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Math Help - Exercises with complex Numbers

  1. #1
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    Exercises with complex Numbers

    Dear Friends,

    I've got 2 exercises I can't solve.

    The first one is :

    If a,b,c ∈ C and they are affix points of the plan.
    Show that the triangle ABC is equilateral only if

    a^2+b^2+c^2 = ab+ bc + ca

    The second one is to demonstrate that :

    X^4 + 2X^3 + 4X^2 + 2X + 3

    has no square root in R.

    I'm looking forward to your answers,

    Many Thanks,

    Fred

    PS : Sorry for my bad english
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  2. #2
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    Hi

    For the first one :
    Let A(a), B(b) and C(c)
    Let I be the middle of [AB]

    ABC being equilateral

    \left(\overrightarrow{AB}+\overrightarrow{CB}\righ  t)\cdot\overrightarrow{AC}=2\:\overrightarrow{IB}\  cdot\overrightarrow{AC}=0

    Therefore
    (b-a) + (b-c) = \pm i\:\sqrt{3}(c-a)

    Squaring gives
    (2b-a-c)^2 = -3(c-a)^2

    Expanding
    4b^2+a^2+c^2-4ab-4bc+2ac = -3a^2-3c^2+6ac

    4a^2+4b^2+4c^2 = 4ab + 4bc + 4ac

    a^2+b^2+c^2 = ab +  bc + ac
    Last edited by running-gag; March 19th 2009 at 02:08 PM. Reason: Correction of an error
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  3. #3
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    Thank you very much !
    Any idea for the second one ?

    Many thanks for this quick answer
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  4. #4
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    f(X)=X^4 + 2X^3 + 4X^2 + 2X + 3

    f'(X)=4X^3 + 6X^2 + 8X + 2

    f"(X)=12X^2 + 12X + 8 = 4(3X^2 + 3X + 2)

    f"(X) \geq 0 therefore f' is increasing over R and f' has one real root \alpha

    f' is negative over ]-\infty,\alpha] and positive over [\alpha,-\infty[

    f is decreasing over ]-\infty,\alpha] and increasing over [\alpha,-\infty[

    We just need to prove that f(\alpha) > 0

    f(X)=X^4 + 2X^3 + 4X^2 + 2X + 3

    f(X)=\left(\frac12 X + \frac14\right)f'(X) + \left(\frac54 X^2 + \frac12 X+\frac14\right)

    f(\alpha)=\frac54 \alpha^2 + \frac12 \alpha + \frac14 which is positive because the discriminant is negative
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  5. #5
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    Thanks !!! Many Thanks.
    But I don't understand how you found this ...
    Can you just detail how you found it ?

    Thanks


    Quote Originally Posted by running-gag View Post
    Hi

    For the first one :
    Let A(a), B(b) and C(c)
    Let I be the middle of [AB]

    ABC being equilateral

    \left(\overrightarrow{AB}+\overrightarrow{CB}\righ  t)\cdot\overrightarrow{AC}=2\:\overrightarrow{IB}\  cdot\overrightarrow{AC}=0
    Where does this comes from ?


    Therefore
    (b-a) + (b-c) = \pm i(c-a)
    And this ?

    Squaring gives
    (2b-a-c)^2 = -(c-a)^2

    Expanding
    4b^2+a^2+c^2-4ab-4bc+2ac = -a^2-c^2+2ac

    a^2+2b^2+c^2 = 2ab + 2 bc

    This is also true when changing a into b, a into c, ...

    a^2+2b^2+c^2 = 2ab + 2 bc
    2a^2+b^2+c^2 = 2ab + 2 ac
    a^2+b^2+2c^2 = 2bc + 2 ac

    Summation gives
    4a^2+4b^2+4c^2 = 4ab + 4 bc + 4ac
    a^2+b^2+c^2 = ab +  bc + ac
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  6. #6
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    Quote Originally Posted by Fred_ View Post
    Thank you very much !
    Any idea for the second one ?

    Many thanks for this quick answer
    I have corrected an error in my first post
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  7. #7
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    Yes I saw it.
    But I still don't understand where does that
    [tex]
    <br />
\left(\overrightarrow{AB}+\overrightarrow{CB}\righ  t)\cdot\overrightarrow{AC}=2\:\overrightarrow{IB}\  cdot\overrightarrow{AC}=0<br />

    and that <br /> <br />
(b-a) + (b-c) = \pm i\:\sqrt{3}(c-a)<br />

    comes from :s

    Thanks
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  8. #8
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    ABC being equilateral the heights and the medians are the same

    \left(\overrightarrow{AB}+\overrightarrow{CB}\righ  t)\cdot\overrightarrow{AC}=\left(\overrightarrow{A  I}+\overrightarrow{IB}+\overrightarrow{CI}+\overri  ghtarrow{IB}\right)\cdot\overrightarrow{AC}=2\:\ov  errightarrow{IB}\cdot\overrightarrow{AC}=0

    \overrightarrow{AB} affix is (b-a)
    \overrightarrow{CB} affix is (b-c)

    Therefore \overrightarrow{AB}+\overrightarrow{CB}=2\:\overri  ghtarrow{IB} affix is (b-a)+(b-c)

    2\:\overrightarrow{IB}\cdot\overrightarrow{AC}=0 means that there exists a rotation of angle \frac{\pi}{2} between (IB) and (AC)

    Therefore
    (b-a) + (b-c) = \pm i\:\alpha(c-a) where \alpha is a real

    ABC being equilateral IB = AC \frac{\sqrt{3}}{2}

    2\:IB = AC \:\sqrt{3}


    |(b-a) + (b-c)| = |\alpha|\: |c-a| = \sqrt{3} \:|c-a| therefore \alpha = \sqrt{3}
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  9. #9
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    Thank you very much for your previous reply's.

    I've got a last question :


    Quote Originally Posted by running-gag View Post
    f(X)=X^4 + 2X^3 + 4X^2 + 2X + 3

    f'(X)=4X^3 + 6X^2 + 8X + 2

    f"(X)=12X^2 + 12X + 8 = 4(3X^2 + 3X + 2) This is the derivative isn't it ?

    f"(X) \geq 0 therefore f' is increasing over R and f' has one real root \alpha Why are you taking the derivative of a derivative ?

    f' is negative over ]-\infty,\alpha] and positive over [\alpha,-\infty[

    f is decreasing over ]-\infty,\alpha] and increasing over [\alpha,-\infty[

    We just need to prove that f(\alpha) > 0

    f(X)=X^4 + 2X^3 + 4X^2 + 2X + 3

    f(X)=\left(\frac12 X + \frac14\right)f'(X) + \left(\frac54 X^2 + \frac12 X+\frac14\right)

    f(\alpha)=\frac54 \alpha^2 + \frac12 \alpha + \frac14 which is positive because the discriminant is negative
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  10. #10
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    Yes f"(X) is the second derivative of f, or the derivative of f'

    To find the variations of f, you need to know the sign of f' but since f' is a cubic function you need to take the second derivative f"
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  11. #11
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    MANY THANKS
    You're a genius
    T'es Francais ?
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  12. #12
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    Quote Originally Posted by Fred_ View Post
    MANY THANKS
    You're welcome

    Quote Originally Posted by Fred_ View Post
    You're a genius
    Surely not

    Quote Originally Posted by Fred_ View Post
    T'es Francais ?
    Oui
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  13. #13
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    Thanks VERY VERY MUCH
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