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. 
Re: Complex equation The quadratic formula works for all quadratics. 
So? 
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You. There is a reasonably short proof using trigonometry, but I don't see any simple method based on the quadratic formula. 
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. 
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. 
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? 
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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). 
Re: Complex equation But how can we say that a and b are real? 
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