Simplifying sum of powers of conjugate pairs The result of summing a conjugate pair of numbers each raised to the power $n$:
$$
(a + bi)^n + (a - bi)^n
$$
Produces a real number where $a + bi$ is a complex number.
Given the result is real, is there a simplified way to express the above expression in terms of $a$ and $b$ involving no imaginary number $i$ in the simplified expression?

By De Moivre's Formula we have, for $a+bi=r(\cos \theta+i\sin \theta)$ 
$$(a+bi)^n+(a-bi)^n=\\r^n(\cos n\theta+i\sin n\theta)+r^n(\cos (-n\theta)+i\sin (-n\theta))=\\2r^n\cos n\theta$$
Since $\cos n\theta=\cos (-n\theta)$ and $\sin n\theta=-\sin(-n\theta)$
Now we use the fact that $$r=\sqrt{a^2+b^2}$$ and 
$$L(a,b)=\cos n\theta=\begin{cases} \cos\theta\Big[1-\frac{(n^2-1^2)\sin^2\theta}{2!}+\frac{(n^2-1^2)(n^2-3^2)\sin^2\theta^4}{4!}-\cdots\Big] \text{n odd}\\1-\frac{n^2\sin^2\theta}{2!}+\frac{n^2(n^2-2^2)\sin^4\theta}{4!}-\cdots \text{n even} \end{cases}$$
where $$\cos\theta=\frac{a}{\sqrt{a^2+b^2}},\sin\theta=\frac{b}
{\sqrt{a^2+b^2}}$$
to obtain the expression $$S(a,b)=2(\sqrt{a^2+b^2})^nL(a,b)$$
which is not dependant on $i$.
