# Simplifying hyperbolic complex function

I am trying to simplify the following fraction:

$$\frac{2\tanh(z)}{1-\tanh(z))}$$

I know it equals: $e^{2z}-1$ from wolframalpha, but I have no idea why.

I imagine I should be using some trig identities, but it looks like it will get too messy: Should I be first putting it as

1) $\tanh(z)=\frac{\sinh(z)}{\cosh(z)}$ and then

2) taking the $\sinh(x+iy) = \sinh(x) +\cos(y)+ i \cosh(x) +\sinh(y)$ and the analog for $\cosh(x+iy)$, and then

3) convert to the $e^{ix}$ form and then to $\sin,\cos$ form, and then

4) simplify?

Will this work? I feel like it is going to become absurdly long.

I suggest using $\sinh z = \frac12 (e^z - e^{-z})$ and $\cosh z = \frac12 (e^z +e^{-z})$. This yields $$\tanh z = \frac{e^z - e^{-z}}{e^z + e^{-z}}$$ So your expression becomes $$\frac{2(e^z - e^{-z})}{(e^z + e^{-z}) (1 - \frac{e^z - e^{-z}}{e^z + e^{-z}})} = \frac{2(e^z - e^{-z})}{e^z + e^{-z} - (e^z - e^{-z})} = \frac{e^z - e^{-z}}{e^{-z}} = e^{2z} - 1$$
• @Identitycrisis Not at all, actually. I only used the two identites I knew by heart ($\sinh z$ and $\cosh z$) and wrote up the solution ad-hoc. – AlexR Mar 29 '15 at 23:10