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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.

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Wait, what? $e^z e^{1/z}$ is not entire. – user53153 Jan 11 '13 at 23:59

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.

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Thanks for your answer. So an entire function can't have an essential singularity at infinity? – Jack Dawkins Jan 12 '13 at 0:18
@user54755 Actually, every entire function that is not a polynomial has an essential singularity at infinity. – mrf Jan 12 '13 at 0:21

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