Complex Analysis Series Question

[RJLiberator]

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

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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?

 

[fzero]

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?

 

[fzero]

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.

 

[RUber]

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?

 

[RUber]

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.

 

[RUber]

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
$\sum_{n=1}^\infty z^n = \frac{1}{1-z}-\frac{1-z}{1-z} = \frac{z}{1-z}$.

 

[RJLiberator]

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)?

 

[RUber]

Yes the hint is wrong.

I am not sure what you did when you 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.

 

[RUber]

Glad to help.