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Math Help - complex roots

  1. #1
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    complex roots

    Can any help me on where I should start to find all 8 solutions to x^8=3-4I . And also not exactly sure how to find the magnitude of each solution .
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    Re: complex roots

    Quote Originally Posted by matthewporter1965 View Post
    Can any help me on where I should start to find all 8 solutions to x^8=3-4I . And also not exactly sure how to find the magnitude of each solution .
    \displaystyle \begin{align*} |3 - 4i| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \end{align*} and \displaystyle \begin{align*} \arg{(3 - 4i)} = -\arctan{\frac{4}{3}} \end{align*}, so \displaystyle \begin{align*} 3 - 4i = 5e^{-i\arctan{\frac{4}{3}}} \end{align*}.

    So if \displaystyle \begin{align*} x^8 = 3 - 4i \end{align*}, then

    \displaystyle \begin{align*} x^8 &= 5e^{-i\arctan{\frac{4}{3}}} \\ x &= \left(5e^{-i\arctan{\frac{4}{3}}}\right)^{\frac{1}{8}} \\ x &= \sqrt[8]{5}\,e^{-\frac{1}{8}i\arctan{\frac{4}{3}}} \end{align*}

    This gives you one of the eighth roots. The others are all evenly spaced around a circle, so have the same magnitude and are all separated by \displaystyle \begin{align*} \frac{2\pi}{8} = \frac{\pi}{4} \end{align*}. Can you evaluate the rest? Remember that \displaystyle \begin{align*} -\pi < \arg{x} \leq \pi \end{align*}.
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    Re: complex roots

    I think I got it but not sure yet. This problem is for a math 142 class, and the way you show is different from anything we have done in class. To find complex roots of something like 3-4I we would first write in polar form z=x+yi =r[cos(θ)+isin(θ)] . I would first find r=√(x^2+y^2) which equals 5. Then find θ since x is positive and y is negative it is in quadrant 3 sin(θ)=y/r =-4/5 . this is where im stuck because im not sure how to find θ in quad 3 whose sin is -4/5. do i convert -4/5 into degree by multiply by 180/π. I also know the magnitude is the same for all 8 solutions but i dont know how to.prove it with a complex in polar form.isnt the magnitude of a complex z=x-iy just √(x^2+y^2) which is 5 in this case. How would i find magnitude for each.solution if they are in polar form.
    Last edited by matthewporter1965; June 6th 2012 at 10:07 AM.
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    Re: complex roots

    The problem states find all 8 solutions t x^8=3-4i. AND prove that they all have the same magnitude by finding the magnitude of each solution. And then plot each solution.
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    Re: complex roots

    Quote Originally Posted by matthewporter1965 View Post
    I think I got it but not sure yet. This problem is for a math 142 class, and the way you show is different from anything we have done in class. To find complex roots of something like 3-4I we would first write in polar form z=x+yi =r[cos(θ)+isin(θ)] . I would first find r=√(x^2+y^2) which equals 5. Then find θ since x is positive and y is negative it is in quadrant 3 sin(θ)=y/r =-4/5 . this is where im stuck because im not sure how to find θ in quad 3 whose sin is -4/5. do i convert -4/5 into degree by multiply by 180/π. I also know the magnitude is the same for all 8 solutions but i dont know how to.prove it with a complex in polar form.isnt the magnitude of a complex z=x-iy just √(x^2+y^2) which is 5 in this case. How would i find magnitude for each.solution if they are in polar form.
    First, 3-4i\in IV not the third.

    Next, 3-4i has no nice polar form. The best you can do is \phi  = \arctan \left( {\frac{{ - 4}}{3}} \right)\left( {\frac{{180}}{\pi }} \right)^o

    Each of the roots has absolute value \sqrt[8]{5} and they are \sqrt[8]{5}\left( {\cos \left( {\frac{{\phi  }}{8}+ 45{k^o}} \right) + i\sin \left( {\frac{{\phi }}{8}}+ 45{k^o} \right)} \right),\;k = 0,1, \cdots ,7
    Last edited by Plato; June 6th 2012 at 10:32 AM.
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    Re: complex roots

    Sorry I meant 4 brain fart I guess. I have lost many hours of sleep over this. Im glad i found this forum. I couldn't find a tutor on campus that could help. If anything confused me even more. I'm going to do it again thanks for the help .
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  7. #7
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    Re: complex roots

    I got it thanks all who helped
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