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I am trying to prove the following trigonometric identity, but I can't.

$$\coth\left(\frac{z}{2}\right) = \frac{\sinh x - i\sin y}{\cosh x-\cos y}$$

Here $z=x+iy$. I tried the following

$$\coth\left(\frac{z}{2}\right) = \frac{\cosh\left(\frac{z}{2}\right)}{\sinh\left(\frac{z}{2}\right)} = \frac{\cosh^2\left(\frac{z}{2}\right)}{\sinh\left(\frac{z}{2}\right)\cosh\left(\frac{z}{2}\right)} = \frac{1+\cosh z}{\sinh z} = \frac{1 + \cosh x\cos y + i\sinh y\sin x}{\sinh x\cos y + i\cosh x\sin y}$$

I don't know if this is the correct way to do the problem or how to continue from here. Can anyone help me?

Thanks.

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  • $\begingroup$ If what you have done is correct, the next step that I would take would be to multiply top and bottom of your rational expression by complex conjugate of the denominator. $\endgroup$ – Tucker Jun 15 '17 at 3:13
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Let $w=z/2 = u+iv$. Then, we can write

$$\begin{align} \coth(z/2)&=\coth(w)\\\\ &=\coth(u+iv)\\\\ &=\frac{\cosh(u+iv)}{\sinh(u+iv)}\\\\ &=\frac{\cosh(u)\cos(v)+i\sinh(u)\sin(v)}{\sinh(u)\cos(v)+i\cosh(u)\sin(v)}\\\\ &=\frac{\cosh(x/2)\cos(y/2)+i\sinh(x/2)\sin(y/2)}{\sinh(x/2)\cos(y/2)+i\cosh(x/2)\sin(y/2)}\\\\ &=\frac{ \sinh(x/2)\cosh(x/2)-i\sin(y/2)\cos(y/2)}{\cosh^2(x/2)-\cos^2(y/2)}\\\\ &=\frac{\frac12 \sinh(x)-i\frac12 \sin(y)}{\frac12 \cosh(x)-\frac12 \cos(y)}\\\\ &=\frac{\sinh(x)-i\sin(y)}{\cosh(x)-\cos(y)} \end{align}$$

as was to be shown!

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  • $\begingroup$ How did you pass from the fifth line to the sixth line? $\endgroup$ – Jack Jun 15 '17 at 15:52
  • $\begingroup$ Jack, from the fifth to sixth lines, I multiplied the numerator and denominator by the complex conjugate of the denominator, which rendered the new denominator equal to the square of the magnitude of the original denominator. $\endgroup$ – Mark Viola Jun 15 '17 at 16:03
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My hint would be that the easiest approach would be to use the defining formulae for $\sin$ and $\sinh$, and $\cos$ and $\cosh$ in terms of exponentials: $$\cos(x) = \frac{1}{2}(e^{ix} + e^{-ix})$$ etc.

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