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How does one show $\tan(nz)$ converges uniformly to $-i$ in the upper half plane on compact sets?

I tried writing out $\tan(nz)$ in terms of exponential functions but I got nowhere.

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Hmm... I'm not convinced the statement is true. Perhaps you mean $$\lim_{n\rightarrow\infty} \tan(nz) = i?$$ Note that the problem is ripe for experimentation, so use your favorite numerical tool to check the value of $\tan(nz)$ for some fixed $z$ and some increasing sequence of $n$s, then you'll see why I'm guessing the limit should be $i$, rather than $-i$.

Is this really the limit for all $z$ in the upper half plane? Well, plug the following into WolframAlpha:

complex expand(tan(n (a + b*i)))

I get $$\frac{\sin (2 a n)}{\cos (2 a n)+\cosh (2 b n)}+i\frac{ \sinh (2 b n)}{\cos (2 a n)+\cosh (2 b n)},$$ and from this form it's not hard to see why the limit is what it is. This expansion for the tangent follows from similar expansions for sine and cosine, which both follow Euler's formula and is really not so hard to work out.


I had asserted that the convergence could not be uniform on the upper half plane. While true, this misses the fact that the problem deals with compact subsets of the upper half plane and I'm quite certain that the convergence is uniform on any compact subset of the upper half plane. One way to prove this is to show that $$\tan(n(a+bi)) - i = \frac{2e^{-2bn}}{\cos(2an) + \cosh(2bn)},$$ a formula I produced with Mathematica. Now, if $K$ is a compact subset of the upper half plane, then $\min\{\Im(z):z\in K\}$ exists and is positive. ($\Im(z)$ denotes the imaginary part of $z$.) This minimum may be taken as lower bound on $b$ so it's now not to hard to show that $|\tan(n(a+bi)) - i|$ can be made small independent of $a$ and $b$.

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I got it to work for i instead of -i – john May 25 '12 at 14:15
@john you should post your solution as an answer to this question. :) – Antonio Vargas May 25 '12 at 14:43
@Mark he's showing it for compact subsets of the open upper half-plane, so he's bounded away from the problematic real axis. – Antonio Vargas May 25 '12 at 14:51
@Antonio Ahh - I missed that, yes! – Mark McClure May 25 '12 at 22:02
Is it still wrong? – john May 25 '12 at 22:27

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