Roots of $\sin^2z=\sinh^2z.$

Question

I'm trying to calculate the spectrum of an operator and after all the simplification, it has boiled down to:

Find the roots of$$\sin^2z=\sinh^2z.$$

I tried to use Mathematica but I get the following error: "This system can't be solved with the methods available to solve".

Can I get a hint on how to solve this?

Edit: As I understand from the comments, the solution set is too broad. So I'll narrow my question further:

Let $$A=\{4z^4:\sin^2z=\sinh^2z\}.$$Find $\sup(A\cap \mathbb{R}).$

Would this be easier than explicitly finding the roots?

Answer 1

The solutions of $\sin^2z=\sinh^2z$ are precisely the points $z=u(1\pm i)$ where $u$ is any of the (infinitely many) real numbers such that $\tan u=\pm\tanh u$. (If you draw the graphs of the tangent and hyperbolic tangent functions, you can see they intersect infinitely many times, once in each interval of width $\pi$.)

To show this, let $z=x+iy$ with $x,y\in\mathbb{R}$. We have

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

and

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

Now suppose $\sin z=\pm\sinh z$. Then, separating the real and imaginary parts, we have two equations:

$$\sin x\cosh y=\pm\sinh x\cos y\quad\text{and}\quad\cos x\sinh y=\pm\cosh x\sin y$$

It's easy to see that if $|x|=|y|=|u|$, the two equations are the same, and can be rewritten as $\tan u=\pm\tanh u$. It remains to show that there are no other solutions (i.e., no solutions with $|x|\not=|y|$).

To show there are no other solutions, let's square the two equations, convert everything to $\cos$ and $\cosh$, and let $a=\cos^2x$, $b=\cos^2y$, $A=\cosh^2x$, and $B=\cosh^2y$, to get

$$(1-a)B=(A-1)b\quad\text{and}\quad a(B-1)=A(1-b)$$

from which we see

$$A+a=Ab+aB=B+b$$

which implies

$$\cosh^2x+\cos^2x=\cosh^2y+\cos^2y$$

But the even function $f(u)=\cosh^2u+\cos^2u$ is increasing on $[0,\infty)$ since

$$f'(u)=2\cosh u\sinh u-2\cos u\sin u=\sinh2u-\sin2u\implies f'(0)=0$$

and $f''(u)=2(\cosh2u-\cos2u)\gt0$ for $u\not=0$. Consequently $\cosh^2x+\cos^2x=\cosh^2y+\cos^2y$ implies $y=\pm x$. And thus we're done.