In the book Asymptotics and Mellin-Barnes integrals by Paris and Kaminski (2001), the following inequality is given for the ratio of two gamma functions (Section 2.1.3, page 34) :
For $b-a\ge 1, a\ge 0$ and $z=x+iy$ with $x>0$, $$\left|\frac{\Gamma(z+a)}{\Gamma(z+b)}\right|\le \frac{1}{|z|^{b-a}}.\tag1 $$
At the end of the proof, the authors state that
A similar procedure can be used to deal with negative ranges of $b − a$.
However it is not clear to me from looking at their proof what the conclusion would be in the case $b-a\le -1$ (my complex analysis knowledge is basically non-existent), and I want to know, is the following inequality true ? (or something "similar", like maybe an absolute value on $b-a$) :
For $b-a\le -1, a\ge 0$ and $z=x+iy$ with $x>0$, $$\left|\frac{\Gamma(z+a)}{\Gamma(z+b)}\right|\le \frac{1}{|z|^{b-a}}.\tag2 $$
UPDATE : In case anyone else is interested/curious, I misunderstood the claim given in the book and inequality $(2)$ is wrong.
What the authors (essentially) said is that if $-n\le b-a < 1-n$ for some integer $n>0$, then by applying inequality $(1)$ with $a\equiv a$, $b\equiv b+n+1$ and applying $n+1$ times the property $\Gamma(\alpha + 1) = \alpha\Gamma(\alpha)$, we get the following, correct inequality :
For $n\in\mathbb N^*$, $-n\le b-a< 1-n, a\ge 0$ and $z=x+iy$ with $x>0$, $$\left|\frac{\Gamma(z+a)}{\Gamma(z+b)}\right|\le \frac{1}{|z|^{b-a}}\cdot\left|\prod_{k=0}^n \frac{(z+b+k)}{z}\right|. $$
Thanks to @Conrad for the help !
Actually what the authors say is something different than what is claimed in the OP - first they get the bound $2.1.17$ for the case $0<b-a<1$ by using $\frac{\Gamma(z+a)}{\Gamma(z+b)}=\frac{(z+b)\Gamma(z+a)}{\Gamma(z+b+1)}$ and using the previous bound for $\frac{\Gamma(z+a)}{\Gamma(z+b+1)}$ since now $b+1 -a>1$ (hence $|\frac{\Gamma(z+a)}{\Gamma(z+b)}| \le \frac{1+b/|z|}{|z|^{b-a}}$ when $0<b-a<1$) and then they say that a similar procedure can be used to deal with $b-a<0$ which means that you keep adding positive integers and use the $\Gamma$ property; so for example if $-1<b-a<0$ one uses $\Gamma(z+b+2)=(z+b+1)(z+b)\Gamma(z+b)$ and the previous bound for $\frac{\Gamma(z+a)}{\Gamma(z+b+2)}$ etc