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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!

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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.

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  • $\begingroup$ I'm early on in my course so our definition of analytic is wherever it has a convergent power series or wherever it is differentiable on some open ball I believe., $\endgroup$ – Aaron Zolotor Feb 14 '16 at 16:01
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    $\begingroup$ In the definition of analyticity, you must specify the point: $f(z)$ is analytic at $z=a$ if the power series centered on $z=a$ converges in a positive radius around $a$. So although the power series of $\tan(z)$ centered on $z=0$ will not converge for $|z| > \pi/2$, you will nonetheless find that its power series centered on any point $z=a$ such that $\cos(a) \ne 0$ has a positive radius of converence. @az89 $\endgroup$ – Lee Mosher Feb 14 '16 at 17:50
  • $\begingroup$ @LeeMosher, you should mention that your comment is addressed to az89 otherwise he won't be notified. $\endgroup$ – Mikhail Katz Feb 14 '16 at 17:51
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    $\begingroup$ I will add that, however, is it not true that the asker of the question is notified of all comments in all answers? $\endgroup$ – Lee Mosher Feb 14 '16 at 17:52
  • $\begingroup$ @LeeMosher, in my experience he is notified of comments to his question and answers to his question. Comments to answers to his question violate transitivity :-) $\endgroup$ – Mikhail Katz Feb 14 '16 at 17:54
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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.

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  • $\begingroup$ I do not follow. $\endgroup$ – Aaron Zolotor Feb 14 '16 at 16:12
  • $\begingroup$ @az89 What, precisely, is the problem? You said in a comment to the other answer that you know differentiability is sufficient (and also necessary) for the function being analytic. $\endgroup$ – egreg Feb 14 '16 at 16:14
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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 $n\geq 2$.

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