Algebra 2 Help!

This is a reapeat question.  You should have sent us with a link, to the original.

Never mind, I do not think you got any answers there.

I played with this for quite a while when you first posted it.

I rarely know the short cuts for complex numbes.

All I could thinkg of doing was getting the 7 answers and adding them up.

If it hadn't been over 1 this would not have been so bad but being over 1 definitely added to the messiness and length of it.

The first obvious z value is   e^(pi *i)     the general z solution is   e^(pi*i -  n*2pi/7)     for n=0 to 6

As shown in the complex plane pic below.

I'm saving now so that I don't lose it.   But I am still going.  I am sure this is the most rediculous way anyone has ever done it ://

Continued,

the end of my table got truncated :/

The entries were

So after all that unnecessary work. 

I get that the sum is undefined becasue the first term in the sum is undefined.

Please give feed back on whether the answer is simply undefined.  OR on how I should have done it. 

Thanks.

Wow! That was very detailed and complicated! Maybe leaving out undefined in the sum is the answer. And the answer to the problem is in the formatting of a fraction, so the answer to the problem shouldn't have any imaginary numbers.

Thank you... 

$\displaystyle sum = \frac{1}{4}+\frac{2}{1+2cos\frac{2\pi}{7}}+\frac{2}{1-2cos\frac{3\pi}{7}}+\frac{2}{1-2cos\frac{\pi}{7}}\\ \\~\\\displaystyle sum \approx 8.3478346790362325$

2/(1+2cos(2pi/7)) = 0.8900837358250465

2/(1-2cos(3pi/7)) = 3.603875471605593

2/(1-2cos(pi/7) = -2.493959207437541

0.25+0.8900837358250465+3.603875471605593+3.603875471605593 = 8.3478346790362325

Ok Tiggsy I have an answer  

Can you please show us how to do it properly now :))

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

Thanks Tiggsy

I can follow all that.  Similar to mine just better :))

$\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}} =\frac{1}{4}\left\{\frac{1}{\sin^{2}(\pi/14)} +\frac{1}{\sin^{2}(3\pi/14)}+\dots +\frac{1}{\sin^{2}(13\pi/14)}\right\}.\\~\\ \displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}} =\frac{1}{4}\left\{\frac{2}{\sin^{2}(\pi/14)} +\frac{2}{\sin^{2}(3\pi/14)}+\frac{2}{\sin^{2}(5\pi/14)}+\frac{1}{\sin^{2}(\pi/2)} \right\}.\\~\\ \displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}} =\frac{1}{4}\left\{\frac{2}{\sin^{2}(\pi/14)} +\frac{2}{\sin^{2}(3\pi/14)}+\frac{2}{\sin^{2}(5\pi/14)}+1 \right\}.\\~\\ \displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}} =\frac{1}{4}+\frac{1}{2}\left\{\frac{1}{\sin^{2}(\pi/14)} +\frac{1}{\sin^{2}(3\pi/14)}+\frac{1}{\sin^{2}(5\pi/14)} \right\}.\\~\\ \displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}} =\frac{1}{4}+\frac{1}{2}\left\{24 \right\}.\\~\\ \displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}=12.25$

LaTex

\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\}.\\~\\
\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}\left\{\frac{2}{\sin^{2}(\pi/14)} +\frac{2}{\sin^{2}(3\pi/14)}+\frac{2}{\sin^{2}(5\pi/14)}+\frac{1}{\sin^{2}(\pi/2)}
\right\}.\\~\\
\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}\left\{\frac{2}{\sin^{2}(\pi/14)} +\frac{2}{\sin^{2}(3\pi/14)}+\frac{2}{\sin^{2}(5\pi/14)}+1
\right\}.\\~\\
\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}+\frac{1}{2}\left\{\frac{1}{\sin^{2}(\pi/14)} +\frac{1}{\sin^{2}(3\pi/14)}+\frac{1}{\sin^{2}(5\pi/14)}
\right\}.\\~\\

\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}+\frac{1}{2}\left\{24
\right\}.\\~\\

\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}=12.25