Use the addition formulae:
$$ \cosh{(x+iy)} = \cosh{x}\cosh{iy}+\sinh{x}\sinh{iy}. $$
Now, $\cosh{iy}=\cos{y}$, and $\sinh{iy}=i \sin{y}$, and the imaginary part is hence $ \sinh{x}\sin{y} $, so in fact the first formula is false.
For the second one, it's definitely wrong (take $z=0$...). Again using the addition formulae,
$$ \sinh{(x+iy)} = \sinh{x}\cosh{iy}+\cosh{x}\sinh{iy} = \sinh{x}\cos{y}+i\cosh{x}\sin{y}. $$
Now everything but the $i$ is real, so we know that
$$ \lvert \sinh{(x+iy)} \rvert^2 = \sinh^2{x}\cos^2{y} + \cosh^2{x}\sin^2{y}. $$
Now, we can decide to use both of $\sin^2+\cos^2=1$, or $\cosh^2-\sinh^2=1$, to find the two equivalent forms
$$ \lvert \sinh{(x+iy)} \rvert^2 = \sinh^2{x} + \sin^2{y} = \cosh^2{x} + \cos^2{y}. $$
Indeed, the method can be salvaged by taking the limits sufficiently carefully. Note that the contour proposed yields the equation
$$-2\sinh^2\pi\int_{-\infty}^\infty\frac{\sin^2x}{\sinh^2 x}dx+\lim_{\epsilon\to 0}\left[\left(\int_{-\infty}^{-\epsilon}+\int^{\infty}_{\epsilon}\right)\frac{\sinh^2\pi}{\sinh^2 x}-\\\int_0^\pi\frac{(\cosh^2\pi+\sinh^2\pi)\sin^2(\epsilon e^{-it})-\sinh^2\pi+i\sinh\pi\cosh\pi\sin (2\epsilon e^{-it})}{\sinh^2 (\epsilon e^{-it})}i\epsilon e^{-it}dt\right]=0$$
Now, we need to understand the leading behavior of the expression in brackets for $\epsilon\to 0$. The first integral is basically a principal value prescription with a double pole at $x=0$ and as such, is of order $\mathcal{O}(1/\epsilon).$ We expect this one to be canceled exactly. To this end we note that by expanding the denominator to leading order for the second term of the indent:
$$\int_0^\pi\frac{\sinh^2\pi}{\sinh^2 \epsilon e^{-it}}i\epsilon e^{-it}dt=-\frac{2\sinh^2\pi}{\epsilon}+o(\epsilon)=\sinh^2\pi\left(\int_{-\infty}^{-\epsilon}+\int^{\infty}_{\epsilon}\right)\frac{dx}{x^2}+o(\epsilon)$$
and we note that it exactly cancels out the divergent part of the PV integral. The first term of the indent vanishes as $\epsilon$ becomes small:
$$\int_0^\pi\frac{(\cosh^2\pi+\sinh^2\pi)\sin^2(\epsilon e^{-it})}{\sinh^2 \epsilon e^{-it}}i\epsilon e^{-it}dt=2(\cosh^2\pi+\sinh^2\pi)\epsilon+o(\epsilon ^3)$$
Finally, the third term tends to a constant:
$$\int_0^\pi\frac{i\cosh\pi\sinh\pi\sin(2\epsilon e^{-it})}{\sinh^2 \epsilon e^{-it}}i\epsilon e^{-it}dt=-2\pi\cosh\pi\sinh\pi+o(\epsilon^2)$$
Collecting everything we can finally evaluate the limit and rewrite the equation as
$$-2\sinh^2\pi \int_{-\infty}^\infty\frac{\sin^2x}{\sinh^2 x}dx+\sinh^2\pi\int_{-\infty}^\infty dx\left(\frac{1}{\sinh^2 x}-\frac{1}{x^2}\right)+2\pi
\cosh\pi\sinh\pi=0$$
Evaluating the integral by using the antiderivative $(\coth x-1/x)'=1/x^2-1/\sinh^2 x$ we have that
$$\int_{-\infty}^\infty dx\left(\frac{1}{\sinh^2 x}-\frac{1}{x^2}\right)=-2$$ and with this we finally recover the result stated above.
Best Answer
$\sinh x=\dfrac {x^1}{1!}+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\cdots$,
so asymptotically $\sinh x\sim x$ as $x\to0$ in $\mathbb R$ or $\mathbb C$.