What is the solution to the complex cosine equation without using logarithms?

[diddy_kaufen]

Homework Statement

Solve the equation

$$cos(\pi e^z) = 0$$

Homework Equations

I am not allowed to use the complex logarithm identities.

$$ \cos z = \frac{e^{iz}+e^{-iz}}{2} $$

$$e^{i\theta}=\cos\theta+i \sin\theta$$

The Attempt at a Solution

All I've gotten is $$\cos(\pi e^z)=0 \iff \pi e^z = \frac{\pi}{2}+p \pi \iff e^z=\frac{1}{2}+p, p\in Z $$

I have no idea how to solve this without resorting to the logarithm.

 

[FactChecker]

Look at the real part and the fact that the imaginary part is 0.

 

[diddy_kaufen]

but wouldn't I then still be using the logarithm? Since then I would have to do 1/2+p = exp( ln( 1/2+p ) ) but I'm not getting all solutions, since p is in Z and I'm not allowed to evaluate a complex logarithm?

 

[FactChecker]

but wouldn't I then still be using the logarithm? Since then I would have to do 1/2+p = exp( ln( 1/2+p ) ) but I'm not getting all solutions, since p is in Z and I'm not allowed to evaluate a complex logarithm?

You have equations in the "Relevant equations" section that can be used.

 

[diddy_kaufen]

Thank you for your help, I'm sorry but I'm still stuck. I used the relevant equations to substitute w = e^(iz)to get w=0 or w=-1, so |w|=e^(-y)=0 or |w|=1, so $$e^{iz}=e^{-y}(\cos x_0 + i \sin x_0)$$ which means that $$ x_0=\pi+n2\pi $$

is this even correct? I'm sorry I'm new to complex numbers

 

[FactChecker]

Continue your first post using e^z = e^x+iy = e^xe^iy = e^x(cos(y) +i sin(y))

 

[diddy_kaufen]

Ok I get $$e^x (\cos y + i \sin y) = \frac{1}{2}+p$$ or $$e^x \cos y = \frac{1}{2}+p$$ can I work around this without using the logarithm?

 

[FactChecker]

Ok I get $$e^x (\cos y + i \sin y) = \frac{1}{2}+p$$ or $$e^x \cos y = \frac{1}{2}+p$$ can I work around this without using the logarithm?

Don't forget about equating the imaginary parts.
I didn't check your work, so I will assume it is ok. x is a logarithm, so I don't see how to express it without using a logarithm.