# A question implying Ganma function

1. Aug 7, 2005

### eljose

A question implying Ganma function....

Let define the function

$$Z(x)=\Gamma(x)2^{1-x}\pi^{-x}cos(x\pi/2)$$

then my question is if [Z(x)]=1=[Z(1-x)] with [z]=a^2+b^^2 then Re(x)=Re(1-x) with []=modulus of the function.

2. Aug 7, 2005

### Hurkyl

Staff Emeritus
Why would it be?

3. Aug 7, 2005

### shmoe

I don't know why you insist on using bizarre non-standard notation. What you call $$Z(x)$$ is what's usually called $$\chi(1-x)$$, though it's usually given as $$\chi(s)=2^s\pi^{s-1}\Gamma(1-s)\sin{s\pi/2}$$ as "x" is almost never used to denote a complex variable. And why not use the usual $$|s|$$ to denote the modulus of a complex variable? I swear you're trying to avoid widespread conventions on purpose.

With these notation changes, you seem to be asking "if $$|\chi(s)|=|\chi(1-s)|=1$$ then must we have the real part of s equal to 1/2?"

The answer is no. $$\chi(-18)=0$$ and $\chi(-19)= \frac{1856156927625}{8\pi^{20}}=26.45...>1$ so we have a real number $$-19<\alpha<-18$$ where $$\chi(\alpha)=1$$ ($$\chi$$ is continuous on (-19,-18)). Since $$\chi(s)\chi(1-s)=1$$ we necessarily have $$\chi(1-\alpha)=1$$ so your conditions are satisfied, yet $$\alpha\neq 1/2$$. If you examine the behavior of $$\chi(s)$$ as s goes to negative infinity along the real axis you can show this sort of thing happens infinitely often.

The converse is true though, if the real part of s=1/2 then $$|\chi(s)|=|\chi(1-s)|=1$$. Apply the reflection principle and use $$\chi(s)\chi(1-s)=1$$.