How to show that $\coth(\frac z2) = \frac{\sinh(x)-i\sin(y)}{\cosh(x)-\cos(y)}$

Question

How would I show that $\coth(\frac z2) = \frac{\sinh(x)-i\sin(y)}{\cosh(x)-\cos(y)}$

What I've done so far is using the definitons for $coth(z), cosh(z)$ and $sinh(z)$ and Euler's identity as follows:

\begin{align} \coth(z/2) &= \frac{\cosh(z/2)}{\sinh(z/2)} \\ \\ & = \frac{\cosh(\frac{x+iy}{2})}{\sinh(\frac{x+iy}{2})} \\ \\ & = \frac{e^{(x+iy)/2}+e^{-(x+iy)/2}}{e^{(x+iy)/2}-e^{-(x+iy)/2}} \\ \\ & = \frac{(e^{x/2}e^{iy/2})+(e^{-x/2}e^{-iy/2})}{(e^{x/2}e^{iy/2})-(e^{-x/2}e^{-iy/2})} \\ \\ & = \frac{(e^{x/2}(\cos(\frac y2)+i\sin(\frac y2))+(e^{-x/2}(\cos(\frac y2)-i\sin(\frac y2))}{(e^{x/2}(\cos(\frac y2)+i\sin(\frac y2)))-(e^{-x/2}(\cos(\frac y2)-i\sin(\frac y2))} \\ \\ & = \frac{\cos(\frac y2)\cosh(\frac x2) + i\sin(\frac y2)\sinh(\frac x2)}{\cos(\frac y2)\sinh(\frac x2) + i\sin(\frac y2)\cosh(\frac x2)} \end{align}

From this point on, I don't see how to factor or cancel terms so that I'm left with the desired expressions.

Answer 1

$$\text{RHS}=\frac{\cos(\frac y2)\cosh(\frac x2) + i\sin(\frac y2)\sinh(\frac x2)}{\cos(\frac y2)\sinh(\frac x2) + i\sin(\frac y2)\cosh(\frac x2)}$$

Keep going, multiply $\cos(\frac y2)\sinh(\frac x2) - i\sin(\frac y2)\cosh(\frac x2)$ on both numerator and denominator,

then use the identity $2\sinh(\frac x2)\cosh(\frac x2)=\sinh(x)$ and $2\sin(\frac x2)\cos(\frac x2)=\sin(x)$, we get

$$\begin{align}\text{RHS}&=\frac{\frac12\cos^2(\frac y2)\sinh(x) +\frac12\sin^2(\frac y2)\sinh(x) + \frac i2\sin(y)\sinh^2(\frac x2)-\frac i2\sin(y)\cosh^2(\frac x2)}{\cos^2(\frac y2)\sinh^2(\frac x2) + \sin^2(\frac y2)\cosh^2(\frac x2)}\\ \\ &=\frac{\frac12\sinh(x)-\frac12i\sin(y)}{\frac{1+\cos (y)}2\cdot \frac{\cosh (x)-1}2+\frac{1-\cos (y)}2\cdot \frac{\cosh (x)+1}2}\\ \\ &=\frac{\frac12\sinh(x)-\frac12i\sin(y)}{-\frac12\cos(y)+\frac12\cosh(x)}\\ \\ &=\frac{\sinh(x)-i\sin(y)}{\cosh(x)-\cos(y)}\end{align}$$