Root of ring - function

Susan123456 · **Posted:** Mon, 5 Sep 2011 22:58:24 UTC

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,

aswoods · **Posted:** Tue, 6 Sep 2011 01:05:33 UTC

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

Shadow · **Posted:** Tue, 6 Sep 2011 03:53:59 UTC

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 $g\ne 0$ , then the discriminant, b^2-4ac< 0

.

Then because you know the quadratic formula you know the roots are ${-b\pm\sqrt{b^2-4ac}\over 2a}$ . From here it is obvious, since two complex numbers are equal iff their real and imaginary parts are, so you immediately get $e={-b\over 2a}$ and $g=\pm{{\sqrt{b^2-4ac}\over 2ai}$ and it's clear that the conjugate is the other root. (note our assumption that $b^2-4ac<0\Rightarrow a\ne 0}$ so this is well-defined.

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

Shadow · **Posted:** Tue, 6 Sep 2011 03:54:16 UTC

Also, this belongs in the algebra forum. Topic moved.

Susan123456 · **Posted:** Tue, 6 Sep 2011 05:29:25 UTC

Thank you for your response. I don't know how you got g = +/-sqrt b^2 -4ac/(2ai). Could you show me?

Thank you

Shadow · **Posted:** Tue, 6 Sep 2011 17:26:40 UTC

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 $-{b\over 2a}$ is real and $\pm {\sqrt{b^2-4ac}\over 2a}$ is purely imaginary, so you just extract the real and imaginary parts of the number.

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 321

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.

This then leads to
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.

Shadow · **Posted:** Tue, 6 Sep 2011 23:30:07 UTC

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.

This then leads to
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 a=1, b=-3, c=2

? 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.

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 321

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.

Susan123456 · **Posted:** Thu, 8 Sep 2011 04:08:00 UTC

thank you

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Root of ring - function

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