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?
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.