How is it "easily checked" that $[1-s(\cos\theta + i \sin \theta)] \sum_{n=0}^\infty s^n [\cos(n\theta)+i \sin(n\theta)] = 1$

Question

This is from a derivation of de Moivre's theorem on p.148 of Grimmett & Stirzaker's Probability and Random Processes.

The setup:

The sequence $a_n = (\cos \theta + i \sin \theta)^n$ has generating function $$G_a(s) = \sum_{n=0}^\infty \left[ s(\cos\theta + i \sin\theta) \right]^n = \frac{1}{1 - s(\cos\theta + i \sin\theta)}$$ if $|s| < 1$; here $i = \sqrt{-1}$.

The part that I don't follow:

It is easily checked by examining the coefficient of $s^n$ that $$\left[1-s(\cos\theta + i \sin \theta) \right] \sum_{n=0}^\infty s^n \left[\cos(n\theta)+i \sin(n\theta) \right] = 1 $$ when $|s| < 1$.

The rest is clear to me, but in case you're interested:

Thus $$\sum_{n=0}^\infty s^n \left[\cos(n\theta)+i \sin(n\theta) \right] = \frac{1}{1-s(\cos\theta + i \sin \theta)}$$ if $|s| < 1$. Equating the coefficients of $s_n$ we obtain the well-known fact that $\cos(n\theta)+i \sin(n\theta) = (\cos\theta + i \sin\theta)^n$.

Thanks for any help!

Answer 1

It follows from the geometric series $\displaystyle \sum_{n = 0}^{\infty} x^n = \dfrac{1}{1-x}, |x| < 1$.

Answer 2

Let's look at the left-hand side. The $s^0$ coefficient is clearly $1$. For $n>0$ the $s^n$ coefficient is$$\cos n\theta+i\sin n\theta-(\cos\theta+i\sin\theta)(\cos(n-1)\theta+i\sin(n-1)\theta),$$which you can verify is zero with compound angle formulae.

Answer 3

You could use geometric series, although this is perhaps not the same as comparing coefficients so it may not be what you want. E.g. we wish to show that $$\left[1-s(\cos\theta + i \sin \theta) \right] \sum_{n=0}^\infty s^n \left[\cos(n\theta)+i \sin(n\theta) \right] = 1.$$ Euler's formula states that $e^{i\theta}=\cos\theta+i\sin\theta$, and De Moivre's theorem basically uses the fact that $(e^x)^n=e^{nx}$ combined with Euler's formula. So we have: $$(1-se^{i\theta})\sum_{n=0}^\infty s^ne^{in\theta}=1.$$ The summand is the same as $(se^{i\theta})^n$, so by geometric series we have $$(1-se^{i\theta})\frac{1}{1-se^{i\theta}}=1,$$ as was to be proved.