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!
 A: The formulation of your question reveals a bit of confusion concerning the issue of analyticity. A function can be analytic beyond its radius of convergence; of course it is not defined there by its power series by rather by analytic continuation. The tangent is a good example of this.  The radius of convergence is the distance to the nearest zero of cosine, namely $\pi/2$, but the function is analytic everywhere except for points where cosine vanishes.
A: It's easier to check where the function is differentiable, and it is in all of its domain, because
$$
\tan'z=\frac{1}{\cos^2z}
$$
The formal rules for differentiation continue to hold for complex functions.
A: 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 $n\geq 2$.
