Bound on Complex Values of Gamma Function

Theorem

 * Let $\Gamma \left({z}\right)$ denote the Gamma function.


 * Then for any complex number number $z=s+it$, we have for $|b|\leq|t|$


 * $ |\Gamma \left({s+it}\right)| \leq \dfrac{|s+ib|}{|s+it|}|\Gamma \left({s+ib}\right)| $

Proof
From the Euler Form of the Gamma Function, we have that:

Because $|b|\leq|t|$, we have that:
 * $ b^2 \leq t^2 $
 * $ \left(\dfrac b n \right)^2 \leq \left(\dfrac t n \right)^2 $
 * $ \left(1 +\dfrac s n \right)^2 + \left(\dfrac b n \right)^2 \leq \left(1 +\dfrac s n \right)^2 + \left(\dfrac t n \right)^2 $


 * $ \dfrac{1}{\left\vert 1 + \frac{s+ib}{n}\right\vert} \geq \dfrac{1}{\left\vert 1 + \frac{s+it}{n}\right\vert}$

Using this we obtain: