zeros of Riemann zeta function. My question is, is $z = 0$ is zero of Riemann zeta function? 
   by putting z $= 0$ in Riemann functional equation 
$$\zeta(z) = [2(2\pi)^{z-1}]\times\zeta(1-z)\times\Gamma(1-z)\times\sin(\frac{\pi z}{2})$$ one can see that sin terms get $0$ and hence R.H.S. becomes zero. so is zeta function has zero at $z = 0$ ?
 A: Method 1: From the formula obtained for even numbers:
\begin{equation}
\zeta(2k)= \frac{(2\pi)^{2k}(-1)^{k+1}B_{2k}}{2(2k)!}
\end{equation}
If we take $k=0$, we get:
\begin{equation}
\zeta(0)= \frac{-B_{0}}{2(0)!}=-\frac{B_0}{2}
\end{equation}
Finally as $B_0=1$
\begin{equation}
\boxed{\zeta(0)=-\frac{1}{2}}
\end{equation}
Method 2: Also, from the formula obtained for negative numbers:
\begin{equation}
\zeta(-k)=(-1)^k\frac{B_{k+1}}{k+1}
\end{equation}
If we take $k=0$, we get:
\begin{equation}
\zeta(0)= B_{1}
\end{equation}
Since $B_{1}=-\frac{1}{2}$, we also obtain:
\begin{equation}
\boxed{\zeta(0)=-\frac{1}{2}}
\end{equation}
Method 3: Also if we take the limit in zero of the functional relation:
\begin{eqnarray*}
\zeta(0)&=&\lim_{s\to 0} \pi^{s-\frac{1}{2}}\frac{\Gamma\left(\frac{1-s}{2}\right)}{\Gamma\left(\frac{s}{2}\right)}\zeta(1-s)\\
&=&\lim_{s\to 0} \pi^{-\frac{1}{2}}\frac{\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{s}{2}\right)}\zeta(1-s)\\
&=&\lim_{s\to 0}\frac{\zeta(1-s)}{\Gamma\left(\frac{s}{2}\right)}\\
\end{eqnarray*}
But $\zeta(s)\approx \frac{1}{s-1}$ lorsque $s\to 1$ (voir \ref{eq_zeta1}) and $\Gamma(s)\approx \frac{1}{s}$ lorsque $s\to 0$ (voir \ref{eq_gamma_zero}). Donc
\begin{eqnarray*}
\zeta(0)&=& \lim_{s\to 0}\frac{\frac{1}{(1-s)-1}}{\frac{2}{s}}\\
&=& -\frac{1}{2}\\
\end{eqnarray*}
We still get:
\begin{equation}
\boxed{\zeta(0)=-\frac{1}{2}}
\end{equation}
