Please help me with this sumation problem, I don't even know where to start with this...

proyaop Feb 10, 2022

#1**+2 **

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.

z | 1+z | |1+z|^2 | ||||

\(e^{-\pi i}\) | \(cos(-\pi)+isin(-\pi)\) | -1+0 | 0 | 0 | ||

\(e^{(\frac{-5\pi i}{7})}\) | \(cos(\frac{-5\pi}{7})+isin(\frac{-5\pi}{7}) \) | 3rd quad | \(-cos(\frac{2\pi}{7})-isin(\frac{2\pi}{7}) \) | \([1-cos(\frac{2\pi}{7})]-isin(\frac{2\pi}{7}) \) | \([1-cos(\frac{2\pi}{7})]^2+sin^2(\frac{2\pi}{7}) \) | \([1-2cos(\frac{2\pi}{7})] \) |

\(e^{(\frac{-3\pi i}{7})}\) | \(cos(\frac{-3\pi}{7})+isin(\frac{-3\pi}{7})\) | 4th quad | \(cos(\frac{3\pi}{7})-isin(\frac{3\pi}{7})\) | \([1+cos(\frac{3\pi}{7})]-isin(\frac{3\pi}{7})\) | \([1-cos(\frac{3\pi}{7})]^2+sin^2(\frac{3\pi}{7}) \) | \([1-2cos(\frac{3\pi}{7})] \) |

\(e^{(\frac{-\pi i}{7})}\) | \(cos(\frac{-1\pi}{7})+isin(\frac{-1\pi}{7})\) | 4th quad | \(cos(\frac{\pi}{7})-isin(\frac{\pi}{7})\) | \([1+cos(\frac{\pi}{7})]-isin(\frac{\pi}{7})\) | \([1+cos(\frac{\pi}{7})]^2+sin^2(\frac{\pi}{7})\) | \([1+2cos(\frac{\pi}{7})]\) |

\(e^{(\frac{\pi i}{7})}\) | \(cos(\frac{1\pi}{7})+isin(\frac{1\pi}{7})\) | 1st quad | \(cos(\frac{1\pi}{7})+isin(\frac{1\pi}{7})\) | \([1+cos(\frac{\pi}{7})]-isin(\frac{\pi}{7})\) | \([1+cos(\frac{3\pi}{7})]^2+sin^2(\frac{3\pi}{7})\) | \([1+2cos(\frac{3\pi}{7})]\) |

\(e^{(\frac{3\pi i}{7})}\) | \(cos(\frac{3\pi}{7})+isin(\frac{3\pi}{7})\) | 1st quad | \(cos(\frac{3\pi}{7})+isin(\frac{3\pi}{7})\) | \([1+cos(\frac{3\pi}{7})]+isin(\frac{3\pi}{7})\) | \([1+cos(\frac{3\pi}{7})]^2+sin^2(\frac{3\pi}{7})\) | \([1+2cos(\frac{3\pi}{7})]\) |

\(e^{(\frac{5\pi i}{7})}\) | \(cos(\frac{5\pi}{7})+isin(\frac{5\pi}{7})\) | 2nd quad | \(-cos(\frac{2\pi}{7})+isin(\frac{2\pi}{7})\) | \([1-cos(\frac{2\pi}{7})]+isin(\frac{2\pi}{7})\) | \([1-cos(\frac{2\pi}{7})]^2+sin^2(\frac{2\pi}{7})\) | \([1-2cos(\frac{2\pi}{7})]\) |

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 ://

Melody Feb 10, 2022

#2**+2 **

Continued,

the end of my table got truncated :/

The entries were

1/|1+z^2| | |

0 | undefined. Can't divide by 0 |

\(1-2cos(\frac{2\pi}{7})\) | |

\(1+2cos(\frac{3\pi}{7})\) | |

\(1+2cos(\frac{\pi}{7})\) | |

\(1+2cos(\frac{\pi}{7})\) | |

\(1+2cos(\frac{3\pi}{7})\) | |

\(1-2cos(\frac{2\pi}{7})\) |

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.

Melody Feb 10, 2022

#3**+1 **

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...

proyaop Feb 10, 2022

#6**+2 **

1-z | |1-z|^2 | |||

1--1 = 2 | 4 | |||

\([1+cos(\frac{2\pi}{7})]+isin(\frac{2\pi}{7})\) | \(1+2cos(\frac{2\pi}{7})\) | |||

\([1-cos(\frac{3\pi}{7})]+isin(\frac{3\pi}{7})\) | \(1-2cos(\frac{3\pi}{7})\) | |||

\([1-cos(\frac{\pi}{7})]+isin(\frac{\pi}{7})\) | \(1-2cos(\frac{\pi}{7})\) | |||

\([1-cos(\frac{\pi}{7})]-isin(\frac{\pi}{7})\) | \(1-2cos(\frac{\pi}{7})\) | |||

\([1-cos(\frac{3\pi}{7})]-isin(\frac{3\pi}{7})\) | \(1-2cos(\frac{3\pi}{7})\) | |||

\([1+cos(\frac{2\pi}{7})]-isin(\frac{2\pi}{7})\) | \(1+2cos(\frac{2\pi}{7})\) | |||

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

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

Melody Feb 11, 2022

#7**0 **

Ok Tiggsy I have an answer

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

Melody Feb 11, 2022

#8**+2 **

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

Tiggsy Feb 12, 2022

#9**+1 **

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

Melody Feb 13, 2022