In problem 33 page 17 -18 on the book "A Hilbert space Problem Book", written by Paul. R Halmos, the author claimed that $e_n(z) = z^n \quad \text{for} \quad |z|=1$ forms an orthonormal basis of $L^2(\mu)$.
I tried to prove that by writting $Z$ in exponential form, but I do not get the desired result. More precisely, $\langle z_n,z_m\rangle= \left\{\begin{matrix} 1 \quad \text{for} \quad m = n\\ 0 \quad \text{for} \quad m \neq n . \end{matrix}\right. $
I did as the following:
Let $z=e^{i\theta}$ then $z^n=e^{in\theta}$, $z^m=e^{im\theta}$.
So, $\langle e_n,e_m\rangle= \int_{0}^{2\pi}e^{in\theta}e^{-im\theta}d\theta=\int_{0}^{2\pi}e^{i(n-m)\theta}d\theta$.
For $n\neq m$, $\langle e_n,e_m\rangle= \int_{0}^{2\pi}e^{i(n-m)}d\theta = 0$,
For $n=m$, $\langle e_n,e_m\rangle= \int_{0}^{2\pi}1d\theta = 2\pi$, but not $1$. .
Could you please help me to find my mistake, please?
Thank you so much for your help