Find all value of $z$ for which each equation holds. $$(a) \sin z=\cosh 4$$
$$(b) \cos z=2$$
$$(c) \sin z=i\sinh 1$$
$$(d) \cosh z=1$$
my answer
(a) $\sin z= \sin x \cosh y+\cos x \sinh y=\cosh 4$
so, $x=2n\pi+{\pi\over2}$ and $y=4$
(b) since maximum value of $\cos z$ is 1, so DNE.
(c) represent, $-e^{iz}+e^{-iz}=e-e^{-1}$
ie) $-e^{-y}(\cos x+i\sin x)+e^y(\cos x-i\sin x)=e-e^{-1}$
then $y=1$ and $x=2n\pi$
(d) $e^x(\cos x+i\sin y)+e^{-x}(\cos x-i\sin y)=2$
ie) $\cos y(e^x+e^{-x})+i\sin y(e^x-e^{-x})=2$
then $y=2n\pi$ and x=0.
$ $
Could you check my answer? and please edit it.

$(d)\displaystyle\cosh(z)=1\implies\cos(iz)=1=\cos0\implies iz=2n\pi$ where $n$ is any integer
$(b)$  $-1\le\cos z\le1$ for real $z$ only
$(a)\displaystyle\sin z=\cosh4=\cos(4i)=\sin\left(\frac\pi2-4i\right)$ 
$\displaystyle z=n\pi+(-1)^n\left(\frac\pi2-4i\right)$
