# Complex Analysis Series Question

• RJLiberator
In summary, the series n=0 to n=infinity Σz^n converges to 1/(1-z) when |z|<1. To solve this problem, you need to replace r\cos(n\theta) with its complex form, then take the sum. After that, you will have something that looks like ##\sum z^n## which you already know how to simplify. The last challenge is extracting the real part of your result. Multiply by the complex conjugate of the denominator and manipulate it into the form you want.
RJLiberator
Gold Member

## Homework Statement

Let 0 < r < 1. Show that

from n=1 to n=∞ of Σ(r^ncos(n*theta)) = (rcos(theta)-r^2)/(1-2rcos(theta)+r^2)

Hint. This is an example of the statement that sometimes the fastest path to a “real” fact is via complex numbers. Let z = reiθ. Then, since r = |z|, and 0 < r < 1, the series n=1 to n=∞ Σz^n converges to 1/(1−z).

## Homework Equations

z = x+iy
Not really sure what else yet...

## The Attempt at a Solution

[/B]
So here, I understand the first step is to look at the sum from n=0 to n=∞ of z^n and that we know converges to 1/(1-z) when |z|<1.
My question is how do I split this up into real and imaginary forms to solve this problem.

Do I split it up in the summation such that from n=0 to n=∞ Σ (x+iy)^n = the sum from n=0 to n=∞ Σx^n + the sum from n=1 to n=∞ Σ(iy)^n and then find the solution that way?

llAlill
Because of the exponential factor, ##z^n = (r e^{i\theta})^n## simplifies in such a way that it is very easy to compute the real and imaginary parts.

RJLiberator
r^n * e^(i*θ*n)

r^n is just r^n
e^(i*θ*n) simplifies to cos(θn)+isin(θn)

And then I'd take the summations of each of these?

RJLiberator said:
r^n * e^(i*θ*n)

r^n is just r^n
e^(i*θ*n) simplifies to cos(θn)+isin(θn)

And then I'd take the summations of each of these?

Yes, you can see that ##r^n\cos n\theta## is what appeared in the original sum.

RJLiberator
RJLiberator said:
r^n * e^(i*θ*n)

r^n is just r^n
e^(i*θ*n) simplifies to cos(θn)+isin(θn)

And then I'd take the summations of each of these?

This exercise wants you to sum the complex form, then take the real part of the result.
## \cos(\theta n ) = \text{Re}\{e^{i\theta n}\}##
## \sum \text{Re}\{ z \} = \text{Re} \left\{ \sum z \right\}##

RJLiberator
Hm.

I'm sort of understanding, yes.

I see now that the Sum of (r^n*cos(n*theta)) is what we originally wanted indeed.
But how do I make the jump now from the sum of (r^ncos(n*theta)) to equalling (rcos(theta)-r^2)/(1-2rcos(theta)+r^2) ?

Is this just how it is defined, am I missing some understanding of series here?

Look back at post #2.
First you need to replace ##r\cos(n\theta)## with its complex form. Then take the sum.

After that, you will have something that looks like ##\sum z^n## which you already know how to simplify.

The last challenge is extracting the real part of your result. Multiply by the complex conjugate of the denominator and manipulate it into the form you want.

RJLiberator
Oh, and as I have been working through this, I noticed that you have the wrong sum formula.
##\sum_{n=0}^\infty z^n = \frac{1}{1-z}##
So starting at n=1, you have to subtract z^0 away.
Giving
##\sum_{n=1}^\infty z^n = \frac{1}{1-z}-\frac{1-z}{1-z} = \frac{z}{1-z}##.

rcos(n*theta) = re^(i*n) as complex form.

So when we replace it with this, we take the sum from n=0 to n=infinity of (re^(i*n)) and see the answer: (-r+r*e^(i*theta)*e^i)/(-1+e^i)

I'm not really sure where the complex conjugate comes into play here.
I understand the strategy you are advising, but it seems that I am missing a basic point in understanding actual series and how to compute them.

Also, based on your last post, is this a typo then in my "hint" of the problem? The series converges to z/(1-z)?

#### Attachments

• Screen Shot 2015-07-14 at 9.29.38 AM.png
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Yes the hint is wrong.

I am not sure what you did when you said:
RJLiberator said:
So when we replace it with this, we take the sum from n=0 to n=infinity of (re^(i*n)) and see the answer: (-r+r*e^(i*theta)*e^i)/(-1+e^i)

Let ##z = r e ^{i\theta}## then put it into the sum formula.
The complex conjugate of ##1-r e^{i\theta}## is ##1-r e^{-i\theta}##.
Once your denominator is real, taking the real part of the fraction is easy.

RJLiberator
Eureka!
A big help was fixing the hint's error to z/(1-z).
Also, the complex conjugate worked beautifully.

Without your help, I would have been lost here.

I now understand this problem pretty well.

Thank you.

RJLiberator

## What is complex analysis?

Complex analysis is a branch of mathematics that studies functions of complex numbers. It deals with properties and behavior of complex-valued functions, such as differentiation, integration, and series.

## What is a complex series?

A complex series is an infinite sum of complex numbers. It is similar to a real series, but the terms involved are complex numbers instead of real numbers.

## What is the difference between a power series and a Laurent series?

A power series is a series in which the terms involve only non-negative integer powers of the variable. A Laurent series, on the other hand, is a series in which the terms involve both positive and negative powers of the variable. In other words, a Laurent series has a negative exponent term while a power series does not.

## What is the radius of convergence of a complex series?

The radius of convergence of a complex series is the distance from the center of the series to the nearest point where the series converges. It is determined by the convergence behavior of the series and can be calculated using the Cauchy-Hadamard formula.

## What applications does complex analysis have?

Complex analysis has various applications in different fields, such as physics, engineering, and economics. It is used to solve problems involving electric fields, fluid flow, and heat distribution. It is also used in signal processing, control theory, and finance.

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