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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.

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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 $$\frac{2(e^z - e^{-z})}{(e^z + e^{-z}) (1 - \frac{e^z - e^{-z}}{e^z + e^{-z}})} = \frac{2(e^z - e^{-z})}{e^z + e^{-z} - (e^z - e^{-z})} = \frac{e^z - e^{-z}}{e^{-z}} = e^{2z} - 1$$

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  • $\begingroup$ Wow, just wow, that is so efficient, how did you know that would work out so quickly? How do you plan ahead? $\endgroup$ – Identity crisis Mar 29 '15 at 23:08
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    $\begingroup$ @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. $\endgroup$ – AlexR Mar 29 '15 at 23:10
  • $\begingroup$ That makes me sad haha. Thank you! $\endgroup$ – Identity crisis Mar 29 '15 at 23:10
  • $\begingroup$ @Identitycrisis After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. $\endgroup$ – AlexR Mar 29 '15 at 23:41

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