Hi I am struggling with this exercise, which may be perceived as simple. so I was trying to write tangents as follows:
$$\tan(z)=-i\frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}$$ and then $$z=a+bi$$, which led me to $$ \tan z=-i\frac{\cos a(e^{-b}-e^{b})+i\sin a(e^{-b}+e^{b})}{\cos a(e^{-b}+e^{b})+i\sin a(e^{-b}-e^{b})}$$, so I guess here I can multiply denominator by conjunction, but this is really a complicated computation on an exam... help appreciated
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It's easier if you don't use sines and cosines. We have $$\overline{e^{iz}+e^{-iz}}=\overline{e^{iz}}+\overline{e^{-iz}},$$ and, for example, $$\overline{e^{iz}}=e^{\overline{iz}}=e^{\overline{ix-y}}=e^{-y}e^{-ix}$$
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HINT
We have that by the standard trick
$$\tan(z)=-i\frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}\cdot\frac{e^{i\bar z}+e^{-i\bar z}}{e^{i\bar z}+e^{-i\bar z}}=\ldots$$
then use the well know identities for $\cos$, $\sin$, $\cosh$ and $\sinh$.
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From the title, you're looking for values $w$ for which $\tan^{-1} w$ exists. So you want to solve the following equation for $z$ :
$$w=\tan z\tag{requires $z\neq\tfrac{(2k+1)\pi}2$}$$ $$w = \frac{(e^{iz}-e^{-iz})/2i}{(e^{iz}+e^{-iz})/2}$$ $$iw=\frac{e^{2iz}-1}{e^{2iz}+1}$$ $$iwe^{2iz}+iw =e^{2iz}-1$$ $$(1-iw)(e^{iz})^2=1+iw$$ $$(e^{iw})^2 = \frac{1+iw}{1-iw}\tag{requires $w\neq-i$}$$ $$e^{iz} = \pm\sqrt{\frac{1+iw}{1-iw}}\tag{both roots included, unambiguous}$$ $$z = -i\log\left( \pm\sqrt{\frac{1+iw}{1-iw}}\right) \tag{requires $w\neq i$; multi-valued}$$
This last is a formula for $\tan^{-1} w$. It is evidently multi-valued and defined except for $w=\pm i$.
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The equation you have to solve is $$ -i\frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}=w $$ that can be rewritten as $$ i-ie^{2iz}=we^{2iz}+w $$ hence $$ e^{2iz}=\frac{i+w}{i-w} $$ The denominator vanishes for $w=i$. The numerator vanishes for $w=-i$. Since $e^{2iz}$ takes on all values except $0$, the equation has solution for $w\ne\pm i$.
It is much easier to deal with this problem using the fact that$$1+\tan^2(z)=\dfrac1{\cos^2(z)}.$$So, which numbers can be written as $\dfrac1{\cos^2(z)}$? Answer: all, except $0$. It follows from this (and from the fact that $\tan$ is an odd function), that the range of $\tan$ is $\mathbb{C}\setminus\{\pm i\}$.