how can I solve sin(z) = cosh(iz)?

Question

I've tried to solve this equation, $\sin(z) = \cosh(iz)$, but I have not found the correct way, any ideas?

I have tried two ways:

Using the relation $\sin(z)=-i\sinh(iz)$

And using that if $z= x + iy$, $\sin(z) = \sin(x)\cosh(y) + i\cos(x)\sinh(y)$

Answer 1

Hint:

Simply use the definitions:

• $\sin z=\dfrac{\mathrm e^{iz}-\mathrm e^{-iz}}{2i}=\dfrac{\mathrm e^{2iz}-1}{2i\,\mathrm e^{iz}}$
• $\cosh iz=\dfrac{\mathrm e^{iz}+\mathrm e^{-iz}}{2}=\dfrac{\mathrm e^{2iz}+1}{2\,\mathrm e^{iz}}$

and set $u=\mathrm e^{iz}$. The equation is then written as $$\frac{u^2-1}{2iu}=\frac{u^2+1}{2u}\iff 2u(u^2-1)=2iu(u^2+1)\iff\begin{cases}u=0&\text{or}\\\dfrac{u^2-1}{u^2+1}=i;\end{cases}$$

Answer 2

As $\cosh(iz)=\cos(z)$, you are looking for roots of $$ 0=\cos(z)-\sin(z)=\sqrt2\cos(z+\frac\pi4)=-\sqrt2\sin(z-\frac\pi4) $$ which has the obvious solutions.