# Thread: The complex gamma function

1. ## The complex gamma function

Hi Havent even got a clue where to start with this one:

If you can answer my previous 3 questions posted plus this one, i will be eternally grateful and quite possible stand a chance at passing my maths degree.

Thanks again

the moolmiester

2. Here is the second one. We need to use the residue theorem here. Look at picture below. The red dots are the poles.

Theorem: Let $n$ be a non-negative integer then $\mbox{res}(\Gamma(z),-n) = \frac{(-1)^n}{n!}$.

Thus, by the residue theorem we have,
$2\pi i \left( \frac{1}{2!} - \frac{1}{1!}+\frac{1}{0!} \right) = \pi i$

3. Originally Posted by moolimanj
Hi Havent even got a clue where to start with this one:

If you can answer my previous 3 questions posted plus this one, i will be eternally grateful and quite possible stand a chance at passing my maths degree.

Thanks again

the moolmiester
The first is:

$\frac{\Gamma(2+2i)}{\Gamma(-1+2i)}$

Now to do this you need a basic recurrence for $\Gamma(z)$, this is:

$\Gamma(1+z)=z\Gamma(z)$.

Using this we have:

$\Gamma(2+2i)= (1+2i)(2i)(-1+2i) \Gamma(-1+2i)$,

so:

$
\frac{\Gamma(2+2i)}{\Gamma(-1+2i)}= (1+2i)(2i)(-1+2i)
$

RonL