# I've animated Riemann zeta function and got a spiral. Why?

So I've obtained this peculiar visualization of Riemann zeta function:

The way I do this: I treat every $nth$ animation frame as a complex plane and plot all values of $\zeta(s)$ for all $s$ such that

$Im(s) = 0.1n - 1$

$-32 \leq Re(s) \leq 8$ (arbitrary numbers: beyond $8$ the values do not stray much from $1 + 0i$ and for reals less than $-32$ the values land outside of the frame)

The white cross marks the point obtained by calculating the zeta for $1/2 + (0.1n - 1)i$ and similarly the pink crosses mark the values obtained for boundaries of the critical strip.

Assuming Riemann Hypothesis, the green curve can only ever intersect the red cross in position of the white cross.

I've also recorded a much, much more detailed and longer visualization here:

To add to this, the spiral looks like it lies flat on the $XZ$ plane for $Im(s) = 0$ and it quickly rotates around $OX$ axis for $Im(s)$ between $0$ and $1$, to be nearly parallel with the $XY$ plane later. The trivial zeroes of Riemann zeta would therefore be intersections of this spiral's branches with the origin, I guess?
• For the green spiral, you'll get the same with any analytic function $f : \mathbb{C} \to \mathbb{C}$ whose argument is not bounded and $\lim_{Re(s) \to +\infty} f(s) = 1$ (so that $\ln f(s)$ is analytic on some half plane $Re(s) > g(Im(s))$) Note that for $Re(s) > \sigma_2 \approx 1.8$ : $\zeta(\sigma_2) = 2$ and $|\zeta(s)-1| = |\sum_{n=2}^\infty n^{-s}| < \sum_{n=2}^\infty n^{-\sigma_2}= 1$ so you have $Re(\zeta(s)) > 0$, and hence $\text{arg } \zeta(s) \in (-\pi,\pi)$. – reuns Sep 4 '16 at 15:54
• And the weird shape at the beginning is due to the pole of $\zeta(s)$ at $s=1$ and the zeros at $s = -2k$ – reuns Sep 4 '16 at 15:59