Derive for $z=x+iy$: $|\Gamma(z)| = \Gamma(x) \Pi_{k=0}^\infty \left[1+\frac{y}{(x+k)^2}\right]^{-1/2}$

Question

What I do know is that $\frac{1}{\Gamma(z)} = ze^{\gamma z} \Pi_{k=1}^\infty \left(1+\frac{z}{k}\right) e^{-z/k}$. But I'm not sure if that could be used here, or if partial fractions could be used.

Answer 1

$$\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod_{k=1}^\infty\frac{e^{\frac{z}{k}}}{(1+\frac{z}{k})}=\frac{e^{-\gamma (x+iy)}}{x+iy}\prod_{k=1}^\infty\frac{e^{\frac{x+iy}{k}}}{(1+\frac{x+iy}{k})}=\frac{e^{-\gamma x}}{x+iy}e^{-\gamma iy}\prod_{k=1}^\infty e^{\frac{x}{k}}(1+\frac{x}{k}+\frac{iy}{k})^{-1}e^{\frac{iy}{k}}$$

Taking absolute value: $$|\Gamma(z)|=\frac{|e^{-\gamma x}|}{|x+iy|}|e^{-\gamma iy}|\prod_{k=1}^\infty |e^{\frac{x}{k}}||(1+\frac{x}{k}+\frac{iy}{k})^{-1}||e^{\frac{iy}{k}}|= \frac{e^{-\gamma x}}{\sqrt{x^2+y^2}}\prod_{k=1}^\infty e^{\frac{x}{k}}((1+\frac{x}{k})^2+(\frac{y}{k})^2)^{-\frac{1}{2}}=\frac{e^{-\gamma x}}{\sqrt{x^2+y^2}}\prod_{k=1}^\infty \frac{e^{\frac{x}{k}}}{\sqrt{(1+2\frac{x}{k}+(\frac{x}{k})^2+(\frac{y}{k})^2)}}$$

$$|\Gamma(z)|=\frac{e^{-\gamma x}}{\sqrt{x^2+y^2}}\prod_{k=1}^\infty \frac{e^{\frac{x}{k}}k^2}{\sqrt{(k^2+2xk+x^2+y^2)}}=\frac{e^{-\gamma x}}{\sqrt{x^2+y^2}}\prod_{k=1}^\infty \frac{e^{\frac{x}{k}}k}{\sqrt{(x+k)^2+y^2}}$$ Could you take it from here?