$\zeta (1/2 + i) = 0$, correct?

Question

Plugging $\zeta (1/2 + i)$ into Wolfram Alpha yields me some complex number, but I was under the understanding that $\zeta (1/2 + it) = 0$, for all $t$ we have yet calculated...

Is Wolfram Alpha just messing with me?

Thanks!

Answer 1

No! Every $s$ such that $\zeta(s)=0$ is assumed to be of the form $1/2 + i t$ yet not every $1/2 + i t$ is a root of $\zeta$.

Recall for example that $\zeta$ is a meromorphic function and as such can only have countably many roots.

Answer 2

Wolfram alpha is correct. Note this for only some $t\in \mathbb{R}$ we have $\zeta(1/2+it)=0$ which are the non trivial zeroes.