$\sum_{m=0}^{\infty} \exp(-ma)\cos(mb)$

Question

This is one of my homework problems:

Let $a,b \in \mathbb{R}$ and $a>0$, show that: $$\sum_{m=0}^\infty \exp(-ma) \cos(mb)= \frac{1-\exp(-a)\cos b}{1- 2 \exp(-a) \cos b + \exp(-2a)} $$

I believe to have an attempt on how to solve this, but most likely diverge during the process.

My approach:

Note that $ \exp(imb) = \cos (mb) + i \sin (mb) $ such that $\cos(mb)=\Re( \exp (imb))$ where $\Re$ represents the real part of the expression. This would lead to that

$$ \exp(-ma)\cos(mb)=\exp(-ma) \cdot \Re ( \exp(imb))=\Re(\exp(-ma+imb)=\Re ( e^{-ma+imb})$$ my idea was to use this method to get it into the form of a geometric Series. $$\Re(e^{ma+imb})=\Re\left(\frac{1}{e^{ma-imb}} \right)=\Re \left(\left(\frac{1}{e^{a-ib}} \right)^m \right)$$ If that is all correct I could make use of the geometric Series $$\Re \left(\sum_{m=0}^\infty \left(\frac{1}{e^{-a+ib}} \right)^m \right)=\Re \left( \frac{e^{a-ib}}{e^{a-ib}-1}\right)=\Re \left(\frac{e^ae^{-ib}}{e^ae^{-ib}-1} \right)= \Re \left( \frac{e^a(\cos b - i \sin (b))}{e^a ( \cos b - i \sin (b))-1} \right)\\=\underbrace{\frac{e^a\cos b}{e^a \cos b-1}}_!$$ Which at least looks to a small degree like the answer, i.e. the $e^a \cos b$ part, however for the rest it's not even near, would someone please highlight my mistakes?

Mistakes:

$ \Re(z_1z_2) \neq \Re (z_1) \cdot \Re (z_2)$

Answer 1

$$\sum_{m=0}^{\infty} \exp(-ma)\cos(mb)=\Re\left(\sum_{m=0}^{\infty} \exp(-ma)\exp(imb)\right)\\=\Re\left(\sum_{m=0}^{\infty}\exp(imb-ma)\right)=\Re\left(\sum_{m=0}^{\infty}\exp(m(ib-a))\right)\\=\Re\left(\frac{1}{1-\exp(ib-a)}\right)$$

Simplfying, you will get the desired answer.

Answer 2

$$\text{Real part of}(z_1z_2) \neq \text{Real part of}z_1 \cdot \text{Real part of}z_2$$ $$\text{Real part of}\left(\dfrac{z_1}{z_2} \right) \neq \dfrac{\text{Real part of}z_1}{\text{Real part of}z_2}$$