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Given $ z = \tan2i $, express $z$ in the form $a + ib$.

I tried this way:

$$ \tan2i = \frac{\sin2i}{\cos2i}=\frac{\frac{e^{-2}-e^{2}}{2i}}{\frac{e^{-2}+e^{2}}{2}} $$

and got stuck with:

$$\frac{e^{-2}-e^{2}}{i\left(e^{-2}+e^{2}\right)}$$

Any ideas will be appreciated.

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    multiply top and bottom by $i$ and you are left with a purely imaginary number – 2017-01-26
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    You practically got stuck with the right solution (Lol)! – 2017-01-26

1 Answers 1

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$\frac{1}{i}=-i.$ You're pretty much done.

$-i\frac{\frac{1}{e^2}-e^2}{\frac{1}{e^2}+e^2}=-i\frac{1-e^4}{1+e^4}=\frac{e^4-1}{e^4+1}i.$