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Math Help - complex numbers in a fourth degree equation

  1. #1
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    complex numbers in a fourth degree equation

    Problem:
    Solve the equation z^4=4+4i and graph the solutions in an argand diagram.


    I'm leaning towards a solution where I use:

    |z|(cos(0)+isin(0))
    z^4=r^4(\cos(4v)+i\sin(4v))

    But it keeps messing up. Help is heavily appriciated.
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  2. #2
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    Quote Originally Posted by Greenb View Post
    Problem:
    Solve the equation z^4=4+4i and graph the solutions in an argand diagram.


    I'm leaning towards a solution where I use:

    |z|(cos(0)+isin(0))
    z^4=r^4(\cos(4v)+i\sin(4v))

    But it keeps messing up. Help is heavily appriciated.
    4 + 4i = 4 \sqrt{2} \text{cis} \, \left( \frac{\pi}{4} + 2 n \pi\right) where n is an integer.

    Compare this with z^4 = r^4 \text{cis} \, (4 \theta) to get the value of r and the values of \theta necesary to write down the four distinct solutions for z = r \text{cis} \, \theta.
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  3. #3
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    Ah you got \frac{\pi}{4} from \tan(\theta)=\frac{b}{a}
    Last edited by Greenb; January 31st 2009 at 03:29 AM. Reason: Ah got it
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  4. #4
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    Quote Originally Posted by Greenb View Post
    I can follow you all the way except how you get the angle \frac{\pi}{4}
    Draw an argand diagram showing 4 + 4i ...... It's in the first quadrant. And \text{Arg} (4 + 4i) = \tan^{-1} \frac{4}{4} = \tan^{-1} 1 ....
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  5. #5
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    Somehow I keep getting reversed solutions..

    Mine are;
    z_1=1.51+0.30i
    z_2=-0.30+1.51i
    z_3=-1.51-0.30i
    z_4=0.30-1.51i

    Answer Key:
    z_1=0.30+1.51i
    z_2=-1.51+0.30i
    z_3=-0.30-1.51i
    z_4=1.51-0.30i

    Is the equation wrong?

    \sqrt[4]{\sqrt{2}*4}(\cos(\frac{\pi+n*2\pi}{4})+i\sin(\fra  c{\pi+n*2\pi}{4}))
    where n=0,1,2,3
    Last edited by Greenb; January 31st 2009 at 04:58 AM. Reason: Missed a +
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  6. #6
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    It must be the angle seeing as the answer is right only reversed, in my opinion. Maybe I'm substituting it wrong?
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  7. #7
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    Quote Originally Posted by Greenb View Post
    Somehow I keep getting reversed solutions..

    Mine are;
    z_1=1.51+0.30i
    z_2=-0.30+1.51i
    z_3=-1.51-0.30i
    z_4=0.30-1.51i

    Answer Key:
    z_1=0.30+1.51i
    z_2=-1.51+0.30i
    z_3=-0.30-1.51i
    z_4=1.51-0.30i

    Is the equation wrong?

    \sqrt[4]{\sqrt{2}*4}(\cos(\frac{\pi+n*2\pi}{4})+i\sin(\fra  c{\pi+n*2\pi}{4}))
    where n=0,1,2,3
    It's not hard to see that solutions in the answer key are wrong. Simply raise them to the power of 4. I get 4 - 4i as the result .....

    Your solutions are the correct fourth roots of 4 + 4i.

    By the way you have a typo in you 'equation', it should be \sqrt[4]{\sqrt{2}*4}\left(\cos \left(\frac{\frac{\pi}{4}+n*2\pi}{4}\right)+i\sin\  left(\frac{\frac{\pi}{4}+n*2\pi}{4}\right)\right)
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  8. #8
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    Darn! Your right, but seriously this has been bugging me all day.... Thanks for your persistence with me hehe
    Last edited by mr fantastic; January 31st 2009 at 12:30 PM. Reason: m --> r
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