The Riemann Zeta function is expressed in complex rectangular form generally.
When it is expressed in polar form as seen in plot of Riemann Zeta function $ \zeta( \frac12 +i \,t) $ along critical line for real $t$, at what $\theta$s do the lines in the line plots go through the origin?
Best Answer
If I get what you're saying, then you want $\zeta(r,\theta)$ instead of $\zeta(x,iy)$ (more commonly written $\zeta(x+iy)$, but you want the $\zeta$ function for the specific line in the complex plane represented as $\frac{1}{2}+iy$
If so, we can derive the equation $r=h csc(\theta-\phi)$, where $h$ is the shortest distance from the origin to the line, and $\phi$ is the arctangent of the slope of the line. Lucky for you, your line is vertical, so its pretty easy to work it out. So, the distance from the origin to your line is $\frac{1}{2}$ because the line is vertical and has no slope to approach the origin.
The slope is either $-\infty$ or $+\infty$, depending on what side of the origin it's on. We're dealing with Quadrant I, so it's on the right side of the origin, meaning it is $-\infty$ we want for this. This makes the linear equation in polar units $\frac{1}{2}csc(\theta+\frac{\pi}{2})$, which can be changed to a simpler $r=\frac{1}{2}sec(\theta)$
This makes our equation $\zeta(r,\theta)$ - but not quite. See, we have to change the equation slightly, so that it's equipped to handle polar units.
To do that, we take advantage of $(r,\theta) = (\sqrt{x^2+y^2},$ arctan$(\frac{y}{x}))$ and $(x,y)=(r$ cos $\theta,r$ sin $\theta)$. Placing that into our equation, we get $\zeta(r,\theta) = \sum \frac{1}{n^{rcos\theta+irsin\theta}}$
Since that's a little hard to see, I'll write it out phonetically. The sum fron $n=1$ to $\infty$ of $1$ over $n$ to the power of ${r cos(\theta) + isin(\theta)}$. That cosine and imaginary sine should ring some bells about complex exponentiation. When combining the linear equation from earlier and the new $\zeta$ function, and assuming the Riemann Hypothesis is true, then we can write the $\zeta$ function in a more specific way, and it's what Riemann used to draw a lot of his proofs.
So, using our new equation and our earlier linear equation in polar units, we can combine the two to probe only the line $\frac{1}{2}+iy$. Because $r$ is written in terms of $\theta$, we can change the power of our new $\zeta$ function to be $(\frac{1}{2}sec(\theta)cos(\theta)+i\frac{1}{2}sec(\theta)sin(\theta))$. Simplifying, we get it to be $\frac{1}{2}+i\frac{1}{2}tan(\theta)$. Hopefully this is what you meant.