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.

Please visit this.

  • $\begingroup$ You can use the Euler product to show $\zeta(1+it) = O(\log t)$. Also from the functional equation and the Phragmén–Lindelöf principle there is the growth $\mu(\sigma) = \max(0,1/2-\sigma)$ function satisfying $|\zeta(s)-\frac{1}{s-1}| \le C + |t|^{\mu(\sigma)+\epsilon}$ which implies $ \zeta(s)-s$ has finitely many zeros for $\Re(s) > -1$. You can use the argument principle to count how many. $\endgroup$ – reuns Feb 6 at 10:53
  • $\begingroup$ @reuns thanks for your response i will definitely try that. Still can you provide something that proves relation 2 leads us to some contradiction. $\endgroup$ – Sujit Bhattacharyya Feb 6 at 11:31

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.

  • $\begingroup$ By a sign change argument, given the $(k! (2\pi)^{-k})^{1\pm \epsilon}$ growth of $\zeta(-k) = (-1)^k\frac{B_{k+1}}{k+1}$, $\zeta(s)-s$ has a zero close to every large enough even negative integer $\endgroup$ – reuns Feb 7 at 17:45

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 $b=0$


wish you all the luck

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

  • $\begingroup$ To improve you answer, can you show us the code use? $\endgroup$ – Luis Felipe Jul 18 at 16:20
  • $\begingroup$ Thank you for your response to my question! Kindly provide your code so we can proceed further to find any pattern. $\endgroup$ – Sujit Bhattacharyya Jul 24 at 2:20

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