Maximum value of tanh Z

Question

$$ f(z) = \tanh (z) = \dfrac{e^z - e^{-z}}{e^z + e^{-z}} $$

Find the point $z$ with $|z| \leq1$ where $|f(z)|$ attain its maximum.

I figured out that the maximum is probably at the edge (concluded it from cauchy integral formula ) but I am not sure what is my next step (In section a,b of the question I found the taylor expansion around $z=0$ and the radius of convergence)

Can someone help ?

Ross Millikan · Accepted Answer · 2014-03-19T12:58:55.110

0

If $|z|=1,$ you can write $\tanh(z)=\frac {e^{e^{i\theta}}-e^{e^{-i\theta}}}{e^{e^{i\theta}}+e^{e^{-i\theta}}}$, then in theory can take absolute value and the derivative with respect to $\theta$. This Alpha plot indicates the maxima are at $z=\pm i$, with $|\tanh z| = \tan 1 \approx 1.5574$

edited Mar 19 '14 at 12:58

answered Mar 18 '14 at 20:39

Ross Millikan

374,822

Thanks but what is the maximum of $$ |tan x| $$? Isnt it undefined ? – sheep Mar 19 '14 at 07:10
I lost a layer of exp there. It should be $x=\exp(i \theta)$ Working. It looks from this Alpha plot to be at $\theta=\pm \frac \pi 2$ – Ross Millikan Mar 19 '14 at 12:49
Still dont get it. The graph of $$ | tanh (z) | $$ is different from the graph you attached. [link] www.wolframalpha.com/input/?i=plot+abs(+(tanh(z))) – sheep Mar 19 '14 at 13:07
Yes, I plotted $|\tanh (\exp(ix))|$, where $x$ is the $\theta$ above. This forces it to the unit circle. – Ross Millikan Mar 19 '14 at 13:09
great, thanks a lot. – sheep Mar 19 '14 at 13:24

Maximum value of tanh Z

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