Making the denominator of a certain fraction real

[TensorCalculus]

Homework Statement

Show that, for |r|<1, $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$

Relevant Equations

$$e^{ix} =\cos(x) + i\sin(x)$$

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers:
Show that, for $|r|<1,$
$$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$

My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above:
$$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$

The sum of this series is just:
$$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$

I'm having some trouble trying to figure out what to multiply the numerator and denominator by in order to make the denominator real (and, thus, be able to split it into real and imaginary parts). If it were just $\frac{(e^{ix})^n-1}{e^{ix} - 1}$ I could just multiply by $e^{-ix/2}$ which would make the bottom just 2x a sine function... but you can't do that for this one because of the r and I'm not sure what to do: I tried expanding $e^{ix}$ out etc but I couldn't find any way to make the denominator real then either. I'm not really sure what to do...

Any hints as to how I can split this into the real and imaginary parts would be much appreciated thanks!

 

[PeroK]

$zz^*$ is always real.

 

[fresh_42]

I would use (Euler and de Moivre)
$$
z^n=r^n(\cos \varphi+i\sin \varphi)^n=r^n(\cos n\varphi + i\sin n\varphi)
$$
which gives you the term for $\cos n\varphi$ by comparison of real and imaginary part. Or simply look up the formula, e.g., https://en.wikipedia.org/wiki/List_of_trigonometric_identities, by using Chebyshev polynomials.

The denominator on the right resembles the cosine theorem, so perhaps there is a geometrical approach.

 

[FactChecker]

I don't know how much complex analysis you have had.
If you just want to multiply by something to make the denominator real, notice that:
$(1+z)(1+\bar{z}) = 1+\bar{z}+z+z\bar{z} = 1+2Re(z)+|z|^2$, which is real for any complex number, $z$.

 

[TensorCalculus]

$zz*$ is always real.

I don't know how much complex analysis you have had.
If you just want to multiply by something to make the denominator real, notice that:
$(1+z)(1+\bar{z}) = 1+\bar{z}+z+z\bar{z} = 1+2Re(z)+|z|^2$, which is real for any complex number, $z$.

Oh right silly me: yes I can just multiply by the complex conjugate of $re^{ix} - 1$ and that should do the trick.
Thank you!

I would use (Euler and de Moivre)
$$
z^n=r^n(\cos \varphi+i\sin \varphi)^n=r^n(\cos n\varphi + i\sin n\varphi)
$$
which gives you the term for $\cos n\varphi$ by comparison of real and imaginary part. Or simply look up the formula, e.g., https://en.wikipedia.org/wiki/List_of_trigonometric_identities, by using Chebyshev polynomials.

The denominator on the right resembles the cosine theorem, so perhaps there is a geometrical approach.

Oh yes: for the numerator right? Useful. Thanks :)