# Determine all entire functions $f(z)$ such that $0$ is a removable singularity of $f(\frac{1}{z})$?

So I am not sure about the answer, but what I did was write $$f(z)$$ in the series form i.e

$$f(z) = a_0 + ... + a_n z^n$$

then I consider $$f(\frac{1}{z})$$ - (and using the fact that in removable singularity principal part is zero)

I get that all $$a_i$$ except $$a_0$$ must be $$0$$! So my answer is coming out to be $$f(z) = c$$, where $$c$$ is some constant!?

• Your answer is correct. – Martin R Jul 12 at 13:21
• Please use MathJax – saulspatz Jul 12 at 13:22

You could also argue that if $$0$$ is a removable singularity of $$f(1/z)$$ then $$f(1/z)$$ is bounded near $$z=0$$, which in turn implies that $$f$$ is a bounded entire function (which is constant according to Liouville's theorem).