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Thread: Complex numbers question

  1. #1
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    Complex numbers question

    Question:
    if the complex number $\displaystyle x + iy $ denoted by z then the complex conjugate number $\displaystyle x-iy $denoted by $\displaystyle a*$.
    a) Express $\displaystyle \left |\ z^* \right | $ and $\displaystyle arg(z^*) $ in terms of $\displaystyle \left |\ z \right | $ and $\displaystyle arg(z) $

    b)if a, b, and c are real numbers prove that if $\displaystyle az^2+bz+c=0$
    then $\displaystyle a(z^*)^2 + b{z^*}+ c=0 $

    c)if p and q are complex numbers and $\displaystyle q \neq 0 $, prove that $\displaystyle \left ( \frac{p}{q} \right )^{*}=\frac{p^*}{q^*} $

    My attempt

    a) $\displaystyle \left |\ z^* \right | = \left |\ x-iy \right | = \sqrt{x^2+(-y)^2} = \sqrt{x^2+y^2} $
    therefore $\displaystyle \left |\ z^* \right | = \left |\ z \right | = \sqrt{x^2+y^2} $
    $\displaystyle arg(z) = \Theta $ and $\displaystyle arg(z^*) = - \Theta$
    therefore $\displaystyle arg(z^*) = - arg(z) $
    [/FONT][/COLOR]
    ) $\displaystyle a(z^*)^2 + b{z^*}+ c=0 $[
    sub $\displaystyle z^* = x-iy $ into the equation
    $\displaystyle a(x-iy)^2+b(x-iy) +c=0 $
    $\displaystyle a(x^2-y^2)+bx+c -iy(2ax-b)=0 $
    I am stuck here
    plus I dont know how to approach part c please can some one help me thank you
    Last edited by bigmansouf; Apr 2nd 2019 at 05:33 PM.
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  2. #2
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    Re: Complex numbers question

    b) $a z^2 + b z + c = 0$

    $z = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

    suppose $Im(z) \neq 0$

    $z = \dfrac{-b \pm i \sqrt{4ac-b^2}}{2a} = -\dfrac{b}{2a} \pm i \dfrac{\sqrt{4ac-b^2}}{2a}$

    It's seen by inspection that both $z$ and $z^*$ are roots

    to cheat a bit less

    $az^2 + bz+c = 0 \Rightarrow a(x^2-y^2)+bx + c = 0,~2axy+b y = 0$ (note $c \in \mathbb{R}$)

    $a (z^*)^2 + b z^* + c = a (x^2 - y^2 )+bx + c + i(-2axy - by) = 0 -i(0) = 0$
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  3. #3
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    Re: Complex numbers question

    c) use $z = |z|e^{i \arg(z)}$ for $p$ and $q$


    Spoiler:

    let

    $p = |p|e^{i \arg(p)},~q = |q|e^{i \arg(q)}$

    $\left(\dfrac p q \right)^* = \\

    \left(\dfrac{|p|}{|q|}e^{i( \arg(p)-\arg(q))}\right)^* = \\

    \dfrac{|p|}{|q|} e^{i(-\arg(p) - (-\arg(q)))} = \\

    \dfrac{|p|e^{-i\arg(p)}}{|q|e^{-i\arg(q)}} =\\

    \dfrac{p^*}{q^*}$
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