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I am studying for complex analysis and am working on the following problem:

For any z in the complex plane with $|\text{Im}(z)| \ge \delta > 0$ and $z = x + iy$,

$$|\tan(z)|^2 <= 1 + \frac{1}{\sinh^2(\delta)}$$

For the problem, I have begun by using the identities for $\sin(z)$ and $\cos(z)$ in terms of the exponential function and expanded, but have not gotten anywhere (in plain language, my result is a big mess). I also tried another approach which was to use the fact that $|\tan(z)|^2 = \frac{|\sin(z)|^2}{|\cos(z)|^2}$, but ended using the angle-sum trig identities and ended again with another mess. Could someone please help in pointing in the right direction with this exercise?

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2 Answers 2

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Using the triangle inequality and assuming $y >0$, we get $$ \begin{align}|\tan(z)| &= |\tan (x+iy)| = \left|\frac{1}{i} \frac{e^{i(x+iy)}-e^{-i(x+iy)}}{e^{i(x+iy)}+e^{-i(x+iy)}} \right| \le \frac{|e^{ix}e^{-y}|+|-e^{-ix}e^{y}|}{\big||e^{ix}e^{-y}|-|e^{-ix}e^{y}|\big|} \\ &= \frac{e^{-y}+e^{y}}{e^{y}-e^{-y}} = \coth(y). \end{align}$$

Therefore, $$ |\tan(z)|^{2} \le \coth^{2}(y) = 1+ \frac{1}{\sinh^{2}(y)}. $$

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    $\begingroup$ (+1) beat me to it :-) This is a very important estimate in many contour integration computations. As $\mathrm{Im}(z)\to\infty$, $\tan(z)\to i$ exponentially. $\endgroup$
    – robjohn
    Dec 9, 2015 at 8:25
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$$ \begin{align} \left|\tan(x+iy)\right| &=\left|\frac{\sin(x)\cosh(y)+i\cos(x)\sinh(y)}{\cos(x)\cosh(y)-i\sin(x)\sinh(y)}\right|\\ &=\left|\coth(y)\right|\left|\frac{\tanh(y)-i\tan(x)}{\coth(y)-i\tan(x)}\right| \end{align} $$ Since $$ \underbrace{\left|\tanh(y)\right|^2}_{\tan(x)=0}\le\left|\frac{\tanh(y)-i\tan(x)}{\coth(y)-i\tan(x)}\right|\le\underbrace{\ \ \ \ \ \ 1\ \ \ \ \ \ }_{\tan(x)=\pm\infty} $$ we have $$ \left|\tanh(y)\right|\le\left|\tan(x+iy)\right|\le\left|\coth(y)\right| $$

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