# New tools for complex analysis and application to the Riemann Zeta function?

I've worked as a graphic artist for the past fifteen years, thus I have no relationship with the academic mathematical community. It is therefore difficult for me to check some results.

1. Tools for complex analysis
Let $\mathit{z}=x+iy$ be a complex number.
I've developed a set of tools which allows to separate the components of any complex function $f(z)$ (i.e: Abs,Re,Im,Arg,etc....) as real values functions depending only of $x$ and $y$.
These tools work equally well for the sum or multiplication as the composition of functions (i.e: $g(f(z))$).
It also allows, among others, to derive or integrate a formula on the x or y axes separately, or both.

2. A concrete example: the Riemann Zeta function
As an example, using these tools and the product definition of the zeta function $\zeta (\mathit{z})=\prod _{k=1}^{\infty } \frac{1}{1-p_k{}^{-\mathit{z}}}$, it is possible to derive the following definition:

Let $\mathit{z}=x+iy$ be a complex number and $p_k{}$ the $k^{nth}$ prime number, then

$$|\zeta (\mathit{z})|=\prod _{k=1}^{\infty } \sqrt{\frac{p_k{}^{2x}}{p_k{}^{2x}-2p_k{}^x\cos \left(y \log \left(p_k\right)\right)+1}}$$ With this definition, we can easily see that there is no particular $z$ satisfying the Riemann hypothesis. However, as it is certainly true, it probably appears as the limit of the multiplication of the denominator terms.

Using the tools, here are the definitions of $\mathfrak{R}\mathfrak{e}( \zeta (\mathit{z}) )$ and $\mathfrak{I}\mathfrak{m}( \zeta (\mathit{z}) )$: $$\mathfrak{R}\mathfrak{e}( \zeta (\mathit{z}) )=\sum _{k=1}^{\infty } \frac{1}{k^x}\cos (y \log (k))$$ $$\mathfrak{I}\mathfrak{m}( \zeta (\mathit{z}) )=-\sum _{k=1}^{\infty } \frac{1}{k^x}\sin (y \log (k))$$

• Note that $\zeta (\mathit{z})=\prod _{k=1}^{\infty } \frac{1}{1-p_k{}^{-\mathit{z}}}$ is valid only for $\text{Re}(z)>1$, so you only show that the zeta function has no zero in that region. – GEdgar Jul 30 '13 at 11:52
• Yes, you're right!... Thanks. – Eddy Khemiri Jul 30 '13 at 15:27
• the set of tools are described in this post: simple tools to extract... I hope someone can answer clearly to the question, as so far it's only commented. – Eddy Khemiri Dec 23 '13 at 10:16

$$\zeta(x+iy)=\sum_{k=1}^{\infty}\frac{1}{k^{x+iy}}=\sum_{k=1}^{\infty}\frac{1}{k^{x}}\frac{1}{k^{iy}}$$ But notice also that $$\frac{1}{k^{iy}}=k^{-iy}=e^{-iy\log{k}}=\cos(y\log{k})-i\sin(y\log{k})$$ Which yields the result you gave immediately.

• Typo: An "i" is missing before the "sin". – John Bentin Jul 30 '13 at 10:25

$$\left|\zeta(z)\right|=\prod _{k=1}^{\infty}\left|\frac{p_k^{z}}{p_k^{z}-1}\right|\\=\prod _{k=1}^{\infty}p_k^x\left|\frac{e^{iy\log(p_k)}}{p_k^xe^{iy\log(p_k)}-1}\right|\\=\prod _{k=1}^{\infty}p_k^x\left|\frac{1}{p_k^x\cos(y\log(p_k))+ip_k^x\sin(y\log(p_k))-1}\right|\\=\prod _{k=1}^{\infty}p_k^x\sqrt\frac{1}{(p_k^x\cos(y\log(p_k))-1)^2+(p_k^x\sin(y\log(p_k)))^2}\\=\prod _{k=1}^{\infty}p_k^x\sqrt\frac{1}{p_k^{2x}- 2p_k^x\cos(y\log(p_k))+1}$$

Unfortunately, I don't see how the Riemann hypothesis follows from here.

• Ian, I proceed in a different way, but I don't understand how do you go from the third to the fourth line... – Eddy Khemiri Jul 30 '13 at 19:49
• about the Riemann hypothesis, as @GEdgard wrote, I missed the fact that $Re(z)$ must be $>1$... – Eddy Khemiri Jul 30 '13 at 19:55
• @Eddy First, I used $|x+iy|=\sqrt{x^2+y^2}.$ There are some properties about modulus, and I used $|1/z|=1/|z|$ after that. How did you proceed? – Ian Mateus Jul 30 '13 at 20:17
• sure! well done, Ian. But the way I proceed refers to my first question: are there general tools working with any multiplication or composition of functions. for example, could you give me a general formula in complex analysis that would give you, let's say as an example, Re(Log(1+z)∗Zeta(z)) as a real valued function? I do have this tool, but i'm currently working on it and... he took me 2 years to "see" it... I would really want to know if the tool already exists in complex analysis! – Eddy Khemiri Jul 31 '13 at 18:19
• @Eddy I'll evaluate $\Re\left(\zeta(z)\log(w)\right)$ to keep things simple. I'll denote $\zeta(z)=\zeta$, $\log(w)=L$, real part in $\color{red}{\text{red}}$ and imaginary part in $\color{blue}{\text{blue}}.$ First, $$\Re\left(\zeta L\right)=\Re\left(\left(\color{Red}{\zeta} +i\color{Blue}{\zeta}\right)\left(\color{Red}{L} +i\color{Blue}{L}\right)\right) =\Re\left(\color{Red}{\zeta}\color{Red}{L} -\color{Blue}{\zeta}\color{Blue}{L} +i(\color{Blue}{\zeta}\color{Red}{L} +\color{Blue}{L}\color{Red}{\zeta})\right)\\ =\color{Red}{\zeta}\color{Red}{L} -\color{Blue}{\zeta}\color{Blue}{L}.$$ – Ian Mateus Jul 31 '13 at 22:11