What is the Gamma Function for negative complex numbers? I know that:
$\Gamma(n) \quad=\quad (n-1)!\quad\forall
n\in\mathbb N$
And that:
$\Gamma(t) \quad=\quad \int_{0}^\infty {x^{t-1}\over e^x}dx,\quad\Re(t)>0$
During some further research, (please correct me if I'm wrong) I learned that the Gamma Function can be defined for a negative value in such manner:
$\Gamma(s)\quad=\quad  \Gamma(s+1)/s\quad\forall s\in\mathbb R^*$  
Therefore (e.g.):
$\Gamma(-{1\over 2})\quad= \quad\Gamma({1\over 2})/(-{1\over 2})\quad=\quad -2\sqrt\pi$
Does this method also work for complex values of s with Real part less than $0$?
Or is there another method?
 A: $\Gamma(z)$ is not going to be defined for all $z$ in $\mathbb{R}^*$. We can show this by doing the following manipulations, and in the process we can also show what values it will be defined for. 

For your first question:
We start with the integral definition of the gamma function:
$$\begin{align*}
 \Gamma(z)&=\int_0^{\infty}e^{-t}t^{z-1}dt \\ 
 &= \int_0^{1}e^{-t}t^{z-1}dt+\int_1^{\infty}e^{-t}t^{z-1}dt\\ 
 &= \int_0^{1}\sum_{n=0}^{\infty}\frac{(-t)^n}{n!} t^{z-1}dt+\int_1^{\infty}e^{-t}t^{z-1}dt\\ 
 &= \sum_{n=0}^{\infty}\int_0^{1}\frac{(-t)^n}{n!} t^{z-1}dt+\int_1^{\infty}e^{-t}t^{z-1}dt\\ 
 &= \sum_{n=0}^{\infty}\frac{(-1)^n}{n!(z+n)}+\int_1^{\infty}e^{-t}t^{z-1}dt\;\;\;\;\;\;(*)
\end{align*}$$
Note that the integral in line $(*)$ is defined $\forall z\in\mathbb{C}$, but the series runs over $n\in\left \{  \;0,\;1,\;2,\;3,\;\cdots\right \}$ and contains $(z+n)$ in the denominator. From this we can see that $\Gamma(z)$ is defined $\forall z\in\mathbb{C}\setminus\left \{  \;0,\;-1,\;-2,\;-3,\;\cdots\right \}$.
So we conclude that the statement $\quad\forall s\in\mathbb {R}^*,\;\;\Gamma(s)= \Gamma(s+1)/s$  is false, as $\Gamma(s)$ is undefined for some $s$ in $\mathbb {R}^*$.

For your second question:
The relation
$$\Gamma(z)=\frac{\Gamma(z+1)}{z}$$
will hold $\forall z\in\mathbb{C}\setminus\left \{  \;0,\;-1,\;-2,\;-3,\;\cdots\right \}$. We see this from the above section.
I am also assuming you are looking for a general way to create closed form representations of $\Gamma(z)$ for $z=iy-x$ where $x\in\mathbb{R}^+\setminus\mathbb{N}$ and $y\in\mathbb{R}$.
As far as I am aware there is no way to further simplify the following any more than shown:
$$\begin{align*}
 \Gamma(iy-x)&=\frac{\Gamma(iy-x+1)}{iy-x} \\ 
 &= \frac{\Gamma(iy-x+2)}{(iy-x)(iy-x+1)}\\ 
 &= \frac{\Gamma(iy-\left \{  x\right \})}{(iy-x)(iy-x+1)\cdots(iy-\left \{ x \right \}-1)}
\end{align*}$$
as there is no other 'nicer' way to represent $\Gamma(iy-\left \{  x\right \})$ where $\left \{  x\right \}$ is the fractional part of $x$. One could also argue that what is shown directly above isn't much of a simplification, but this does show that the relation holds for $z$ such that $\Re(z)<0$.
