Hi Melody.
My method is pretty much the same as yours, though I did arrive at a different result, and I do have a suggestion for taking it further.
\(\displaystyle \text{If } z^{7}=-1=1.\text{cis}(\pi), \text{then} \\ z = \{1.\text{cis}(\pi+2k\pi)\}^{1/7}=\text{cis}(\pi/7+2k\pi/7), \;k=0,1,2,\dots,6.\)
Now, if
\(\displaystyle z=\cos(\theta)+i\sin(\theta), \text{ then}\\ 1-z=1-\cos(\theta)-i\sin(\theta) \text{ so}\\ \mid1-z\mid^{2}=(1-\cos(\theta))^{2}+\sin^{2}(\theta)=2-2\cos(\theta) =2(1-\cos(\theta)).\)
Then, using the trig identity \(\displaystyle \cos2A = 1-2\sin^{2}A,\)
\(\displaystyle \mid1-z\mid^{2}=4\sin^{2}(\theta/2).\)
So,
\(\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}=\frac{1}{4}\left\{\frac{1}{\sin^{2}(\pi/14)} +\frac{1}{\sin^{2}(3\pi/14)}+\dots +\frac{1}{\sin^{2}(13\pi/14)}\right\}.\)
Summing the terms inside the curly bracket gets you the number 49.
That's just too much of a coincidence, 49 being 7 squared.
I checked it out for z^3 = -1 and z^5 = -1 and the results were 9 and 25.
So, it would seem that this is a standard result, one the I don't recall seeing before, and for the moment don't see how to prove.
I'll see if I can take it further
Tiggsy
!
