\(z^{n}+\overline{z}^{n}\) is a real number for all positive \(n\).
Proof 1
Treat \(z=re^{i\theta}\) and you can show that the sum equals to \(2r^{n}\cos(n\theta)\).
Proof 2
Let \(z=a+ib\) and invoke the Binomial Theorem:
\begin{equation*}
\left(a+ib\right)^{n}=\sum_{k=0}^{n}\dbinom{n}{k}a^{n-k}b^{k}i^{k}
\end{equation*}
\begin{equation*}
\left(a-ib\right)^{n}=\sum_{k=0}^{n}\dbinom{n}{k}a^{n-k}b^{k}i^{k}\left(-1\right)^{k}
\end{equation*}
Adding the two, we get:
\begin{equation*}
\sum_{k=0}^{n}\dbinom{n}{k}a^{n-k}b^{k}i^{k}\left(1+\left(-1\right)^{k}\right)
\end{equation*}
The term in parenthesis is 0 when \(k\) is odd. When \(k\) is even, the term is real as \(i^{k}\) for even k is real.
Proof 3
The above two proofs are not the most intuitive. Let’s go back to the geometric form \(z=re^{i\theta}\) and treat it as a vector. Note that \(\overline{z}\) is of the same length but opposite angle. Clearly, adding the two cancels out the \(y\) axis (i.e. the imaginary portion).
Now when you take the exponent, you are merely scaling and rotating. The key things to realize are:
- You are scaling both vectors by the same amount.
- However much \(z\) is rotated, \(\overline{z}\) is rotated the same amount in the opposite direction. This leaves the resulting y-coordinates of each still exactly the opposite of each other. Adding the two vectors still cancels out the y-component.