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Math Help - Complex Number

  1. #1
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    Complex Number

    Can help me to solve this?

    1. Given that (1+5i)p-2q=3+7i, find the values of p and q when p and q are respectively a complex number and its conjugate.

    2. Given that the complex number z and its conjugate z* satisfy the equation zz*+2zi=12+6i, find the possible values of z.

    3. Given that z=x+yi and w= \frac {z+8i}{z-6}. If w is totally imaginary, show that x^2+y^2+2x-48=0.


    Answers:

    1. p=2-i, q=2+i
    2. 3-i, 3+3i
    Last edited by mr fantastic; June 18th 2009 at 05:08 AM. Reason: Questins moved from original thread
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  2. #2
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    Quote Originally Posted by cloud5 View Post
    3. Given that z=x+yi and w= \frac {z+8i}{z-6}. If w is totally imaginary, show that x^2+y^2+2x-48=0.
    We solved a problem like this one in this thread: http://www.mathhelpforum.com/math-he...ex-number.html


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  3. #3
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    Going a bit backwards here:

    Quote Originally Posted by cloud5 View Post
    2. Given that the complex number z and its conjugate z* satisfy the equation zz*+2zi=12+6i, find the possible values of z.
    zz*+2zi=12+6i

    If z = a + bi, then z* = a - bi:

    \begin{aligned}<br />
(a + bi)(a - bi) + 2(a + bi)i &= 12+6i \\<br />
a^2 + b^2 + 2ai -2b &= 12 + 6i \\<br />
(a^2 + b^2 - 2b) + 2ai &= 12 + 6i<br />
\end{aligned}

    Now equate the real and imaginary coefficients:
    \begin{aligned}<br />
a^2 + b^2 - 2b &= 12 \\<br />
2a &= 6 \\<br />
a &= 3 \\<br />
9 + b^2 - 2b &= 12 \\<br />
b^2 - 2b -3 &= 0 \\<br />
(b - 3)(b + 1) &= 0 \\<br />
b &= 3\;\;{\color{red}or} \\<br />
b &= -1<br />
\end{aligned}

    So the answers, in the form of a + bi, are
    3 + 3i and
    3 - i.

    Quote Originally Posted by cloud5 View Post
    1. Given that (1+5i)p-2q=3+7i, find the values of p and q when p and q are respectively a complex number and its conjugate.
    This looks like the same problem as #2. Substitute a + bi for p and a - bi for q. Set the real and imaginary coefficients equal to each other and solve for a and b.


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