Is there a formula for $(1+i)^n+(1-i)^n$?

Let us avoid the binomial theorem.

• Since $1-\mathrm i$ is the conjugate of $1+\mathrm i$, the number $\color{red}{x_n=(1+\mathrm i)^n+(1-\mathrm i)^n}$ is twice the real part of $(1+\mathrm i)^n$.

• Since $\frac{1+\mathrm i}{\sqrt2}=\mathrm e^{\mathrm i\pi/4}$, $(1+\mathrm i)^n$ is $(\sqrt2)^n$ times $\mathrm e^{\mathrm in\pi/4}$.

• The real part of $\mathrm e^{\mathrm in\pi/4}$ is $\cos(n\pi/4)$.

Thus, $\color{red}{x_n=\ldots}$

First, notice that $1+i = \sqrt2(\cos\frac{\pi}{4} + i \sin\frac{\pi}{4}$) and $1-i = \sqrt2(\cos(-\frac{\pi}{4}) + i \sin(-\frac{\pi}{4}))$.

Using de Moivre's formula and the fact that $\sin$ is odd and $\cos$ is even, we get

\begin{align*} (1+i)^n + (1-i)^n &= 2(\sqrt2)^n\cos\frac{n\pi}{4} \end{align*}

$(1+i)^n+(1-i)^n = 2 \Re ((1+i)^n)$. Now expand $(1+i)^n$ using the binomial theorem. The real part is formed by the odd-numbered terms.

Another approach is to note that $(1+i)/\sqrt{2}$ is an 8-th root of unity and so $(1+i)^n$ depends only on $n \bmod 8$ except for a power of $\sqrt{2}$.