if $f$ is entire then show that $f(z)f(1/z)$ is also entire This is again for an old exam.
Let $f$ be an entire function, show that f(z)f(1/z) is entire.
How do I go about showing the above.
Do I use the definition of analyticity?.,
Call g: f(z)f(1/z) and show that it is complex differentiable everywhere?
Edit: Well the original question was.
Let $f$ be entire and suppose $f(z)f(1/z)$ is bounded on $\mathbb C$, then
$f(z)=az^n$ for some $a\in \mathbb C$.
I was trying to show that $f(z)f(1/z)$ is entire and then use Louiville's theorem.
:). I hope this makes sense.

As Pavel already mentioned, this is not true. In fact, the only entire functions that satisfy the stated conclusion are $f(z) = cz^n$, where $c\neq 0$.
First of all, $f$ must be a polynomial, otherwise $f(1/z)$ has an essential singularity at $z=0$. If $\deg f = n$, then $f(1/z)$ has a pole of order $n$ at the origin, so to cancel this, $f$ itself must have a zero of order $n$ at $z=0$,
Edit incidentally, the above should help you with the edited question too.
