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Math Help - Help in justifying complex number question

  1. #1
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    Help in justifying complex number question

    Given Z1 and Z2 are two distinct complex numbers.
    If |Z1| = |Z2|, then Re  \frac{z1+ z2}{z1 - z2} = 0
    Is the above statement true or false?
    Show working to justify your answer.
    i managed to worked it out by letting z1= a+jb and z2= x+jy and then sub it in to work out and got true, not sure if i'm correct. Help is much appreciated
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  2. #2
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    Re: Help in justifying complex number question

    Quote Originally Posted by tempq1 View Post
    Given Z_1 and Z_2 are two distinct complex numbers.
    If |Z_1|=|Z_2|, then Re  \frac{z_1+ z_2}{z_1 - z_2} = 0
    Is the above statement true or false?
    Show working to justify your answer.
    i managed to worked it out by letting z1= a+jb and z2= x+jy and then sub it in to work out and got true, not sure if i'm correct. Help is much appreciated
    It is true. Messy to show.
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  3. #3
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    Lightbulb Re: Help in justifying complex number question

    Quote Originally Posted by tempq1 View Post
    Given Z1 and Z2 are two distinct complex numbers.
    If |Z1| = |Z2|, then Re  \frac{z1+ z2}{z1 - z2} = 0
    Is the above statement true or false?
    Show working to justify your answer.
    i managed to worked it out by letting z1= a+jb and z2= x+jy and then sub it in to work out and got true, not sure if i'm correct. Help is much appreciated
    Representing z1 & z2 in exponential form we have:

    \text{z1}\text{:=}r e^{i \theta }
    \text{z2}\text{:=}r e^{i \omega }

    Then: \Re\left(\frac{\text{z1}+\text{z2}}{\text{z1}-\text{z2}}\right) = \left(\sin ^2(\theta )+\cos ^2(\theta )-\sin ^2(\omega )-\cos ^2(\omega )\right)/\[-2 (-1+\text{Cos}(\theta -\omega ))]= 0
    Last edited by MaxJasper; September 3rd 2012 at 01:25 PM.
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    Re: Help in justifying complex number question

    Quote Originally Posted by MaxJasper View Post
    Representing z1 & z2 in exponential form we have:
    \text{z_1}\text{:=}r e^{i \theta }
    \text{z_2}\text{:=}r e^{i \omega }
    Then: \Re\left(\frac{\text{z1}+\text{z2}}{\text{z1}-\text{z2}}\right) = \left(\sin ^2(\theta )+\cos ^2(\theta )-\sin ^2(\omega )-\cos ^2(\omega )\right)/\[-2 (-1+\text{Cos}(\theta -\omega ))]= 0
    You can note that:
    \frac{e^{i \theta }+ e^{i \omega}}{e^{i \theta }- e^{i \omega}}\cdot\frac{e^{-i\theta }- e^{-i\omega}}{e^{-i \theta }- e^{-i \omega}}=\frac{e^{i(\omega-\theta)} - e^{i(\theta- \omega)}}{|e^{i \theta }- e^{i \omega}|^2} and \text{Re}\left(e^{i(\omega-\theta)} - e^{i(\theta- \omega)}\right)=0
    I said it is messy.
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  5. #5
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    Re: Help in justifying complex number question

    Hello, tempq1!

    \text{Given: }\,z_1\text{ and }z_2\text{ are two distinct complex numbers.}

    \text{If }|z_1| = |z_2|,\,\text{ then }Re\left(\frac{z_1+z_1}{z_1-z_2}\right) \:=\:0

    \text{Is the above statement true or false?}

    \text{Let: }\:\begin{Bmatrix}z_1 \:=\:a+bj \\ z_2 \:=\:c+dj \end{Bmatrix}


    \text{We are told: }\:|z_1| \,=\,|z_2| \quad\Rightarrow\quad \sqrt{a^2+b^2} \,=\,\sqrt{c^2+d^2}

    . . . . a^2+b^2 \:=\:c^2+d^2 \quad\Rightarrow\quad (a^2+b^2) - (c^2+d^2) \:=\:0 .[1]


    \text{Consider: }\:\frac{z_1+z_2}{z_1-z_2} \;=\;\frac{(a+c) + (b+d)j}{(a-c) + (b-d)j}

    \text{Rationalize: }\:\frac{(a+c) + (b+d)j}{(a-c)+(b-d)j}\cdot\frac{(a-c)-(b-d)j}{(a-c)-(b-d)j}

    . . =\;\frac{(a+c)(a-c) - (a+c)(b-d)j + (a-c)(b+d)j + (b+d)(b-d)}{(a-c)^2 - (b-d)^2}

    . . =\;\frac{(a^2-c^2) -(ab-ad+bc-cd)j + (ab + ad - bc - cd)j + (b^2-d^2)}{(a-c)^2 - (b-d)^2}

    . . =\;\frac{(a^2-c^2 + b^2-d^2) + 2(ad - bd)j}{(a-c)^2 - (b-d)^2}

    . . =\;\frac{(a^2+b^2)-(c^2+d^2)}{(a-c)^2 - (b-d)^2} + \frac{2(ad-bc)}{(a-c)^2 - (b-d)^2}j


    \text{The real component is: }\:\frac{(a^2+b^2) - (c^2+d^2)}{(a-c)^2 - (b-d)^2}

    . . which, according to [1], equals 0.

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    Re: Help in justifying complex number question

    Quote Originally Posted by Soroban View Post
    Hello, tempq1!
    \text{Let: }\:\begin{Bmatrix}z_1 \:=\:a+bj \\ z_2 \:=\:c+dj \end{Bmatrix}
    \text{We are told: }\:|z_1| \,=\,|z_2| \quad\Rightarrow\quad \sqrt{a^2+b^2} \,=\,\sqrt{c^2+d^2}
    . . . . a^2+b^2 \:=\:c^2+d^2 \quad\Rightarrow\quad (a^2+b^2) - (c^2+d^2) \:=\:0 .[1]
    \text{Consider: }\:\frac{z_1+z_2}{z_1-z_2} \;=\;\frac{(a+c) + (b+d)j}{(a-c) + (b-d)j}
    \text{Rationalize: }\:\frac{(a+c) + (b+d)j}{(a-c)+(b-d)j}\cdot\frac{(a-c)-(b-d)j}{(a-c)-(b-d)j}
    . . =\;\frac{(a+c)(a-c) - (a+c)(b-d)j + (a-c)(b+d)j + (b+d)(b-d)}{(a-c)^2 - (b-d)^2}
    . . =\;\frac{(a^2-c^2) -(ab-ad+bc-cd)j + (ab + ad - bc - cd)j + (b^2-d^2)}{(a-c)^2 - (b-d)^2}
    . . =\;\frac{(a^2-c^2 + b^2-d^2) + 2(ad - bd)j}{(a-c)^2 - (b-d)^2}
    . . =\;\frac{(a^2+b^2)-(c^2+d^2)}{(a-c)^2 - (b-d)^2} + \frac{2(ad-bc)}{(a-c)^2 - (b-d)^2}j
    \text{The real component is: }\:\frac{(a^2+b^2) - (c^2+d^2)}{(a-c)^2 - (b-d)^2}
    . . which, according to [1], equals 0.
    Well I did say it is messy, but not that messy.
    That is a notation nightmare.
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