Show $\Gamma(z)y^{-z}$ vanishes as $z\to\pm i\infty$

Question

I need to show that $\Gamma(z)y^{-z}$ vanishes as $z\rightarrow\pm i\infty$ if $|\arg(y)|< \frac{\pi}{2}$. Note that $z$ is purely imaginary.

Currently my approach is to use Stirling's Formula to get a representation for this product for large $|z|$. This gives $\sqrt{2\pi}z^{z}e^{-z}y^{-z}$.

I am struggling to make an observation which would cause this this product to vanish for large imaginary $z$. I tried rewriting this as $$\sqrt{2\pi}e^{z \log(z)}e^{-z}e^{-z\log(y)} = \sqrt{2\pi}e^{z\log(z)- z - z\log(y)} = \sqrt{2\pi}e^{z(\log(z) - \log(y) -1)}$$

Am I on the right track here?

Answer 1

We have $\Gamma(z)y^{-z}=\Gamma(ix)e^{-ix\operatorname{Log}y}$ and $\operatorname{Log}y=\log|y|+i\arg y$ so $$|\Gamma(z)y^{-z}|=|\Gamma(ix)||e^{x\arg y}|<\sqrt{\frac{\pi e^{\pi x}}{x\sinh\pi x}}\to0$$ as $|x|\to\infty$ using Euler's reflection formula.