Show that $\forall n \in \mathbb{N} \left ( \left [(2+i)^n + (2-i)^n \right ]\in \mathbb{R} \right )$ Show that  $\forall n \in \mathbb{N} \left ( \left [(2+i)^n + (2-i)^n  \right ]\in \mathbb{R} \right )$
My Trig is really rusty and weak so I don't understand the given answer:
$(2+i)^n + (2-i)^n $
$= \left ( \sqrt{5} \right )^n \left (\cos n\theta + i \sin n\theta  \right ) + \left ( \sqrt{5} \right )^n \left (\cos (-n\theta) + i \sin (-n\theta)  \right ) $
$= \left ( \sqrt{5} \right )^n \left ( \cos n\theta + \cos (-n\theta) + i \sin n\theta + i \sin (-n\theta) \right ) $
$= \left ( \sqrt{5} \right )^n 2\cos n\theta$
Could someone please explain this?
 A: There are two ways to write a complex number: rectangular form, e.g., $x+iy$, and polar form, e.g., $re^{i\theta}$. The conversion between them uses trig functions: $$re^{i\theta}=r\cos\theta+ir\sin\theta\;.\tag{1}$$ Going in the other direction, $$x+iy=\sqrt{x^2+y^2}\,e^{i\theta}\;,$$ where $\theta$ is any angle such that $$\cos\theta=\frac{x}{\sqrt{x^2+y^2}}\;\text{ and }\sin\theta=\frac{y}{\sqrt{x^2+y^2}}\;.$$ The important thing for your argument is that $r=\sqrt{x^2+y^2}$.
The $r$ corresponding to $2+i$ is therefore $\sqrt{2^2+1^2}=\sqrt5$, and that corresponding to $2-i$ is $\sqrt{2^2+(-1)^2}=\sqrt5$ as well. The angles for $2+i$ is an angle $\theta$ whose cosine is $\frac2{\sqrt5}$ and whose sine is $\frac1{\sqrt5}$, while the angle for $2-i$ is an angle whose cosine is $\frac2{\sqrt5}$ and whose sine is $-\frac1{\sqrt5}$. It doesn’t matter exactly what they are; the important thing is that if we let the first be $\theta$, the second is $-\theta$, since $$\cos(-\theta)=\cos\theta\;\text{ and }\sin(-\theta)=-\sin\theta\;.$$
Substituting into $(1)$ gives you $$2+i=\sqrt5\cos\theta+i\sqrt5\sin\theta=\sqrt5(\cos\theta+i\sin\theta)=\sqrt5 e^{i\theta}$$ and $$2-i=\sqrt5\cos(-\theta)+i\sqrt5\sin(-\theta)=\sqrt5(\cos\theta-i\sin\theta)=\sqrt5 e^{-i\theta}\;.$$
Now use the fact that it’s easy to raise an exponential to a power:
$$\begin{align*}
(2+i)^n+(2-i)^n&=(\sqrt5)^n\left(e^{i\theta}\right)^n+(\sqrt5)^n\left(e^{-i\theta}\right)^n\\
&=(\sqrt5)^n\left(e^{in\theta}+e^{-in\theta}\right)\\
&=(\sqrt5)^n\Big(\big(\cos n\theta+i\sin n\theta\big)+\big(\cos(-n\theta)+i\sin(-n\theta)\big)\Big)\\
&=(\sqrt5)^n\Big(\cos n\theta+i\sin n\theta+\cos n\theta-i\sin n\theta\Big)\\
&=(\sqrt5)^n 2\cos n\theta\;.
\end{align*}$$
A: If you believe that complex conjugation respects products (hence also powers), then the simple way is:
$$
\overline{x}=\overline{(2+i)^n+(2-i)^n}=(\overline{2+i})^n+(\overline{2-i})^n=(2-i)^n+(2+i)^n=x.
$$
So $\overline{x}=x$, and hence $x$ is real.

The binomial formula gives an alternative route:
$$
x=(2+i)^n+(2-i)^n=\sum_{k=0}^n{n\choose k}2^ki^{n-k}+\sum_{k=0}^n{n\choose k}2^ki^{n-k}(-1)^{n-k}.
$$
Here the terms where $n-k$ is odd cancel each other, so we get
$$
x=2\sum_{k=0,\ k\equiv n\pmod2}^n{n\choose k}2^ki^{n-k}.
$$
Here everywhere $i^{n-k}$ is real, because $(n-k)$ is even in all the terms remaining in the sum.
A: Hint $\ $ Scaling the equation by $\sqrt{5}^{\:-n}$ and using Euler's $\: e^{{\it i}\:\!x} = \cos(x) + {\it i}\: \sin(x),\ $ it becomes
$$\smash[b]{\left(\frac{2+i}{\sqrt{5}}\right)^n + \left(\frac{2-i}{\sqrt{5}}\right)^n} =\: (e^{{\it i}\:\!\theta})^n + (e^{- {\it i}\:\!\theta})^n $$ 
But
$$\smash[t]{ \left|\frac{2+i}{\sqrt{5}}\right| = 1\ \Rightarrow\ \exists\:\theta\!:\ e^{{\it i}\:\!\theta} = \frac{2+i}{\sqrt{5}} \ \Rightarrow\ e^{-{\it i}\:\!\theta} = \frac{1}{e^{i\:\!\theta}} = \frac{\sqrt{5}}{2+i} = \frac{2-i}{\sqrt 5}}$$
Remark $\ $ This is an example of the method that I describe here, of transforming the equation into a simpler form that makes obvious the laws or identities needed to prove it. Indeed, in this form, the only nontrivial step in the proof becomes obvious, viz. for complex numbers on the unit circle, the inverse equals the conjugate: $\: \alpha \alpha' = 1\:\Rightarrow\: \alpha' = 1/\alpha.$
