+0  
 
+1
246
9
avatar+993 

 

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

 Feb 10, 2022
 #1
avatar+118117 
+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+000 
\(e^{(\frac{-5\pi i}{7})}\)\(cos(\frac{-5\pi}{7})+isin(\frac{-5\pi}{7}) \)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 ://

 

 

 Feb 10, 2022
 #2
avatar+118117 
+2

Continued,

 

the end of my table got truncated :/

 

The entries were

 

 1/|1+z^2|
0undefined. 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.

 Feb 10, 2022
edited by Melody  Feb 10, 2022
 #4
avatar+230 
+1

It's 1 - z, not 1 + z isn't it ? That removes the undefined term.

Tiggsy  Feb 11, 2022
 #5
avatar+118117 
0

Bummer,  you are right Tiggsy!

 

Do you knopw how to do this an easier way?

Melody  Feb 11, 2022
 #3
avatar+993 
+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... smiley

 Feb 10, 2022
 #6
avatar+118117 
+2

 

 

 

 

1-z|1-z|^2   
1--1 = 24   
\([1+cos(\frac{2\pi}{7})]+isin(\frac{2\pi}{7})\)\(1+2cos(\frac{2\pi}{7})\)   
\([1-cos(\frac{3\pi}{7})]+isin(\frac{3\pi}{7})\)\(1-2cos(\frac{3\pi}{7})\)   
\([1-cos(\frac{\pi}{7})]+isin(\frac{\pi}{7})\)\(1-2cos(\frac{\pi}{7})\)   
\([1-cos(\frac{\pi}{7})]-isin(\frac{\pi}{7})\)\(1-2cos(\frac{\pi}{7})\)   
\([1-cos(\frac{3\pi}{7})]-isin(\frac{3\pi}{7})\)\(1-2cos(\frac{3\pi}{7})\)   
\([1+cos(\frac{2\pi}{7})]-isin(\frac{2\pi}{7})\)\(1+2cos(\frac{2\pi}{7})\)   
     

 

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

 

 

 

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

 Feb 11, 2022
 #7
avatar+118117 
0

Ok Tiggsy I have an answer  wink

 

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

 Feb 11, 2022
 #8
avatar+230 
+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

 Feb 12, 2022
 #9
avatar+118117 
+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

 Feb 13, 2022

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