Analyticity of $\tan(z)$ and radius of convergence Define $\tan(z)=\dfrac{\sin(z)}{\cos(z)}$
Where is this function defined and analytic?
My answer:
Our function is analytic wherever it has a convergent power series. Since (I am assuming) $\sin(z)$ and $\cos(z)$ are analytic the quotient is analytic wherever $\cos(z) \not\to 0$???
Is there more detail to this that I am missing? Without Cauchy is there a way to determine analyticity etc...
Further more, once we have found the first few terms as I have 
$$z+\frac{z}{3}+\frac{2}{15}z^5$$
How do we make an estimate of the radius of convergence? Do I make an observation that the terms are getting close to $0$ and say perhaps the $nth$ root is heading there as well, $\therefore$ $R=\infty$. I'm fairly lost on this part. 
Thanks for your help!

The Laurent series of $\tan z$ for $z\in \mathbb{C}$ is complicated:
$\displaystyle \tan z=z+\frac{z^{3}}{3}+\frac{2}{15}z^{5}+\ldots +
(-1)^{n-1}\frac{2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}z^{2n-1}+\ldots \:, \: |z|<\frac{\pi}{2}$
where $B_{n}$ is Bernoulli number.  The radius of convergence can be justified by $$4\sqrt{n\pi} \left( \frac{n}{\pi e} \right)^{2n} < (-1)^{n-1} B_{2n} <5\sqrt{n\pi} \left( \frac{n}{\pi e} \right)^{2n}$$ for $n\geq 2$.
