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Topic: Complex roots (Read 2157 times) 

Sameer
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Complex roots
« on: Sep 17^{th}, 2007, 11:21pm » 
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Another one from my book: Find all the roots of the equation: (x  1)^{n} = x^{n} where n is a positive integer.

« Last Edit: Sep 17^{th}, 2007, 11:22pm by Sameer » 
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Aryabhatta
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Re: Complex roots
« Reply #1 on: Sep 18^{th}, 2007, 1:33am » 
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Clearly x = 0 is not a root. Divide by x^{n} and we are done, right?


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towr
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Re: Complex roots
« Reply #2 on: Sep 18^{th}, 2007, 9:44am » 
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So x = 1/(1 r), where r is any of the nth roots of 1. i.e., if memory serves me right, r= e^{i 2}^{ k/n} with k=0..n1

« Last Edit: Sep 18^{th}, 2007, 9:46am by towr » 
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pex
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Re: Complex roots
« Reply #3 on: Sep 18^{th}, 2007, 2:02pm » 
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on Sep 18^{th}, 2007, 9:44am, towr wrote:So x = 1/(1 r), where r is any of the nth roots of 1. i.e., if memory serves me right, r= e^{i 2}^{ k/n} with k=0..n1 
 Yes, except that k=0 won't work, obviously.


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Sameer
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Re: Complex roots
« Reply #4 on: Sep 18^{th}, 2007, 9:46pm » 
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on Sep 18^{th}, 2007, 1:33am, Aryabhatta wrote:Clearly x = 0 is not a root. Divide by x^{n} and we are done, right? 
 Yep. on Sep 18^{th}, 2007, 2:02pm, pex wrote: Yes, except that k=0 won't work, obviously. 
 Well in this case k = 0 does work!! (I think!!!)


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Aryabhatta
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Re: Complex roots
« Reply #5 on: Sep 18^{th}, 2007, 11:40pm » 
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on Sep 18^{th}, 2007, 9:46pm, Sameer wrote: Yep. Well in this case k = 0 does work!! (I think!!!) 
 The original equation has exactly n1 roots (counting multiplicity) as it is a polynomial of degree n1. So we have to reject one root.


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pex
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Re: Complex roots
« Reply #6 on: Sep 19^{th}, 2007, 5:40am » 
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on Sep 18^{th}, 2007, 11:40pm, Aryabhatta wrote: The original equation has exactly n1 roots (counting multiplicity) as it is a polynomial of degree n1. So we have to reject one root. 
 Yes, and k = 0 leads to r = 1, leaving x = 1/(1  r) undefined.


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Grimbal
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Re: Complex roots
« Reply #7 on: Sep 19^{th}, 2007, 5:52am » 
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Now compute Re(x)


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pex
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Re: Complex roots
« Reply #8 on: Sep 19^{th}, 2007, 6:12am » 
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on Sep 19^{th}, 2007, 5:52am, Grimbal wrote:That's interesting! I get that Re(x) = 1/2 for all of the roots. Derivation:hidden:  x_{k} = 1 / (1  exp(2k pi i / n)) = (1  exp(2k pi i / n)) / (2  exp(2k pi i / n)  exp(2k pi i / n)) = (1  cos(2k pi / n) + i sin(2k pi / n)) / (2  2cos(2k pi / n)) = 1/2 + 1/2 i [ sin(2k pi / n) / (1  cos(2k pi / n)) ], and clearly, the real part is 1/2, independent of k. 

« Last Edit: Sep 19^{th}, 2007, 6:13am by pex » 
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Sameer
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Re: Complex roots
« Reply #9 on: Sep 19^{th}, 2007, 9:18am » 
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Yep, I just thought infinity could be one of the roots!! Maybe not.. Yep I get the real part to be 1/2 too!! My original answer was in a reduced form 1/2(1+icotk/2) k=1..n1 which easily showed what the real part was.


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srn437
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Re: Complex roots
« Reply #10 on: Sep 19^{th}, 2007, 6:38pm » 
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If n is even, x can be 1/2.


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Sameer
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Re: Complex roots
« Reply #11 on: Sep 19^{th}, 2007, 6:55pm » 
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on Sep 19^{th}, 2007, 6:38pm, srn347 wrote:If n is even, x can be 1/2. 
 The question asks for general solution over all n.


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srn437
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Re: Complex roots
« Reply #12 on: Sep 23^{rd}, 2007, 3:29pm » 
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It wuold have to be one where x=x1, so x=infinity.


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towr
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Re: Complex roots
« Reply #13 on: Sep 23^{rd}, 2007, 11:16pm » 
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on Sep 23^{rd}, 2007, 3:29pm, srn347 wrote:It wuold have to be one where x=x1, so x=infinity. 
 And yet again you fail. There are n1 solutions that don't tend to infinity. And since infinity isn't a complex number it can't be a solution, so only those other n1 remain.


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srn437
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Re: Complex roots
« Reply #14 on: Sep 24^{th}, 2007, 6:25pm » 
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It has to be complex? Take the n root and x=x1, therefore it must be infinity(positive or negative).


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Sameer
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Re: Complex roots
« Reply #15 on: Sep 24^{th}, 2007, 8:24pm » 
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on Sep 24^{th}, 2007, 6:25pm, srn347 wrote:It has to be complex? Take the n root and x=x1, therefore it must be infinity(positive or negative). 
 ok to make it easier to understand, if you have complex set available, what are all the roots of x^{3} = 1?


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towr
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Re: Complex roots
« Reply #16 on: Sep 25^{th}, 2007, 1:26am » 
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on Sep 24^{th}, 2007, 6:25pm, srn347 wrote:Complex numbers include the real numbers, FYI. Infinity is neither a real number, nor an imaginary number nor a combination. Hence it can't be a solution. Quote:Take the n root and x=x1, therefore it must be infinity(positive or negative). 
 Wrong. By simply taking the nth root at both sides, you lose n1 solutions (and there are only n1 solutions in the first place) If e.g. x^{2}=(x1)^{2}, then you don't get the solution by taking x=(x1); so why would you think this is allowed for other n? For x^{2}=(x1)^{2}, just simplify the equation: (x1)^{2} = x^{2} 2x + 1, therefore x^{2}=x^{2} 2x + 1, or x=1/2. You even suggested x=1/2 yourself for even n!

« Last Edit: Sep 25^{th}, 2007, 1:26am by towr » 
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srn437
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Re: Complex roots
« Reply #17 on: Sep 25^{th}, 2007, 5:07pm » 
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The roots of x^{3}=1 are 1, 1/2 + sqrt(3)(i)/2, and 1/2 sqrt(3)(i)/2


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Sameer
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Re: Complex roots
« Reply #18 on: Sep 25^{th}, 2007, 6:12pm » 
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on Sep 25^{th}, 2007, 5:07pm, srn347 wrote:The roots of x^{3}=1 are 1, 1/2 + sqrt(3)(i)/2, and 1/2 sqrt(3)(i)/2 
 Good, now look at towr's example and what do you learn?


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srn437
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Re: Complex roots
« Reply #19 on: Sep 25^{th}, 2007, 8:04pm » 
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That there are no real solutions for all natural number exponents(which I already new).


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Obob
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Re: Complex roots
« Reply #20 on: Sep 25^{th}, 2007, 8:30pm » 
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Except whenever n is even, x=1/2 is a solution...


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Sameer
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Re: Complex roots
« Reply #21 on: Sep 25^{th}, 2007, 8:44pm » 
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on Sep 25^{th}, 2007, 8:04pm, srn347 wrote:That there are no real solutions for all natural number exponents(which I already new). 
 What part of the title of this problem confuses you? Are you genuinely trying to learn or just wasting my time?


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srn437
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Re: Complex roots
« Reply #22 on: Sep 26^{th}, 2007, 7:23pm » 
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As already stated, you need to have all the n's answered. x would have to equal n1.


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Sameer
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Re: Complex roots
« Reply #23 on: Sep 26^{th}, 2007, 9:39pm » 
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Note to Mods: Can you please delete posts that are irrelevant to this thread? You get my drift!!


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srn437
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Re: Complex roots
« Reply #24 on: Sep 26^{th}, 2007, 9:48pm » 
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Try asking a question with an answer!


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