Let polynomial $p(z)=z^2+az+b$ be such that $a$and $b$ are complex numbers and $|p(z)|=1$ whenever $|z|=1$. Prove that $a=0$ and $b=0$.
I could not make much progress. I let $z=e^{i\theta}$ and $a=a_1+ib_1$ and $b=a_2+ib_2$
Using these values in $P(z)$ i got $|P (z)|^2=1=(\cos (2\theta)-a_2\sin (\theta)+a_1\cos (\theta)+b_1)^2+(\sin(2\theta)+a_1\sin (\theta)+a_2\cos (\theta)+b_2)^2$
But i dont see how to proceed further neither can i think of any other approach any other approach. So, someone please help. I dont know complex analysis so it would be more helpful if someone can provide hints/solutions that dont use complex analysis.