Proving that $f$ is unitary iff there exists an orthonormal basis of Eigenvectors

Question

I'm having some trouble understanding one step of the proof for the following statement:

Theorem:

Let $V$ be a Complex Vector space with dimension $n>0$ and with the usual inner product and let $f: V\to V$ be an endomorphism.

Then, $f$ is unitary $\iff$ there exists an orthonormal basis $B$ of eigenvectors of $f$ and all eigenvalues are elements of the set $\{\lambda \in \mathbb C: |\lambda|=1\}$.

To prove this we used induction, and my teacher stated the following to prove the case $n = 1$ and start the induction:

Let $f$ be unitary. Then if $\dim(V)=1$ the Theorem is valid.

And then continued with the proof. This is the only part of the proof that I did not understand. Why is this true?

Answer 1

In the one-dimensional case $f(c)=cz$ for some $z$. The fact that $f$ is an isometry implies that $|z|=1$ and $z$ is the only eigen value of $f$.