Using complex exponentials to prove 1+acostheta

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Homework Statement


Use complex exponentials to prove 1 + acos(theta) + a^2cos(2theta) + a^3cos(3theta)... = (1 - acos(theta))/(1 - 2acos(theta) + a^2)


Homework Equations



euler's e^itheta/2 +e^-itheta/2=2cos(2theta)

The Attempt at a Solution



a^(n)cos(ntheta) = e^nitheta = e^-nitheta

from there i got the series

(a^n(e^itheta)^n)/2

now from here I think I setup the summation formula but this is where I get stuck. Any help is greatly apprecited.
 
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Is |a| < 1?
 
Yes, a is a real constant and |a| < 1. sorry about that
 
<br /> a^n \cos n\theta = a^n\frac {e^{in\theta} + e^{-in\theta}} {2}<br /> = \frac {a^ne^{in\theta} + a^ne^{-in\theta}} {2}<br /> = \frac {p^n + q^n} {2}<br /> \\ p = (\ln a)e^{i\theta}, \ |p| &lt; 1<br /> \\ q = (\ln a)e^{-i\theta}, \ |q| &lt; 1<br />

What is the sum of p^n and q^n?
 
That makes sense. That will then give me a real and an imaginary result of which I take the real I believe. Also, I apologize for the typo on the first post.
 
Well, you can take the real part, but the sum is real anyway.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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