I am working on an exercise and at the current stage, I want to show that (perhaps this is wrong)
For $z\in\mathbb{C}$, $\cos(z)=0$ only has real solution.
However, I had some short attempt but did not know how to proceed.
For instance, writing $z=x+iy$, we know that $$\cos(z)=0\implies \dfrac{e^{iz}+e^{-iz}}{2}=0\implies e^{ix}e^{-y}=-e^{-ix}e^{y},$$ but how could I argue from here to conclude that we must have $y=0$?
A similar argument is that $$\cos(x+iy)=\cos(x)\cos(iy)-\sin(x)\sin(iy)=0,$$ gives us $$\cos(x)\cos(iy)=\sin(x)\sin(iy),$$ again how could I use this to show $y=0$ must be true?
Thank you so much!
Best Answer
Building on your second approach: $$ \cos(x+iy)=\cos(x)\cos(iy)-\sin(x)\sin(iy) =\cos(x)\cosh(y)- i\sin(x)\sinh(y) $$ so that $$ |\cos(x+iy)| ^2 = \cos^2(x)\underbrace{\cosh^2(y)}_{1 + \sinh^2(x)} + \underbrace{\sin^2(x)}_{1 - \cos^2(x)}\sinh^2(y) = \cos^2(x) + \sinh^2(y) $$ and therefore $$ \begin{align} \cos(x+iy) = 0 &\iff \cos(x) = 0 \text{ and } \sinh(y) = 0 \\ &\iff \cos(x) = 0 \text{ and } y = 0 \, . \end{align} $$