I was looking at a contour integration where the claim had been made that the following function $1/\sinh^2 z $ goes to zero along the following lines in complex plane $(-\infty, 0)$ to $(-\infty, i\pi)$ and $(\infty, 0)$ to $(\infty, i\pi)$ - both lines running parallel to the imaginary axis.
Now with $z= x + iy$, $1/\sinh^2{z}$ becomes $$\frac{1}{(\sinh^2x \cos^2y -\cosh^2x\sin^2y -2i\sinh x \cosh x \sin y \cos y)}$$
It's not obvious to me that this goes to $0$ when $x\rightarrow \pm \infty$, because of the opposing signs of the first two terms. It would be helpful if someone could clarify. Thanks.
We have $$ \lvert \sinh{(x+iy)} \rvert^2 = \lvert \sinh{x}\cos{y}+i\cosh{x}\sin{y} \rvert^2 = \sinh^2{x}\cos^2{y}+\cosh^2{x}\sin^2{y}, $$ using that $\cosh{iy}=\cos{y}$ and so on, which can be written as $$ \sinh^2{x}\cos^2{y}+\cosh^2{x}\sin^2{y} = \sinh^2{x}(1-\sin^2{y})+(1+\sinh^2{x})\sin^2{y}\\ = \sinh^2{x}+\sin^2{y} $$ using the Pythagorean identities. This tends to $\infty$ as $x \to \pm\infty$, so $1/$ it will tend to zero.