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Gustav December 5th, 2010 05:09 AM

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.

DLowry December 5th, 2010 07:41 AM

Re: Complex equation
 
The quadratic formula works for all quadratics.

skipjack December 5th, 2010 08:07 AM

So?

DLowry December 5th, 2010 12:44 PM

Re:
 
Quote:

Originally Posted by skipjack
So?

Are you asking me or him?

skipjack December 5th, 2010 09:23 PM

You. There is a reasonably short proof using trigonometry, but I don't see any simple method based on the quadratic formula.

Gustav December 5th, 2010 10:11 PM

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.

skipjack December 5th, 2010 11:19 PM

Let where "cis" means "cos + i sin" and and are real, then
p and q
Now it's fairly easy to show that p/q > 0.

Gustav December 6th, 2010 01:29 AM

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?

DLowry December 6th, 2010 05:35 AM

Re:
 
Quote:

Originally Posted by skipjack
Let where "cis" means "cos + i sin" and and are real, then
p and q
Now it's fairly easy to show that p/q > 0.

Ah, this is much shorter than the way I had in mind. I yield for your victory.

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).

Gustav December 6th, 2010 08:31 AM

Re: Complex equation
 
But how can we say that a and b are real?


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