 treetopssan

2022-01-27

Find the set of values, which are taken by $\mathrm{tan}z$
I was trying to write tangents as follows:
$\mathrm{tan}\left(z\right)=-i\frac{{e}^{iz}-{e}^{-iz}}{{e}^{iz}+{e}^{-iz}}$
and then
z=a+bi
, which led me to
$\mathrm{tan}z=-i\frac{\mathrm{cos}a\left({e}^{-b}-{e}^{b}\right)+i\mathrm{sin}a\left({e}^{-b}+{e}^{b}\right)}{\mathrm{cos}a\left({e}^{-b}+{e}^{b}\right)+i\mathrm{sin}a\left({e}^{-b}-{e}^{b}\right)}$ izumrledk

Expert

It is much easier to deal with this problem using the fact that
$1+{\mathrm{tan}}^{2}\left(z\right)=\frac{1}{{\mathrm{cos}}^{2}\left(z\right)}$
So, which numbers can be written as $\frac{1}{{\mathrm{cos}}^{2}\left(z\right)}$? Answer: all, except 0. It follows from this (and from the fact that $\mathrm{tan}$ is an odd function), that the range of Jude Carpenter

Expert

HINT
We have that by the standard trick
$\mathrm{tan}\left(z\right)=-i\frac{{e}^{iz}-{e}^{-iz}}{{e}^{iz}+{e}^{-iz}}\cdot \frac{{e}^{i\stackrel{―}{z}}+{e}^{-i\stackrel{―}{z}}}{{e}^{i\stackrel{―}{z}}+{e}^{-i\stackrel{―}{z}}}=\dots$
then use the well know identities for

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