Solving $\sin z=\sinh z$

Question

I guess that the only $z\in \mathbb C$ such that $\sin z=\sinh z$ is $z=0$. Well, I cannot see immediately a proof.

Answer 1

I can provide a solution for all $z \in \Re$.

The rate of change of $\sinh x$ with respect to $x$ is $\cosh x$, which is always greater than one (except at $x=0$), and the rate of change of $\sin x$ is $\cos x$, which is never greater than one. Because $\sinh x=\sin x$ when $x=0$ but the rate of change of $\sinh x$ always remains greater than that of $\sin x$, they do not intersect again for $x \gt 0$. Since both $\sin x$ and $\sinh x$ are odd functions (they are symmetric about the origin), they also do not intersect for $x \lt 0$, showing that $(0,0)$ is the only point of intersection.

Answer 2

For $z\in\mathbb R$, we see that

$$|\sin(z)|\le|z|\le|\sinh(x)|$$

which means $z=0$ is the only real solution.

For $z\in\mathbb C$, we let $z=x+iy$ to get

$$\sin(x+iy)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)$$

$$\sinh(x+iy)=\sinh(x)\cos(y)+i\cosh(x)\sin(y)$$

By equating parts, we find

$$\begin{cases}\sin(x)\cosh(y)=\sinh(x)\cos(y)\\\cos(x)\sinh(y)=\cosh(x)\sin(y)\end{cases}$$

Both equations have infinitely many solutions. Particularly, there exist many solutions along $x=y$.

$$\begin{cases}\sin(x)\cosh(x)=\sinh(x)\cos(x)\\\cos(x)\sinh(x)=\cosh(x)\sin(x)\end{cases}$$

This reduces both lines to the same problem, so we're faced with

$$\tanh(x)=\tan(x)$$

And since $\tanh(x)$ is bounded in $\pm1$ and continuous while $\tan(x)$ is periodic every $\pi$ and goes through all real numbers in each interval of $\pi$, there are infinitely many solutions. Solving $\tan(x)=\pm1$ will then give the approximate form of the solutions, mainly $x\approx\pm\frac\pi4+\pi k$ for $k\in\mathbb Z$.

There also exists other solutions, but they are harder to point out.

Answer 3

Observe by Euler's Formula that:

$$ \sin(z) = \frac{e^{ix} - e^{-ix}}{2i} $$ $$ \sinh(z) = \frac{e^{x} - e^{-x}}{2}$$

So our equation is asking us for solutions to:

$$ \frac{e^{ix} - e^{-ix}}{2i} = \frac{e^{x} - e^{-x}}{2} $$

That is

$$ \frac{e^{ix + x} - e^{x}}{i} = e^{ix + x} - e^{ix} $$

Dividing out the $i$ yields:

$$ -ie^{ix + x} + ie^{x} = e^{ix + x} - e^{ix} $$

balancing terms

$$ (1+i)e^{ix + x} - ie^x - e^{ix}= 0 $$

Let $u = e^{x}$ then this bcomes

$$ (1+i)u^{i+1} - iu - u^i = 0$$

And this will have infinitely many solutions in $\mathbb{C}$, each one giving rise to a solution to the original equation by $x = \ln(u)$.

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Why infinitely many solutions?

Lets begin by looking at solutions to

$$ x^w = 1 $$

What this can be interpreted as (via Euler's formula) is looking for points $x$ such that there is some integer $k$ such that $k\frac{2*i*\pi}{w} $ radians counterclockwise from the point $x$ lands you at the point $1$. If $w$ is rational, then there will only be finitely many such points. If $w$ is irrational real, then there will be infinitely many such points.

Now if $w$ is complex, is natural to assume that is a generalization of the irrational case (BUT the notion of traveling around the circle doesn't make sense for complex numbers, so i'm not yet fully comfortable with this analogy).

I'll ponder on this a bit more,

Answer 4

We employ Newton-Raphsons iteration for complex argument directly as there is no closed form for transcendental equation.

Just as in real arguments we have

$$ x_{n+1}= x_n- \frac{f}{f^{'}} $$

we can have complex $f$

$$ z_{n+1}= z_n- \frac{f}{f^{'}} =z_n- \frac{\sin z- \sinh z}{ \cos z -\cosh z}$$

Seed values can be used as real asymptotic solutions as indicated by others, resulting in an infinite set of complex roots( just as inverse trig function solutions have multi-valued roots in the reals) using a CAS. A sufficiently close interval spaced complex number in AP also supplies needed complex solutions.