Need help showing Riemann's Functional equation for negative numbers and complex numbers

Riemann's Functional equation: $$\zeta(-z)$$=$${-2*z!\over(2\pi)^{z+1}}sin({\pi z\over2})\zeta(z+1)$$

This formulas expresses $$\zeta(-z)$$ in terms of $$\zeta(z+1)$$
Note: I read that the author said, suppose $$z=x+iy$$ is a complex number in the right half plane, meaning $$x \ge 0$$ then $$-z$$ is in the left half plane. Since $$z+1$$ is in the right half plane, Riemann's functional equation allows us to indirectly define $$\zeta(-z)$$. For example, if we take $$z=2+3i$$, the point $$z+1=3+3i$$ is located in the half plane where $$x \ge 1$$ but we already know how to compute $$\zeta(z+1)$$ using the zeta series and a computer and it will show that $$\zeta(3+3i)\approx 0.94+0.008i$$

My first question is how do I compute $$\zeta(z+1)$$ using the zeta series or a computer for fact checking?
My 2nd question is how do I compute Riemann's functional equation when $$z$$ is equal to a negative number because I can solve this when $$z \ge 0$$

For example, when $$z=2$$ in the Riemann functional equation we have:
$$\zeta(-2)$$=$${-2*2!\over(2\pi)^3}sin({2\pi \over2})\zeta(3)$$
Here we have $$sin\pi=0$$ which indicates that $$\zeta(-2)=0$$
What if we had $$z=-2$$, how do I solve this?
Can someone show me in detail when $$z=-2$$ because the equation would now look like:

$$\zeta(2)$$=$${(-2)(-2!)\over(2\pi)}sin({-2\pi \over2})\zeta(-1)$$

• The functional equation is $\zeta(s) = \chi(s) \zeta(1-s)$ where $\chi(s) = \pi^{s-1} 2^s \Gamma(1-s) \sin(\pi s/2)$. This is an equality of meromorphic functions. From the zeros-poles locations of $\Gamma(1-s)$ you know those of $\chi(s)$. It is always true that $\zeta(s) = \lim_{z \to s}\chi(z) \zeta(1-z)$ and $\zeta(1-s) = \lim_{z \to s}\frac{1}{\chi(z)} \zeta(z)$ and depending if $\chi$ has no pole or zero at $s$ then you can remove the $\lim_{z \to s}$ and use direct evaluation $\zeta(s) = \chi(s) \zeta(1-s),\zeta(1-s) = \frac{1}{\chi(s)} \zeta(s)$ – reuns Jan 23 at 0:10
• $\chi(s)$ has a zero at $s=-2$ so $\zeta(-2) = \chi(-2) \zeta(3) = 0$. And $\frac{1}{\chi(s)}$ has a pole at $s=-2$ so $\zeta(3) = \lim_{s \to -2}\zeta(1-s)=\lim_{s \to -2} \frac{\zeta(s)}{\chi(s)}$, as the pole is simple the latter $=\frac{\zeta'(-2)}{\chi'(-2)}$ whose only the denominator has a closed-form – reuns Jan 23 at 0:16

Your problem boils down to finding $$\zeta(2k)$$. In fact, it can be shown that $$\zeta(2k)=(-1)^{k+1}\frac{B_{2k} 2^{2k-1}\pi^{2k}}{(2k)!}$$ Here are some proofs. No known values of the zeta function exist except for the even positive integers and negative integers.
EDIT: $$\zeta(z)$$ is equal to $$0$$ at the notorious non-trivial zeroes as well as negative even integers.
• okay. I was just fact checking for all real numbers(negative numbers included) in the functional equation for $z$ – DDP Jan 22 at 22:45