opepayflarpws

2022-06-24

Find all z such that $|\mathrm{tan}z|=1$

Jaylee Dodson

Expert

$\mathrm{tan}z=\frac{\mathrm{sin}z}{\mathrm{cos}z}=\frac{{e}^{iz}-{e}^{-iz}}{i\left({e}^{iz}+{e}^{-iz}\right)}$
So
$|\mathrm{tan}z|=|\frac{{e}^{iz}-{e}^{-iz}}{{e}^{iz}+{e}^{-iz}}|=1$
$⇒|{e}^{iz}+{e}^{-iz}|=|{e}^{iz}-{e}^{-iz}|$
$⇒|{e}^{2iz}+1{|}^{2}=|{e}^{2iz}-1{|}^{2}$
$⇒\left({e}^{2iz}+1\right)\overline{\left({e}^{2iz}+1\right)}=\left({e}^{2iz}-1\right)\overline{\left({e}^{2iz}-1\right)}$¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
$⇒\left({e}^{2iz}+1\right)\left({e}^{-2iz}+1\right)=\left({e}^{2iz}-1\right)\left({e}^{-2iz}-1\right)$
$⇒1+{e}^{2iz}+{e}^{-2iz}+1=1-{e}^{2iz}-{e}^{-2iz}+1$
$⇒{e}^{2iz}+{e}^{-2iz}=0$
$⇒\mathrm{cos}\left(2z\right)+i\mathrm{sin}\left(2z\right)+\mathrm{cos}\left(2z\right)-i\mathrm{sin}\left(2z\right)=0$
$⇒\mathrm{cos}\left(2z\right)=0$
So
$2z=n\pi +\frac{\pi }{2}$

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