I have calculated the real and imaginary parts of $\dfrac{\sin z}z.$ I've obtained
$\begin{eqnarray} \frac{\sin z}z&=&\frac{\sin(x+iy)}{(x+iy)}\\ &=& \frac{\sin(x)\cosh(y)+i\cos(x)\sinh(y)}{x+iy}\\ &=& \frac{x\sin(x)\cosh(y)+y\cos(x)\sinh(y)}{x^2+y^2}+i\frac{x\cos(x)\sinh(y)-y\sin(x)\cosh(y)}{x^2+y^2}. \end{eqnarray}$
Before calculating it, I hoped I could use this in a solution to a certain problem, but I would need to differentiate the real and imaginary parts of $\dfrac{\sin z}z$ with regard to $x$ and $y$. But what I'm getting is very complicated. I understand that it could be that that's just the way it is. But maybe there are some simplifications to be done either in the functions or in their derivatives? I've never really used hyperbolic functions before so I'm not sure I'm not missing something.
Added. The problem I'm working on is the following.
Prove that the solutions of the equation $\tan z=z$ are all real.
I will sketch the solution I know here.
It can be calculated that
$\tan(x+iy)=\frac{\sin(2x)}{\cos(2x)+\cosh(2y)}+i\frac{\sinh(2y)}{\cos(2x)+\cosh(2y)}.$
Plugging this to the equation, we get
$\begin{eqnarray} x&=&\frac{\sin(2x)}{\cos(2x)+\cosh(2y)}\\ y&=&\frac{\sinh(2y)}{\cos(2x)+\cosh(2y)}. \end{eqnarray}$
From this we can get
$\sin(2x)\cdot2y=\sinh(2y)\cdot2x.$
By plotting the functions $\sin,$ $\mathrm{id}$ and $\sinh$ (and proving the right inequalities we can see from the plots), we can see that this is only possible when $xy=0$. But when we plug $x=0$ to the second equation, we can show that the only solution is $y=0$, which ends the proof.
What strikes me in this solution is the fact that $\dfrac{\sin z}z$ is used. (Not explicitly in my statement.) This function has another connection to the problem:
The non-zero solutions of the equation $\tan z=z$ are exactly the zeros of the derivative of $\dfrac{\sin z}z.$
This is easy to prove. I wondered if there shouldn't be a solution to the problem which uses this fact, given that $\dfrac{\sin z}z$ appears in the solution I know.