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While learning about entire (complex) functions, I had to prove why the following function is entire (or in other words, prove that it is holomorphic for all complex $z$):

$f(z) =\frac{\sqrt3\cos(z)}{z^4} -\frac{\sin (z\sqrt3)}{z^5}$

This is not evident to me, as I would think there is a singularity for $z=0$. My teacher stated that $f(z)$ is entire, whereas Mathematica suggests that the function is not entire. Could someone explain this to me?

Joep Nieuwdorp
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  • Can you determine the behavior at $z=0$? If a function is analytic in a punctured neighborhood of $0$ and bounded there, then it is analytic even at $0$ (extends uniquely). – GEdgar Oct 24 '19 at 20:24
  • Write the Taylor series for $z\sqrt 3 \cos z- \sin(\sqrt 3 z)$ and note that it is $z^5f(z)$ so the ratio is entire – Conrad Oct 24 '19 at 20:28
  • Thanks all! Didn't see that at once, it's clear now. – Joep Nieuwdorp Oct 24 '19 at 21:42

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It should be clear that the function is holomorphic outside $z = 0$.

Now at $z = 0$, you should expand it in Laurent series. This will look like: $$f(z) = \sqrt{3}z^{-4}(1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots) - z^{-5}(\sqrt{3}z - \frac{(\sqrt{3}z)^3}{3!} + \frac{(\sqrt{3}z)^5}{5!} - \cdots). $$

You will see that the negative powers of $z$ are cancelled. So indeed $f$ is entire.

WhatsUp
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