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.
 A: $(d)\displaystyle\cosh(z)=1\implies\cos(iz)=1=\cos0\implies iz=2n\pi$ where $n$ is any integer
$(b)$  $-1\le\cos z\le1$ for real $z$ only
$(a)\displaystyle\sin z=\cosh4=\cos(4i)=\sin\left(\frac\pi2-4i\right)$ 
$\displaystyle z=n\pi+(-1)^n\left(\frac\pi2-4i\right)$
A: You'd better use general formulas for the inverse trigonometric functions, together with the periodicity of the complex logarithm. http://en.wikipedia.org/wiki/Hyperbolic_function#Inverse_functions_as_logarithms, http://en.wikipedia.org/wiki/Hyperbolic_function#Hyperbolic_functions_for_complex_numbers,http://en.wikipedia.org/wiki/Complex_logarithm.
A: These are really simple to solve, if thought about in the right way. We only need one fact, $$e^{z}=e^{w}\ \ \ \ \ \text{if and only if}\ \ \ \ \ z-w=2\pi i n\ \ \ \ \ \text{for some}\ \ \ \ \ n\in\mathbb{Z}.$$ Consider case two, $$\cos(z)=2.$$ Write, $$\frac{e^{iz}+e^{-iz}}{2}=2,$$ cancel  fractions, $$(2e^{iz})\frac{e^{iz}+e^{-iz}}{2}= (2e^{iz})2,$$ isolate the quadratic, $$(e^{iz})^2-4e^{iz}+1=0,$$ solve the quadratic, $$e^{iz}=2\pm \sqrt{3}=e^{\ln(2\pm\sqrt{3})},$$ hence, $$zi-\ln(2\pm \sqrt{3})=2\pi i n,$$ thus, $$z=-\ln(2\pm\sqrt{3})i+2\pi n.$$ These are all the solutions and the only solutions for $\cos(z)=2$. Compare with wolframalpha. Notice all the answers are non-real, this follows because (like you said) $-1\le \cos(r)\le 1$ holds for all real numbers. In general, all answers to trigonometric functions look like $\theta+2\pi n$ for some angle $\theta$. For hyper-trigonometric functions all answers look like $\theta+2\pi i n$ for some angle $\theta$. When solving Trigonometric or hyper-trigonometric equations, it is always best to use the exponential definitions, because of the property mention at the top. In this way, all these problems can be similarly solved.  
