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Math Help - Complexe

  1. #1
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    Complexe

    sorry Guys but my math teachers gave 30 problems For us he thinks that i don't have only Math to do it , i don't have a lot of times i'll wake up today to study lol.

    anyway i stopped in 3 question - Hard For me

    z is a complex number, prove if
    {1 - iz
    -------} = 1, then z is real
    1 + iz


    my second question:
    that is to say A and B are two points of the plan of respective affixes zA, zB
    will
    a.Prove that [zA-ZB] = AB
    B. find all the numbers complexes z such as the triangle whose tops are the images of 1, Z, z^2 is equilateral


    and my last question:

    Z and z' are two complex numbers unspecified.
    Show that [z+z'] ^2 + {z-z'] ^2 = 2 {[Z} 2 + {z'} ^2}


    P.s: i don't know how to write them so [ ] =




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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by iceman1 View Post
    z is a complex number, prove if
    {1 - iz
    -------} = 1, then z is real
    1 + iz
    I'll do better than that:
    If
    \frac{1 - iz}{1 + iz} = 1

    1 - iz = 1 + iz

    Thus
    -z = z

    Thus z = 0. Yeah, it's a real number!

    -Dan
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  3. #3
    MHF Contributor red_dog's Avatar
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    1). I think it's about \displaystyle\left|\frac{1-iz}{1+iz}\right|=1.
    Then |1-iz|=|1+iz|. (1)
    Let z=x+yi\Rightarrow 1-iz=(1+y)-ix, \ 1+iz=(1-y)+ix.
    Then the equality (1) becomes \sqrt{(1+y)^2+x^2}=\sqrt{(1-y)^2+x^2}.
    Square both sides: 1+2y+y^2+x^2=1-2y+y^2+x^2\Rightarrow 4y=0\Rightarrow y=0.
    So, z=x\in\mathbf{R}.

    2)
    a) Let A(x_A,y_B),B(x_B,y_B)
    Then z_A=x_A+y_Ai, \ z_B=x_B+y_Bi.
    |z_A-z_B|=|x_A-x_B+(y_A-y_B)i|=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}=AB.

    b) Let A(1),B(z),C(z^2)
    Then ABC is equilateral \Leftrightarrow AB=AC=BC\Leftrightarrow |z-1|=|z^2-1|=|z^2-z| (2)
    The equalities (2) can be written as |z-1|=|z-1|\cdot |z+1|=|z|\cdot |z-1|.
    Suppose that z\neq 1, otherwise the triangle degenerates to a point. Then |z-1|\neq 0, so we can divide all three members by |z-1|.
    We get
    1=|z+1|=|z|.
    Let z=x+yi.
    Then \left\{\begin{array}{ll}(x^2+y^2=1\\(x+1)^2+y^2=1\  end{array}\right..
    Solving the system, we get \displaystyle x=-\frac{1}{2}, \ y=\pm\frac{\sqrt{3}}{2}.

    3) We'll use the relation z\cdot\overline{z}=|z|^2.
    |z+z'|^2+|z-z'|^2=(z+z')(\overline{z+z'})+(z-z')(\overline{z-z'})=(z+z')(\overline{z}+\overline{z'})+(z-z')(\overline{z}-\overline{z'})=
    =z\cdot\overline{z}+z\cdot\overline{z'}+z'\cdot\ov  erline{z}+z'\cdot\overline{z'}+z\cdot{z}-z\cdot\overline{z'}-z'\cdot\overline{z}+z'\cdot\overline{z'}=
    =2(z\cdot\overline{z}+z'\cdot\overline{z'})=2(|z|^  2+|z'|^2)
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