cosh (X) =0

what values of x give cosh (x) =0 ?

$\begin{array}{rcl} \cosh{(x)} &=& \dfrac { e^x + e^{-x} } {2} = 0 \\ \dfrac { e^x + e^{-x} } {2} &=& 0 \\ e^x + e^{-x} &=& 0 \\ e^x + \dfrac{ 1 } { e^{x} } &=& 0 \qquad | \qquad \cdot e^x\\ e^{2x} + 1 &=& 0 \qquad | \qquad \boxed{~ \text{Euler's identity: } e^{i\pi} + 1 = 0 ~}\\ e^{2x} + 1 &=& e^{i\pi} + 1 \\ e^{2x} &=& e^{i\pi}\\ 2x &=& i\pi \\ x &=& \dfrac{ i\pi } {2} \end{array}$

There aren't any Real values for x that satisfy this equality:

The following Imaginary numbers satisfy it:  i(2n-1)*pi/2  where n is an integer.

All Solutions for x:

$\begin{array}{rcl} \cosh{(x)} &=& \dfrac { e^x + e^{-x} } {2} = 0 \\ \dfrac { e^x + e^{-x} } {2} &=& 0 \\ e^x + e^{-x} &=& 0 \\ e^x + \dfrac{ 1 } { e^{x} } &=& 0 \qquad | \qquad \cdot e^x\\ e^{2x} + 1 &=& 0 \qquad | \qquad \boxed{~ \text{Euler's identity: } e^{i\pi} + 1 = 0 ~}\\ e^{2x} + 1 &=& e^{i (\pi \pm 2k\pi) } + 1 \\ e^{2x} &=& e^{i(\pi \pm 2k\pi)}\\ 2x &=& i(\pi \pm 2k\pi) \\ x &=& \dfrac{ i(\pi \pm 2k\pi) } {2}\\ \mathbf{x} & \mathbf{=}& \mathbf{ \dfrac{ i\pi }{2} \pm k\pi i } \qquad k \in N \qquad | \qquad \dfrac{ i\pi }{2} - \pi i = -\dfrac{ i\pi }{2} \\ \mathbf{x} &\mathbf{=}& \mathbf{ -\dfrac{ i\pi }{2} \pm k\pi i \qquad k \in N } \\ \end{array}$