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


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


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.

  • $\begingroup$ Since $f(z)=\frac{\sin z}{z}$ is harmonic, if you write $g(x,y)=f(x+iy)$ then $\frac{dg}{dy} = i\frac{dg}{dx}$. So you only need to compute half of the partial derivatives, at least. $\endgroup$ Commented Nov 28, 2012 at 22:09
  • $\begingroup$ @ThomasAndrews Thank you. Yes, I realize this, but still the formulas I'm getting aren't manageable at all... $\endgroup$
    – Bartek
    Commented Nov 28, 2012 at 22:11
  • $\begingroup$ Also, $dg/dx$ is $df/dz$ evaluated at $x+iy$, and $df/dz = \frac{z\cos z - \sin z}{z^2}$. Might or might not make it easier to substitute $x+yi$ there. $\endgroup$ Commented Nov 28, 2012 at 22:14
  • $\begingroup$ In complex analysis it's mostly not helpful to switch back to real number interpretations. It may be easier to solve your problem directly in $\mathbb{C}$ in this case as well. $\endgroup$
    – WimC
    Commented Nov 29, 2012 at 18:36
  • $\begingroup$ @WimC Unfortunately the problem is explicitly about the real and imaginary parts of a certain function. I know how to solve the problem, but the solution I have is ugly. The function in the problem is $\tan (z)-z$, and $\dfrac{\sin z}z$ appears in the solution. I hoped looking at the latter function would give me a nicer solution. $\endgroup$
    – Bartek
    Commented Nov 30, 2012 at 11:38

1 Answer 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 $f$.)

  • $\begingroup$ Wow, thanks! This is way, way above my head, but I will try to study these definitions and the theorem to understand your answer. Is there any connection here with the $\dfrac{\sin z}z$ appearing in the solution I gave? $\endgroup$
    – Bartek
    Commented Dec 1, 2012 at 0:41
  • $\begingroup$ @Bartek, my pleasure. Are you referring to the appearance of $\sin z/z$ in the solution to the problem that you provided? I can't say I see one, but I won't rule out the possibility. $\endgroup$ Commented Dec 1, 2012 at 0:56
  • $\begingroup$ Yes, that's what I meant. Thank you very much for the answer. I'll wait with accepting it and awarding the bounty because some more exposure can't hurt. $\endgroup$
    – Bartek
    Commented Dec 1, 2012 at 2:14

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