$\zeta(2 + it) = \zeta(2-it)$

Question

Let $\zeta(s)$ denote the Riemann zeta-function. Show that $\zeta(2 + it) = \zeta(2-it)$ for all real t.

Give some hints how to do this one.Thanks in advance.

Answer 1

The problem should read "Show that $$ \overline{\zeta(2+it)}=\zeta(2-it)."$$ The 2 is arbitrary; it will hold for real part of $s$ any $\sigma>1$. Complex conjugation is continuous, so you may pass it through the infinite series defining the zeta function. (In fact this holds for all $\sigma$ once we have an analytic continuation.)

Answer 2

As others have pointed out, the claim is not correct. These two numbers are in fact complex conjugates.

If $f$ is any meromorphic function which takes real values on the real line, then $\overline{f(z)} = f(\overline{z})$ for all $z$. This follows from the Schwartz reflection principle and from the identity theorem.

Answer 3

Only the real part of $\zeta(2 + it)$ is symmetric over $t=0$

For a visual, see this plot where $-2 < t < 2$

Like the others have said, something is missing, whether it be some kind of conjugation, a re(), or a modulus.