Why does the imag. part of the graph of $\zeta(n^{ix})$ resemble the tangent function?

Question

If you input $\zeta(n^{ix})$ into the Wolfram Alpha search bar, in the plot, you get an infinitely repeating sinusoidal curve, which resembles the real part, and you get an infinitely repeating tangent curve, which resembles the imaginary part. Yet, plotting $n^{ix}$ alone does not give a tangent-like imaginary curve.

So, why does a tangent-like curve appear?

Note: the constant $n$ is any arbitrary real number.

Link for confirmation.

Answer 1

Two facts explain the qualitive picture from your link. First: $\zeta(z)\;$ has a pole at $z=1= e^{i\cdot 0}\;$ with $\zeta(e^{ix})=-\frac{i}{x} + \dots\;$ for $x\approx 0,\;$ second: $e^{ix}=e^{i(x+2\pi)},\;$ this gives the periodic structure. Further: Since $\zeta(-1)=-\frac{1}{12}\;$ you have $\Im \zeta(e^{i\pi})=0.$