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December 5th, 2010, 05:09 AM  #1 
Newbie Joined: Oct 2010 Posts: 27 Thanks: 0  Complex equation
I need a help with this. Prove that if complex roots and of equation have their absolute values equal ( then is a real number. Thanks for any advices. 
December 5th, 2010, 07:41 AM  #2 
Senior Member Joined: Nov 2010 Posts: 502 Thanks: 0  Re: Complex equation
The quadratic formula works for all quadratics.

December 5th, 2010, 08:07 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038 
So?

December 5th, 2010, 12:44 PM  #4  
Senior Member Joined: Nov 2010 Posts: 502 Thanks: 0  Re: Quote:
 
December 5th, 2010, 09:23 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038 
You. There is a reasonably short proof using trigonometry, but I don't see any simple method based on the quadratic formula.

December 5th, 2010, 10:11 PM  #6 
Newbie Joined: Oct 2010 Posts: 27 Thanks: 0  Re: Complex equation
And could you please show me both way to do it? Especially trigonometry one. Common quadratic formula works, but from the result I am unable to work out the proof. 
December 5th, 2010, 11:19 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038 
Let where "cis" means "cos + i sin" and and are real, then p and q² Now it's fairly easy to show that p²/q² > 0. 
December 6th, 2010, 01:29 AM  #8 
Newbie Joined: Oct 2010 Posts: 27 Thanks: 0  Re: Complex equation
But I have to prove that p/q is real number, not that p*p/q*q > 0 Or did I understand badly to it? 
December 6th, 2010, 05:35 AM  #9  
Senior Member Joined: Nov 2010 Posts: 502 Thanks: 0  Re: Quote:
To Gustav, consider what we now have. If p^2/q^2 = a > 0, then we also have that p^2/q^2  a = 0. (I chose this form because I think it's easiest to see).  
December 6th, 2010, 08:31 AM  #10 
Newbie Joined: Oct 2010 Posts: 27 Thanks: 0  Re: Complex equation
But how can we say that a and b are real?


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