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Math Help - Complex numbers equation with conjugate.

  1. #1
    Newbie TriForce's Avatar
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    Complex numbers equation with conjugate.

    I'm having a bit problem solving the following equation:

    c * conjugate(z) = i where c = 3 + 5i and z is a complex number (a + bi)

    I tried following

    conjugate(z) = i/c
    But have no idea how to remove the conjugate from z.

    Any ideas?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TriForce View Post
    I'm having a bit problem solving the following equation:

    c * conjugate(z) = i where c = 3 + 5i and z is a complex number (a + bi)

    I tried following

    conjugate(z) = i/c
    But have no idea how to remove the conjugate from z.

    Any ideas?
    The conjugate of the conjugate of z is z itself. So what is the conjugate of i/c?

    -Dan
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    Quote Originally Posted by TriForce View Post
    I'm having a bit problem solving the following equation:

    c * conjugate(z) = i where c = 3 + 5i and z is a complex number (a + bi)
    If z=a+b\mathif{i} then \overline{z}=a-b\mathif{i}.
    So (3+5\mathif{i})(a-b\mathif{i})=(3a+5b)+(-3b+5a)\mathif{i}.

    So solve 3a+5b=0~\&~(-3b+5a)=1
    Last edited by Plato; April 29th 2011 at 12:59 PM.
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    Newbie TriForce's Avatar
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    Oh, thanks so much, i really need to practice more...
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    Quote Originally Posted by TriForce View Post
    @topsquark Well that's kind of the problem, i can't find anything about conjugating a complex number fraction...
    If you insist upon this: \frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}.

    Then \overline{~\frac{1}{z}~} = \frac{z}{{\left| z \right|^2 }}.

    But I find the system approach clearer.
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  6. #6
    Newbie TriForce's Avatar
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    Quote Originally Posted by Plato View Post
    So solve 3a+5b=0~\&~(-3b+5a)=1
    Could you please explain why you put real numbers = 0, and complex = 1, and where did complex i go?
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  7. #7
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    Quote Originally Posted by TriForce View Post
    Could you please explain why you put real numbers = 0, and complex = 1, and where did complex i go?
    If a+bi=3-4i then a=3~\&~b=-4.

    So if a+bi=i=0+i then a=0~\&~b=1.
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