So I have to solve this equation:
$$\cos z=0$$
Solution:
$$\cos z=\frac{e^{iz}+e^{-iz}}{2}=0$$ $$e^{iz}+e^{-iz}=0$$ $$e^{i2z}=-1$$
Here I get stuck. The answer is obviously $\pi/2+k\pi, k \in \mathbb{Z}$ But I can't seem to figure what to do here to proove it but the thing that worries me more is that $$\cosh z=\cos z$$ but $$\cosh z \neq \cos z$$ $$\cosh z=i(k\pi+\pi/2), k \in \mathbb{Z}$$
Why is that true, how two functions which look the same way can have different values ?!
How do I prove that $\cos z=\pi/2+k\pi$
$\endgroup$ 31 Answer
$\begingroup$Expand using $$ \cos z = \cos \left( x + i y \right) = \cos x \cos (iy) - \sin x \sin (iy) = \cos x \cosh y + i \sin x \sinh y $$ Enforce $$ \cos z = 0 + 0 i $$
For the real part, the constraint is $$ \cos x = 0 \qquad \Rightarrow \qquad x = \frac{\pi}{2} + k \pi, \quad k\in\mathbb{Z} $$
For the imaginary part, the constraint $$ \sin x \sinh y = 0 \qquad \Rightarrow \qquad y = 0 $$
The solution to $$ \boxed{ \cos z = 0 \qquad \Rightarrow \qquad z = \left( \frac{\pi}{2} + k \pi \right) + 0i, \quad } $$
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