Fixed point of Riemann Zeta function

Question

I have been looking for fixed points of Riemann Zeta function and find something very interesting, it has two fixed points in $\mathbb{C}\setminus\{1\}$.

The first fixed point is in the Right half plane viz. $\{z\in\mathbb{C}:Re(z)>1\}$ and it lies precisely in the real axis (Value is : $1.83377$ approx.).

Question: I want to show that Zeta function has no other fixed points in the right half complex plane excluding the real axis, $D=\{z\in\mathbb{C}:Im(z)\ne 0,Re(z)>1\}$.

Tried: In $D$ the Zeta function is defined as, $\displaystyle\zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s}$. If possible let it has a fixed point say $z=a+ib\in D$. Then, $$\zeta(z)=z\\ \implies\sum_{n=1}^\infty\frac{1}{n^z}=z\\ \implies \sum_{n=1}^\infty e^{-z\log n}=z\\ \implies \sum_{n=1}^\infty e^{-(a+ib)\log n}=a+ib$$ Equating real and imaginary part we get, $$\sum_{n=1}^\infty e^{-a\log n}\cos(b\log n)=a...(1) \\ \sum_{n=1}^\infty e^{-a\log n}\sin(b\log n)=-b...(2)$$ Where $b\ne 0, a>1$.

Problem: How am I suppose to show that the relation (2) will NOT hold at any cost?

Any hint/answer/link/research paper/note will be highly appreciated. Thanks in advance.

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Answer 1

I do not think your statement about fixed points in the plane to be true - it may be true for $Re(z)>1$ in the sense of being just one fixed point there, but otherwise $(s-1)\zeta(s)$ is an entire function of order 1 and maximal type (by the usual properties of the critical strip zeros - eg their ~$T\log(T)$ density and general stuff about entire functions of finite order - the usual notion of density of zeros for entire functions and the one for $\zeta$ differ a little but they have the same order of magnitude) and subtracting a polynomial like $s(s-1)$ doesn't change order 1 or maximal type as those depend on the Taylor coefficients at infinity for any entire function, so in particular $(s-1)\zeta(s) - s(s-1)$ is entire of order 1 and maximal type and those have lots of zeros - either they have density growing faster than T at infinity or the conditional sum of their reciprocals is not convergent by a theorem of Lindelof. Maximal type is crucial because obviously exponentials of linear polynomials have order 1 and arbitrary finite type.

Note that the reciprocal of the Gamma function is order 1 and maximal type but has the density of zeros ~T (say on the disk of radius T centered at the origin) as its zeros are just the negative numbers (so, in particular, the conditional sum of their reciprocals is not convergent, so it's possible the number of fixed points of $\zeta$ to be of order T only sure; similar considerations apply to any equation of the type $\zeta(s)=Polynomial(s)$ by multiplying with s-1 and reducing to considerations about entire functions of order 1 and maximal type.

Answer 2

Hmm...i did a run in my computer cause i found you question of fixed points interesting so..

the only result i got is this for $a=1.8337719154395\cdots$ and for $b=0$

$\zeta(1.8337719154395\cdots)=1.8337719154395\cdots$

note: this is an engineer approach i'm no mathematician

I made this for you https://www.desmos.com/calculator/hoyjifa8wn

second version added (3/5/2021): https://www.desmos.com/calculator/t52rv9aevc

also the code for the real part when a>1 and b=0 for ζ(a)=a or in the other program I made was a trick of input and output so if ζ(χ0)=χ1 then repeat this rule ζ((x0+x1)/2)=x2 and again put the output as input this way Ζ((((χ0+χ1)/2)+χ2)/2)=x3 on and on n times the more the better because it minimizes the ''error'' to have ζ(s)=s , s ≈ 1.83377265168027...= a to be more accurate , you have to find a mathematical theorem that does that for you tho.

cant find the code I had made but if i remember correctly it was like this https://i.sstatic.net/0ZLud.jpg

for starting condition of a=1 and b=1 , using the above rule ζ(a+bi) =z ≈ -0.295905005575214 gives this solution but it turns out again b=0 after many n times

for starting conditions of a=2 and b=1 ζ(z) =z ≈1.83377265168027 it makes b=0 after many n times again.

After watching it many times with other numbers it shows that introducing the b≠0 makes the system rotate anyway from the initial conditions so there is no way to have an input with a>1 and b≠0 to have it stay somewhere close with fixed points, trying to minimize the error between input and output makes the b=0 so it will have θ=0 , z ≈1.83377265168027 and θ=180 degrees , z ≈ -0.295905005575214 there are also infinite solutions of this when b=0 and all are a straight line on the x-axis as long as a>1

Conclusions i made is that if you are going to try to find fixed points with z=a+bi and a>1, b=0 then it has infinite solutions and are all real numbers on the x-axis but, if b≠0 then ζ(z)≠z due to rotation in the system, so we have ζ(z)=w another coplex number and w-z≠0 its also a complex number with real and imaginary part, so using it in the denominator of a fraction like this 1/(ζ(z)-z) is no problem as long as b≠0 and z is not real in general!!!