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Math Help - complex no. modulus

  1. #1
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    complex no. modulus

    p289 q20
    question: For any complex number z, let \bar z and |z| be its conjugate and modulus respectively. suppose z_1 and z_2 are complex numbers such that |z_1|=|z_2|=|z_1+z_2|=1
    (a)By using the property z \bar z = |z|^2, find the value of (z_1+z_2) (\frac 1 z_1 + \frac 1 z_2)
    (b)Using the result of (a) , show that z_1 ^2+z_1 z_2 +z_2^2 = 0. Hence deduce that z_1^3=z_2^3

    my working:
    =\frac {(z_1+z_2)^2} {z_1 z_2}
    =\frac {z_1^2 + z_2^2 + 2z_1 z_2 }{z_1 z_2}
    = 2 +\frac {z_1}{ z_2} + \frac{ z_2 }{z_1}
    stuck here and don't know how to do. thanks!
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by afeasfaerw23231233 View Post
    p289 q20
    question: For any complex number z, let \bar z and |z| be its conjugate and modulus respectively. suppose z_1 and z_2 are complex numbers such that |z_1|=|z_2|=|z_1+z_2|=1
    (a)By using the property z \bar z = |z|^2, find the value of (z_1+z_2) (\frac 1 z_1 + \frac 1 z_2)
    Remember that generally, if you have a complex number in a denominator, multiply by its conjugate

    \frac{1}{z_1}=\frac{\bar z_1}{z_1 \bar z_1}=\frac{\bar z_1}{|z_1|^2}=\boxed{\bar z_1}

    The same way, we get :
    \boxed{\frac{1}{z_2}=\bar z_2}

    \implies (z_1+z_2)\left(\frac{1}{z_1}+\frac{1}{z_2}\right)=  (z_1+z_2)(\bar z_1+\bar z_2)

    Noticing that \bar z_1+\bar z_2 is the conjugate of z_1+z_2, what can you conclude ?


    (it's equal to 1 - I put it because we need it for the following)
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  3. #3
    Moo
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    Quote Originally Posted by afeasfaerw23231233 View Post
    (b)Using the result of (a) , show that z_1 ^2+z_1 z_2 +z_2^2 = 0.

    my working:
    =\frac {(z_1+z_2)^2} {z_1 z_2}
    =\frac {z_1^2 + z_2^2 + 2z_1 z_2 }{z_1 z_2}

    [snip]
    Here, your working will help

    1=(z_1+z_2)\left(\frac{1}{z_1}+\frac{1}{z_2}\right  )=\frac {z_1^2 + z_2^2 + 2z_1 z_2 }{z_1 z_2}
    This is what you've written.

    \begin{aligned}1 &=\frac{z_1 z_2+{\color{blue}z_1^2+z_2^2+z_1 z_2}}{z_1 z_2} \\<br />
{\color{red}1} &={\color{red}1}+\frac{{\color{blue}z_1^2+z_2^2+z_  1 z_2}}{z_1 z_2} \end{aligned}

    \implies \frac{{\color{blue}z_1^2+z_2^2+z_1 z_2}}{z_1 z_2}=0

    What can you conclude ?


    Hence deduce that z_1^3=z_2^3
    a^3-b^3=(a-b)(a^2+ab+b^2)
    Last edited by Moo; May 19th 2008 at 01:56 AM. Reason: math tags missing
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  4. #4
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    thank you very much! next time can i treat \frac 1 z = \bar z as an identity and write it directly ?
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  5. #5
    Moo
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    Quote Originally Posted by afeasfaerw23231233 View Post
    thank you very much! next time can i treat \frac 1 z = \bar z as an identity and write it directly ?
    No, it's only true if |z|=1 ...

    But in general, when you have a complex number in the denominator, multiply it by its conjugate, because it'll always yield a real number, and it's easier to deal with it
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