \frac{dx}{(x^2 + a^2)^4}

Homework Equations

Given

The Attempt at a Solution

a) I gather that 5+i is the center of the disk of convergence? Doing the ratio test I get,

|4(z-(5+i))^2| < 1

I'm a bit lost how to solve this from here.

b) I don't know how to go about expanding this as a Laurent series. If it were a Taylor series, I would factor out a 1/-2^3 from 1/(z-2)^3 and then expand the remaining 1/(1-z/2) and cube it. But this gives me the expansion about z = 0.c) You can extend this integral to the complex plane and write

∫(closed) 1/(z^2+a^2)^4 dz
where singularities would be z = +or- i a
And choosing the upper half of the semi circle contour, I only have to deal with the +'ve i a

Then using the Residue equation for poles of higher order,
I find that the integral is 2∏(0) = 0.

But I'm not sure its correct.

 

[haruspex]

|4(z-(5+i))^2| < 1

What would the relationship be between |z²| and |z|?

b) I don't know how to go about expanding this as a Laurent series.

Since you need to expand about z=2, I would substitute w=z-2. This should make it more obvious.

I find that the integral is 2∏(0) = 0.

I don't. Pls post your working.

 

[karan4496]

a) The relationship would be,

|z| = \sqrt{x^2 + y^2} = r
&
|z^2| = |z|^2 = r^2


b) Okay, I got the series expanded by using the substitution.

f(z) = \frac{1}{2w^3} - \frac{1}{4w^2} + \frac{1}{8w} - ...
So, 1/8 is the residue.


c)

I found my mistake on part c)

I took the limit before taking the derivative in the formula for the residues of higher order poles.

Now I get

Res(a i) = 5/(32a^7i)

And setting the integral from -inf to inf equal to 2PI i * Res(a i)
I get 5PI/16a^7

which becomes 5PI/32a^7 since I'm taking integral from 0 to inf instead of -inf to inf and can multiply by 1/2 since its an even function in the integrand.

 

[haruspex]

|z^2| = |z|^2

Right, so apply that to |4(z-(5+i))²|
Your b) and c) answers look right.