I really can't figure out how to do this at all. I've been trying to show this for nearly 4 hours now. I've tried working from $\tanh(z)=\frac{\sinh(z)}{\cosh(z)}$ and expanding the top and bottom, but that just becomes a mess that, after trying so hard to put into the desired form, didn't work. I also tried working from the identity


but I wasn't able to put that into the desired form as well. I've tried working from the right side to the left, using every formula for $\sin(2y)$, $\sinh(2x)$, $\cosh(2x)$, and $\cos(2y)$ I could derive/find. I even tried multiplying the right-side by $\coth(z)$ and working it out to show that it's equal to $1$, but that didn't work.

Could you please give me some hints?

  • $\begingroup$ In $\tanh(z) = \frac{\sinh(z)}{\cosh(z)}$, write out the ratio in terms of exponentials, and then multiply and divide by the conjugate of the denominator (i.e $\cosh(\overline{z})$), $\endgroup$ – vnd Nov 5 '15 at 3:42
  • $\begingroup$ @vnd Yeah I tried that and it worked. I just saw that somebody had already worked out the solution in an answer (below)! $\endgroup$ – Arturo don Juan Nov 5 '15 at 4:07

Let me try. We have $$\tanh (z) = \frac{\sinh z}{\cosh z} = \frac{e^z - e^{-z}}{e^z + e^{-z}} = \frac{e^{2z} -1 }{e^{2z}+1} = \frac{e^{2x}(\cos(y) + i\sin(y))^2 -1}{e^{2x}(\cos(y) + i\sin(y))^2 + 1} = \frac{e^{2x}\cos 2y -1 + ie^x\sin 2y}{e^{2x}\cos 2y + 1 + ie^{2x}\sin 2y} = \frac{(e^{2x}\cos 2y -1 + ie^{2x}\sin 2y)(e^{2x}\cos 2y +1 - ie^{2x}\sin 2y)}{(e^{2x}\cos 2y +1)^2 + e^{4x}\sin^22y} = \frac{e^{4x}\cos^2 2y - (1-ie^{2x}\sin 2y)^2}{e^{4x}+1 + 2e^{2x}\cos 2y} = \frac{e^{4x}-1 + 2ie^{2x}\sin 2y}{e^{4x}+1 + 2e^{2x}\cos 2y} = \frac{e^{2x}-e^{-2x} + 2i\sin 2y}{e^{2x}+e^{-2x} + 2\cos 2y} = \frac{\sinh 2x + i\sin 2y}{\cosh 2x + \cos 2y}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.