Can the real and imaginary parts of $\dfrac{\sin z}z$ be simplified?

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.

Solution 1:

The function $f$ defined by

$$ f(z) = \frac{\sin z}{z} $$

is nonconstant, entire, of order $1$, and of genus either $0$ or $1$. Since $f(z)$ is real for real $z$ and all zeros of $f$ are real, Laguerre's theorem on separation of zeros implies that all zeros of $f'$ are real (and separated by the zeros of $f$).

I figured I'd give a more self-contained answer, essentially replicating the proof of Laguerre's theorem.

We have

$$ f(z) = \frac{\sin z}{z} = \prod_{n \neq 0} \left(1 - \frac{z}{\pi n}\right) e^{z/(\pi n)}. $$

From this we get

$$ \frac{f'(z)}{f(z)} = \sum_{n \neq 0} \left(\frac{1}{z - \pi n} + \frac{1}{\pi n}\right), $$

so that, if $z = x+iy$,

$$ \operatorname{Im}\left(\frac{f'(z)}{f(z)}\right) = -y \sum_{n \neq 0} \frac{1}{(x-\pi n)^2+y^2}, $$

which cannot vanish unless $y = 0$. Thus $f'$ has only real zeros.

(Additionally, by showing that $[f'(x)/f(x)]' < 0$ where $f(x) \neq 0$, one can see that $f'/f$ has exactly one zero between each pair of zeros of $f$.)