Help with a tan(arccos(z)) question

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I need help describing the relationship of tan(arccos(z)) when z=(x-1)/(x+1) as the core piece of this equation here:
http://3.bp.blogspot.com/-5UhMF-uGw...AFQ/oDdl_oSXPM0/s1600/prime-+squares+edit.png
and here notice the square root of pimes x, hoizontal vector towards the y axis. You may have to zoom in:
http://2.bp.blogspot.com/_u6-6d4_gs...AAAE8/_hov_b0sno4/s1600/prime-+square+12a.png

Now in iy, right?
http://upload.wikimedia.org/wikiped...mental_relationship_to_Circle_(and_Helix).gif

and its relation ship to roots of unity:
http://www.katjaas.nl/rootsofunity/rootsofunity.html
root systems:
http://en.wikipedia.org/wiki/Root_system

Does it have a relation to “(x-1)/(x+1)^(-x/2) = e” as well?
Any help wood be greatly appreciated. Trying to teach myself…
 
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Ok, I'm working on an animation of what I'm trying to equate. here is what I have so far:
http://i98.photobucket.com/albums/l267/alienearcandy/sqrt-1.gif
It seems to be related to the harmonic series, Fourier transform and the inverse-square law.
Is there already an equation that defines this animation?

Again it releates twards primes...
http://3.bp.blogspot.com/-5UhMF-uGw...AFQ/oDdl_oSXPM0/s1600/prime-+squares+edit.png

and

http://4.bp.blogspot.com/_u6-6d4_gs.../bdPIJMIFTLE/s1600/prime-+square+12a+zoom.png

I know you guys have to be sick of seeing these images...
 
I'm seeing these sequences in a triangular sequence related to angular momentum: t(n,m)=4*(n*(n-1)-m*(m-1)). :
http://oeis.org/A152420
 
I'm trying to put this in complex form as it relates to the roots of unity.
http://i98.photobucket.com/albums/l267/alienearcandy/sqrt-1.gif
In this equation the root values lie on the y axis. I'm not using 2pi/n, so the radians intervals are not equal, infact they are releated to the complementary numbers of the base (x+1) where the roots are of the form tan(arccos((x-d)/(x+1))) where d eqauls intergers from 1 to x-1. I need help defining this root system.
 
tan(arccos(z)) = sin(arccos(z))/cos(arccos(z)) = \frac{\sqrt{1-z^2}}{z} = \frac{\sqrt{1-((x-1)/(x+1))^2}}{(x-1)/(x+1)} = \frac{\sqrt{(x+1)^2-(x-1)^2}}{x-1} = \frac{\sqrt{4x}}{x-1} = \frac{2\sqrt{x}}{x-1}

with sign depending on the quadrant the angle z lies in.
 
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