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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!?

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

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).

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