Riemann surface

This figure is piling squares which are prime numbers.


The pattern includes complex plane, so I think that I can describe Riemann surface.



θ=2nπ　(n=0,±1,±2........) and arg Z=θ+2nπ (-π＜θ≦π).
Therefore, arg Z change according to n.

n=0 →　-π＜arg Z≦π
n=1 →　π＜arg Z≦3π
n=5 →　9π＜arg Z≦11π
n=-5 → -11π＜arg Z≦-9π

∴(2n-1)π＜arg Z≦(2n+1)π

Complex planes Z depend on n, and these are piled infinitely.

－Π＜θ≦Π

I define how to expand circles in complex plane.




Then, this is apparent.

(n=0,±1,±2.......)

Therefore



－Π＜θ≦Π

Z keep circling because of Z=X+iY

Maclaurin's expansion

It is impossible to find all prime numbers, so Maclaurin's expansion is a good example to show convergence.

If x had a certain value, this mathematical formula would be convergence.
Then, you see complex number Z.



This complex plane don't include point at infinity, so it must be absolute convergence.



Z=X0 means expansion of the circle, and X0 is a radius of each circles.

Absolute Convergence

We tend to think that infinity must be the realm of philosophy. Our body can’t stand eternity. It is just the ideal. Riemann’s zeta function must be almost correct, but nobody can prove this difficulty because of infinity. You can see the goal, but it moves forward with you. You never reach it.
Therefore, we need to know absolute convergence to ignore infinity, but it includes complex plane.



In order to prove absolute convergence, you should accept Z=Z0 and |Z-a|＜|Z0-a|.
M must be very huge values.


(n=0.1,2....)







Then you see that |Z0-a| is the radius of the circle.
∴ |Z-a|＜|Z0-a|

This means expansion because of Σ, but it must be absolute convergence.