The formula is this
$$(z^*_m)^2= (z^*_{m-1})^2+\frac{1}{m^{2a}}-\frac{2z^*_{m-1}*cos(\pi-(j+b))}{m^{a}}$$
$$B= acos(\frac{(z^*_{m-2})^2-(z^*_{m-1})^2+\frac{1}{m^{2a}}}{z^*_{m-1}*m^{a}})$$
$$J=ln(\frac{m}{m-1})x$$ where Z*_m is the magnitue of the zeta function at a given sum m
Here is how I arrived at it if we look at $$\sum_{n=0}^m \frac{1}{n^s}$$
EDIT
S=a+ix
end EDIT
(note this all only works for values of M>3)
for values where it is defined, it can be potted on a complex plain.
it must then have a radial magnitude associated with it. Z*_m
if M is greater than 2 we can plot the values which add to it.
which means with law of cosines
$$(z^*_m)^2= (z^*_{m-1})^2+\frac{1}{m^{2a}}-\frac{2z^*_{m-1}*cos(k)}{m^{a}}$$
we can find K if we look at how we added to z*_{m-1}
$$(z^*_m)^2= (z^*_{m-1})^2+\frac{1}{m^{2a}}-\frac{2z^*_{m-1}*cos(\pi-(j+b))}{m^{a}}$$
we can find b with the law of cosines
$$B= acos(\frac{(z^*_{m-2})^2-(z^*_{m-1})^2+\frac{1}{m^{2a}}}{z^*_{m-1}*m^{a}})$$
and we can find J with as
$$J=ln(\frac{m}{m-1})x$$
Update: I wrote a python program to check whether it works or not. you need to keep x and b small to avoid rounding erros. making m to hi also can creat rounding errors. you can find a link hear https://drive.google.com/open?id=0B8YKu_MJDZkGUW5qcVZZekctNXc
Update2: I was asked in the coments to explain why I thought this was interesting the thought is i can say
given $$(z^*_m)^2= (z^*_{m-1})^2+\frac{1}{m^{2a}}-\frac{2z^*_{m-1}*cos(\pi-(j+b))}{m^{a}}$$ set z_m to 0 and use the quadrac formual on Z_(m-1) $$z*_(m-1)=(\frac{2*cos(\pi-(j+b))}{m^{a}}+/-\sqrt{\frac{4*cos(\pi-(j+b))^2-4}{m^{2a}}})/2$$ z_(m-1) is a scailer and cannot be imigary therefor cos(\pi-(j+b))=1 or -1 therefor $$z*_{(m-1)}=+/-\frac{1}{m^{a}}$$ therefor if z_m=0 $$z*_m=(0 or_\frac{4}{m^{2a}})/2$$ and its zero becuse of corse it is. i forgot the 1/2 in the quadric and they messed me up but i can say
$$z*_{(m-1)}=\frac{1}{m^{a}}$$ and J+B=pi+3pi*n Whitch you can see by the picture. i feel kinda silly. I still think it kinda cool anyway!