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Consider the function $$f(z)=\frac{\tan(z)}{z}, \quad z \in \mathbb{C}$$

I want to find it's singular points, determine which are removable and classify the smallest positive non-removable singular point.

I know that $f$ has a singular point at $z=0$, which is removable since $\lim_{z \to 0}f(z) \neq0$. How do I find any others and then classify the smallest non removable one though?

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    It should be obvious that the only other singular points of f are singular points of $tan(x)= \frac{sin(x)}{cos(x)}$. What can you say about that? – 2017-01-14

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HINT:

Note that $\tan(z)=\frac{\sin(z)}{\cos(z)}$ and $\cos(z)=0$ for $z=(n+1/2)\pi$ where $n$ is any integer.

Then, note that $\lim_{z\to (n-1/2)\pi}\frac{z-(n-1/2)\pi}{\cos(z)}=(-1)^n$

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    Thank you, that makes sense. How do I use the limit you gave in order to show that $\lim_{z \to (n+1/2)\pi} (z-(n+1/2)\pi) \tan(z)/z \neq 0$? – 2017-01-14
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    What is the limit of the sine function? – 2017-01-14
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    As $z \to (n+1/2)\pi$ it's $(-1)^n$? – 2017-01-14
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    Well done! You have it now. Pleased to see that the hint was useful. -Mark – 2017-01-14
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    So the smallest positive non-removable singularity is $z=\pi /2$? – 2017-01-14
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    What do you mean by smallest? There are poles at $\pm \pi/2$. – 2017-01-15