Why $\cosh(y)$ cannot be zero in this example? "  Find the points $z \in \mathbb{C}$ at which the function $g(z) = \cos(\bar{z})$ satisﬁes the Cauchy-Riemann equations. "
So I've written $g(z) = g(x,y) = \cos(x-iy)=\cos(x)\cosh(y)+i\sin(x)\sinh(y)$. So that $u(x,y)=\cos(x)\cosh(y)$ and $v(x,y)=\sin(x)\sinh(y)$.
From the first C-R equation I get $-\sin(x)\cosh(y) = \sin(x)\cosh(y)$ which is true when $\sin(x)\cosh(y)=0$.
Now here I thought this implied $\sin(x) = 0$ or $\cosh(y) = 0$. However the solutions give only $\sin(x) = 0$, why?
Also, when doing the second equation, you get $\cos(x)\sinh(y) = 0$, but this time the solutions split the case into $\cos(x)=0$ and $\sinh(y)=0$ as I did above.
Why this?
EDIT
Fair enough, I've found this document explaining why it is not zero, as it was said in the comments. 
Hyperbolic Functions
 A: 
Now here I thought this implied $\sin(x) = 0$ or $\cosh(y) = 0$. However the solutions give only $\sin(x) = 0$, why?

$$ \cosh x = \frac{e^x + e^{-x}}2$$
So if $\cosh x = 0$, then $e^x + e^{-x} = 0$.  But this can't happen because both $e^x$ and $e^{-x}$ are positive, no matter what real value $x$ has.  And the sum of two positive numbers can never be zero.  Therefore $\cosh x$ can never be zero for real values of $x$.

Also, when doing the second equation, you get $\cos(x)\sinh(y) = 0$, but this time the solutions split the case into $\cos(x)=0$ and $\sinh(y)=0$ as I did above.
Why this?

$$ \sinh x = \frac{e^x - e^{-x}}2$$
So if $\sinh x = 0$, then $e^x - e^{-x} = 0$.  And this can happen because the difference of two positive numbers can be zero.  And the solution is
\begin{align*}
  e^x - e^{-x} &= 0\\
  e^x &= \frac1{e^x}\\
  e^{2x} &= 1\\
  2x &= \ln 1\\
  x &= 0
\end{align*}
A: Hint:
$$\cosh^2(y)=1+\sinh^2(y)$$
What happens if $\cosh(y)=0$?
A: The $\cosh$ of a real number is never zero.
