What happens to $\tan(u+iv)$ as $u^2+v^2\to \infty$ via a path where $v\neq 0$ and $u,v\in \mathbb R$? How can I tell?
Behaviour of $\tan$
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complex-analysis
trigonometry
1 Answers
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You may begin by applying the definition $\tan z=\frac{\sin z}{\cos z}$ getting : $\tan(u+iv)=\frac 1i \frac{e^{iu-v}-e^{-iu+v}}{e^{iu-v}+e^{-iu+v}}=\frac 1i \frac{e^{2iu-2v}-1}{e^{2iu-2v}+1}=\frac 1i \frac{e^{-2v}-e^{-2iu}}{e^{-2v}+e^{-2iu}}$ You may try to rewrite this further (expanding $e^{-2iu}$, multiplying by the conjugate and so on) but I'll stop here and observe :
- a periodicity of $\pi$ for $u$ : $\tan(u+k\pi+iv)=\tan(u+iv)$
- $\tan(u+iv) \sim i\ $ as $v\to +\infty$
- $\tan(u+iv) \sim -i\ $ as $v\to -\infty$
A picture of the imaginary part of $\tan(u+iv)$ could perhaps help too : $
An here is the real part :
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0@Joha$n$na: Thanks! It's an earlier version o$f$ MuPAD at work. – 2012-02-20