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 Post subject: Root of ring - functionPosted: Mon, 5 Sep 2011 22:58:24 UTC
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Joined: Fri, 18 Jul 2008 01:21:42 UTC
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Hello,

Problem statement: Assume that e+gi is a root of f(x) = ax^2 + bx +c. Prove that e-gi is also a root of f(x). (a,b,c,e,g in Reals).

I have so far - f(e+gi)= ae^2 + ag^2 +be+c+(2aeg+bg) i=0, and thus ae^2 + ag^2+be+c=0 and 2aeg+bg=0.

f(e-gi) = ae^2 + ag^2 +be+c+(2aeg+bg) i=0. Now I need to show that f(e-gi) = 0. Would someone kindly help me?

Thank you,

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 Post subject: Re: Root of ring - functionPosted: Tue, 6 Sep 2011 01:05:33 UTC
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Joined: Mon, 23 Feb 2009 23:20:33 UTC
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Quote:
f(e+gi)= ae^2 + ag^2 +be+c+(2aeg+bg) i=0
f(e-gi) = ae^2 + ag^2 +be+c+(2aeg+bg) i=0

Should be:
f(e+gi) = (ae²-ag²+be+c)+(2aeg+bg)i = 0 + 0i
f(e-gi) = (ae²-ag²+be+c)-(2aeg+bg)i

Subtracting the second from the first:
f(e+gi)-f(e-gi) = 2 (2aeg+bg)i
0-f(e-gi) = 2(0)i
f(e-gi) = 0

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 Post subject: Re: Root of ring - functionPosted: Tue, 6 Sep 2011 03:53:59 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12071
Location: Austin, TX
Susan123456 wrote:
Hello,

Problem statement: Assume that e+gi is a root of f(x) = ax^2 + bx +c. Prove that e-gi is also a root of f(x). (a,b,c,e,g in Reals).

I have so far - f(e+gi)= ae^2 + ag^2 +be+c+(2aeg+bg) i=0, and thus ae^2 + ag^2+be+c=0 and 2aeg+bg=0.

f(e-gi) = ae^2 + ag^2 +be+c+(2aeg+bg) i=0. Now I need to show that f(e-gi) = 0. Would someone kindly help me?

Thank you,

The case is trivial, so let us assume that , then the discriminant, .

Then because you know the quadratic formula you know the roots are . From here it is obvious, since two complex numbers are equal iff their real and imaginary parts are, so you immediately get and and it's clear that the conjugate is the other root. (note our assumption that so this is well-defined.

Edit: changed = under square root to -. Nothing material.

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 Post subject: Re: Root of ring - functionPosted: Tue, 6 Sep 2011 03:54:16 UTC
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Also, this belongs in the algebra forum. Topic moved.

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 Post subject: Re: Root of ring - functionPosted: Tue, 6 Sep 2011 05:29:25 UTC
 S.O.S. Oldtimer

Joined: Fri, 18 Jul 2008 01:21:42 UTC
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Thank you for your response. I don't know how you got g = +/-sqrt b^2 -4ac/(2ai). Could you show me?

Thank you

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 Post subject: Re: Root of ring - functionPosted: Tue, 6 Sep 2011 17:26:40 UTC
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Susan123456 wrote:
Thank you for your response. I don't know how you got g = +/-sqrt b^2 -4ac/(2ai). Could you show me?

Thank you

Two complex numbers are the same iff their real and imaginary parts are the same.

It's clear that is real and is purely imaginary, so you just extract the real and imaginary parts of the number.

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 Post subject: Re: Root of ring - functionPosted: Tue, 6 Sep 2011 23:05:26 UTC
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Joined: Fri, 1 Jul 2011 01:17:26 UTC
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There is an alternate approach. If e + ig is a root and r + is is the other root, then a(x - e - ig)(x - r -is)=ax^2 + bx + c.

x^2 ] a=a
x ] a(-e -r - i(g + s)) = b, since b is real, s = -g
constant ] a(rc -gs -i(rg + es)) = c, since c is real, rg +es = 0 and since s = -g, r=e.

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 Post subject: Re: Root of ring - functionPosted: Tue, 6 Sep 2011 23:30:07 UTC
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mathematic wrote:
There is an alternate approach. If e + ig is a root and r + is is the other root, then a(x - e - ig)(x - r -is)=ax^2 + bx + c.

x^2 ] a=a
x ] a(-e -r - i(g + s)) = b, since b is real, s = -g
constant ] a(rc -gs -i(rg + es)) = c, since c is real, rg +es = 0 and since s = -g, r=e.

That is the same approach; that's how one derives the quadratic formula.

There's also a problem with this approach, what if ? Then the roots are both real, and the other root is NOT the complex conjugate, so again you need to assume the discriminant is negative to get your version. In this case you conclude the desired result by noting that g=0, so trivially e-gi is a root. It is vital that the question is phrased as "[. . .] e-gi is also a root." and NOT as "[. . .]e-gi is the OTHER root", since that is false.

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 Post subject: Re: Root of ring - functionPosted: Wed, 7 Sep 2011 23:00:26 UTC
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The last line of my derivation does not hold if both roots are real.
rg + es = 0 does not imply r = e if g and s are 0.

Therefore my proof does require an assumption that the given root is not real.

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 Post subject: Re: Root of ring - functionPosted: Thu, 8 Sep 2011 04:08:00 UTC
 S.O.S. Oldtimer

Joined: Fri, 18 Jul 2008 01:21:42 UTC
Posts: 147
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thank you

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