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Math Help - Solving complex Equation

  1. #1
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    Solving complex Equation

    Q1.
    Solve equa for z=a+bi where a and b are real numbers.
    (z-zbar)^2=2z+6i

    The ans is -18-3i but do not know how they get this.
    Last edited by heatly; July 27th 2010 at 05:17 PM.
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  2. #2
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    z-\overline{z}=2\text{Im}(z)i
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  3. #3
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    for z=a+bi

    (z-\bar{z})^2=2z+6i is

    (a+bi-(a-bi))^2=2(a+bi)+6i

    Now expand this expression out.
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    Thanks for that ,But i did that and and got -2b^2=a+bi+3i
    The ans is supposed to be -18-3i
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  5. #5
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    I get

    (a+bi-(a-bi))^2=2(a+bi)+6i

    (a+bi-a+bi)^2=2a+2bi+6i

    (2bi)^2=2a+2bi+6i

    4b^2(-1)=2a+2bi+6i

    -4b^2=2a+2bi+6i

    -4b^2-2a-2bi=0+6i

    Now equating coefficients

    -2b=6 , -4b^2-2a=0\implies b = -3 ,-4(-3)^2-2a=0

    \implies -4\times 9-2a=0\implies -36-2a=0, a=-18
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  6. #6
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    Thanks for that.
    I have another one Iam having difficulty with too iff your interested.

    z^2 = 3i
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    z^2 = 3i

    (a+bi)^2 = 0+3i

    (a+bi)(a+bi) = 0+3i

    a^2+2abi+(bi)^2 = 0+3i

    a^2+2abi-b^2 = 0+3i

    a^2-b^2 = 0, 2abi=3i

    Can you solve it form here?
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  8. #8
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    3i = 3\left(\cos{\frac{\pi}{2}} + i\sin{\frac{\pi}{2}}\right)

     = 3e^{\frac{\pi i}{2}}.


    So z^2 = 3i

    z^2 = 3e^{\frac{\pi i}{2}}

    z = \left(3e^{\frac{\pi i}{2}}\right)^{\frac{1}{2}}

    z = \pm\sqrt{3}e^{\frac{\pi i}{4}}

    z = \pm\sqrt{3}\left(\cos{\frac{\pi}{4}} + i\sin{\frac{\pi}{4}}\right)

    z = \pm\sqrt{3}\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}i}{2}\right)

    z = \frac{\sqrt{6}}{2} + \frac{\sqrt{6}i}{2} or z = -\frac{\sqrt{6}}{2} - \frac{\sqrt{6}i}{2}.
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  9. #9
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    'pickslides' Thats actually where I got up to and then I got stuck in finding a and b.
    if you can go a bit futher would be good.
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  10. #10
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    a^2-b^2=0 \implies a^2=b^2 \implies a=b \implies 2b^2= 3
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  11. #11
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    Thanks,the ans in my text book says..... +-sqrt(3/2)*(1+i)
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    Quote Originally Posted by pickslides View Post
    a^2-b^2=0 \implies a^2=b^2 \implies a=b \implies 2b^2= 3
    Actually a^2 = b^2 \implies |a| = |b|...
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  13. #13
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    Quote Originally Posted by heatly View Post
    Thanks,the ans in my text book says..... +-sqrt(3/2)*(1+i)
    Which is the same as what I put, I however have rationalised the denominator.
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