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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. $p,q \in C,\ q \neq 0$ Prove that if complex roots $x_1$ and $x_2$ of equation $x^2 + px + q^2= 0$ have their absolute values equal ( $|x_1|=|x_2|$ then $\frac{p}{q}$ 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
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Re:

Quote:
 Originally Posted by skipjack So?
Are you asking me or him?

 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 $x_{\small1}\,=\,r\,\text{cis}(a),\ x_{\small2}\,=\,r\,\text{cis}(b),$ where "cis" means "cos + i sin" and $a$ and $b$ are real, then p $=\,-r(\text{cis}(a)\,+\,\text{cis}(b))$ and q² $=\,r^{\small2}\text{cis}(a)\text{cis}(b)\,=\,r^{\s mall2}\text{cis}(a+b).$ 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
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Re:

Quote:
 Originally Posted by skipjack Let $x_{\small1}\,=\,r\,\text{cis}(a),\ x_{\small2}\,=\,r\,\text{cis}(b),$ where "cis" means "cos + i sin" and $a$ and $b$ are real, then p $=\,-r(\text{cis}(a)\,+\,\text{cis}(b))$ and q² $=\,r^{\small2}\text{cis}(a)\text{cis}(b)\,=\,r^{\s mall2}\text{cis}(a+b).$ Now it's fairly easy to show that p²/q² > 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).

 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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