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Math Help - complex number

  1. #1
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    complex number

    given z=x+yi and  w=(z-4i)/(z+3), where z is not -3, has no real parts, show that
    x^2+y^2+3x-4y=0

    what does no real parts mean?
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  2. #2
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    Quote Originally Posted by z1llch View Post
    given z=x+yi and  w=(z-4i)/(z+3), where z is not -3, has no real parts, show that Mr F says: You mean that the real part is equal to zero.
    x^2+y^2+3x-4y=0

    what does no real parts mean?
    w = \frac{x + iy - 4i}{x + iy + 3} = \frac{x + i(y - 4)}{(x + 3) + iy}.

    Now express w in the form w = a + ib in the usual way. Now put a = 0 since Re(w) = a.
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  3. #3
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    i multiply the denominator with its conjugate? i got a very very long equation in the end..
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  4. #4
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    Quote Originally Posted by z1llch View Post
    i multiply the denominator with its conjugate? Mr F says: Yes.

    i got a very very long equation in the end.. Mr F says: So start digging.
    ..
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  5. #5
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    Quote Originally Posted by z1llch View Post
    i multiply the denominator with its conjugate? i got a very very long equation in the end..
    Yes. Just keep going...
    \begin{aligned}<br />
w &= \frac{x + (y - 4)i}{(x + 3) + yi} \\<br />
&= \frac{x + (y - 4)i}{(x + 3) + yi} \times \frac{(x + 3) - yi}{(x + 3) - yi} \\<br />
&= \frac{x(x + 3) - xyi + (y - 4)(x + 3)i + y(y - 4)}{(x + 3)^2 + y^2} \\<br />
&= \frac{x^2 + 3x - xyi + xyi + 3yi - 4xi - 12i + y^2 - 4y}{(x + 3)^2 + y^2}<br />
\end{aligned}

    \begin{aligned}<br />
&= \frac{x^2 + 3x - xyi + xyi + 3yi - 4xi - 12i + y^2 - 4y}{(x + 3)^2 + y^2} \\<br />
&= \frac{x^2 + y^2 + 3x - 4y + 3yi - 4xi - 12i}{(x + 3)^2 + y^2} \\<br />
&= \frac{(x^2 + y^2 + 3x - 4y) + (3y - 4x - 12)i}{(x + 3)^2 + y^2} \\<br />
&= \frac{x^2 + y^2 + 3x - 4y}{(x + 3)^2 + y^2} + \left(\frac{3y - 4x - 12}{(x + 3)^2 + y^2}\right)i<br />
\end{aligned}

    Remember, this is w. The first big fraction is the real part, and the second big fraction is the imaginary part. You said that w has no real part, so the numerator of the first fraction must be zero, or x^2 + y^2 + 3x - 4y = 0.


    01


    EDIT: Oops, sorry to let the cat out of the bag so early.
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  6. #6
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    i got
    [x^2+y^2+3x-4y-4xi+3yi+12i] / [ (x+3)^2+y^2 ]
    no real parts meaning the real part is equal to zero? i'm a little bit confused.
    Last edited by z1llch; June 14th 2009 at 05:16 AM. Reason: typo error
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  7. #7
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    Remember, this is w. The first big fraction is the real part, and the second big fraction is the imaginary part. You said that w has no real part, so the numerator of the first fraction must be zero, or x^2 + y^2 + 3x - 4y = 0.


    01

    understood thanks a lot!
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  8. #8
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    Quote Originally Posted by z1llch View Post
    i got
    [x^2+y^2+3x-4-4xi+3yi+12i] / [ (x+3)^2+y^2 ]
    no real parts meaning the real part is equal to zero? i'm a little bit confused.
    The real part is \frac{x^2 + y^2 + 3x - 4{\color{red}y}}{(x + 3)^2 + y^2} and this will be zero if the numerator is equal to zero.
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  9. #9
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    It may be easier to use this form: \frac{{z - 4i}}{{z + 3}} = \frac{{z\overline z  - 4i~\overline z  + 3z - 12i}}{{\left| {z + 3} \right|^2}}.

    Note that \left| {z + 3} \right|^2  = \left( {x + 3} \right)^2  + y^2 and z\overline z  = x^2  + y^2 .
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