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Math Help - Complex Numbers

  1. #1
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    Complex Numbers

    Find the set of real values k, k ≠ -1, for which the roots of the equation x2 + 4x - 1 + k(x2 + 2x + 1) = 0 are complex with positive real part.

    So far, I have found the discriminant of the equation and solved for k when
    Δ < 1, giving me k < -5/4. But this is only the complex(?) part of the solution. How do I find the "positive real part"?


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  2. #2
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    Re: Complex Numbers

    Quote Originally Posted by Fratricide View Post
    Find the set of real values k, k ≠ -1, for which the roots of the equation x2 + 4x - 1 + k(x2 + 2x + 1) = 0 are complex with positive real part.

    So far, I have found the discriminant of the equation and solved for k when
    Δ < 1, giving me k < -5/4. But this is only the complex(?) part of the solution. How do I find the "positive real part"?


    Hello,

    the complete solution of this equation is:

    x = \underbrace{-\frac{k+2}{k+1}}_{\text{real part}} \pm  \sqrt{4k+5}

    That means you have to find those values of k for which

    -\frac{k+2}{k+1} >0 under the condition that k <-\frac54

    -\frac{k+2}{k+1} >0~\implies~\frac{k+2}{k+1} <0

    A quotient is negative if numerator and denominator have different signs:

    \left( \left(\underbrace{k+2>0}_{\text{positive}}\ \wedge \ \underbrace{k+1 <0}_{\text{negative}} \right)~\vee~\left( k+2<0\ \wedge \ k+1 >0 \right) \right)~\wedge~k<-\frac54 \\ \left(-2<k<-1~\vee~\emptyset \right)~\wedge~k<-\frac54

    So you'll get the final solution as: k\in \left]-2 , -\frac54 \right[
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    Re: Complex Numbers

    Thanks for the reply; however, it's still not clear to me why you why the solution is -2 < k < -5/4. (Unfamiliar notation, perhaps.)
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    Re: Complex Numbers

    Quote Originally Posted by Fratricide View Post
    Thanks for the reply; however, it's still not clear to me why you why the solution is -2 < k < -5/4. (Unfamiliar notation, perhaps.)
    I cannot read post 2 for some reason so this post may be entirely repetitive.

    First off, your logic in post 1 was perfect.

    $x^2 + 4x - 1 + k(x^2 + 2x + 1) = 0 \implies (k + 1)x^2 + (2k + 4)x + (k - 1) = 0 \implies x = \dfrac{-\ (2k + 4) \pm \sqrt{(2k + 4)^2 - 4(k + 1)(k - 1)}}{2(k + 1)} \implies$

    $x = \dfrac{(-2)(k + 2) \pm \sqrt{4k^2 + 16k + 16 - 4k^2 + 4}}{2(k + 1)} = \dfrac{-k - 2 \pm \sqrt{4k + 5}}{(k + 1)}.$

    Now if $4k + 5 \ge 0$, then x has no imaginary part. So for x to have an imaginary part $\implies 4k + 5 < 0 \implies 4k < - 5 \implies k < -\ \dfrac{5}{4}.$ Well done.

    So consider x if k < - (5/4).

    $x = \dfrac{-k - 2 \pm \sqrt{(5 - 4k)(-\ 1)}}{k + 1} = \left(- \dfrac{k + 2}{k + 1}\right) \pm \left(\dfrac{\sqrt{5 - 4k}}{k + 1}\right) * i.$ You with me?

    So what determines if the real part of x is positive? k + 2 and k + 1 must have opposite signs. Does that make sense?

    OK. But $1 < 2 \implies k + 1 < k + 2.$ So the only way that they can have opposite signs is if k + 2 is positive and k + 1 is negative. Stop me if this is unclear.

    $k + 2 > 0 \implies k > -\ 2.$ And $k + 1 < 0 \implies k < - 1 = -\ \dfrac{4}{4}.$ But we have already determined that $k < -\ \dfrac{5}{4} < -\ \dfrac{4}{4}.$

    And so to meet both conditions, an imaginary part and a positive real part, $\implies -\ 2 < k < -\ \dfrac{5}{4}.$

    Excuse me if this just repeats post 2, which does not render for me. Let us know if you have further questions.
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    Re: Complex Numbers

    Quote Originally Posted by JeffM View Post
    So what determines if the real part of x is positive? k + 2 and k + 1 must have opposite signs. Does that make sense?

    OK. But $1 < 2 \implies k + 1 < k + 2.$ So the only way that they can have opposite signs is if k + 2 is positive and k + 1 is negative. Stop me if this is unclear.
    If k + 2 and k + 1 have opposite signs, won't that make the quotient negative? Also, I'm still not clicking with the second line. Why can't k + 1 be positive? (Forgive me if this is obvious, it's quite late where I am.)
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    Re: Complex Numbers

    Quote Originally Posted by Fratricide View Post
    If k + 2 and k + 1 have opposite signs, won't that make the quotient negative? Also, I'm still not clicking with the second line. Why can't k + 1 be positive? (Forgive me if this is obvious, it's quite late where I am.)
    there's a negative sign in front of the quotient. That's why the quotient itself needs to be negative.

    it must be late...

    $(k+1)>0\Rightarrow (k+2)=(k+1)+1 >0$

    so if one of them is negative it must be (k+1)
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