Find the zeros of $\cos(z)+\sinh(iz)$

Question

As part of a worksheet I am currently going through, a question asks to

Find the zeros and their order of the following function: $\cos(z)+\sinh(iz)$

My attempt:

Rewrite $\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$ and $\sinh(z)=\frac{e^{z}-e^{-z}}{2}$ and we can write $\sinh(iz)=i\sin(z)=\frac{e^{iz}-e^{-iz}}{2}$

Then equating the two gives us $$e^{iz}+e^{-iz}=-(e^{iz}-e^{-iz}) \iff e^{iz}=-e^{iz}\iff e^{iz}=0$$

This, of course, cannot be true and so there is no solution i.e. there are no zeros of this equation.

I find it hard to believe that an exercise sheet would ask us to find the zeros and their order of a function that has none.

Is this simply an open shut case of "this function has no zeros" or is there something more complex that I am missing?

Answer 1

Note that $\sinh(iz)=i\sin z$ and therefore that$$\cos z+\sinh(iz)=0\iff\cos z+i\sin z=0\iff e^{iz}=0.$$So, yes, there are no roots.

Answer 2

HINT.

$$\sinh(iz) = i\sin(z)$$