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

Question

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.

Do these tools already exist?

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 $\mathfrak{I}\mathfrak{m}( \zeta (\mathit{z}) )$: $$ \mathfrak{R}\mathfrak{e}( \zeta (\mathit{z}) )=\sum _{k=1}^{\infty } \frac{1}{k^x}\cos (y \log (k)) $$ $$ \mathfrak{I}\mathfrak{m}( \zeta (\mathit{z}) )=-\sum _{k=1}^{\infty } \frac{1}{k^x}\sin (y \log (k)) $$

Do these definitions already exist?

thank you in advance, Eddy.

Answer 1

$$\zeta(x+iy)=\sum_{k=1}^{\infty}\frac{1}{k^{x+iy}}=\sum_{k=1}^{\infty}\frac{1}{k^{x}}\frac{1}{k^{iy}}$$ 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.

Answer 2

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

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