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Math Help - Complex number

  1. #1
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    Complex number

    Find the complex number w, lying in the first quadrant, and having the largest possible real part, which satisfies the equation:

    w^6=-6-6*I*sqrt(3)

    write your answer for w in polar form

    Can u please help me with this question? the answer is 12^(1/6)*exp(2/9*I*Pi)

    Thanks!
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  2. #2
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    Quote Originally Posted by Rine198 View Post
    Find the complex number w, lying in the first quadrant, and having the largest possible real part, which satisfies the equation: w^6=-6-6*I*sqrt(3); the answer is 12^(1/6)*exp(2/9*I*Pi)
    If W=-6-6\sqrt{3}\mathfi{i} then |W|=12 and \text{Arg}(W)=-\frac{2\pi}{3}.
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  3. #3
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    yeah, i got that part, but i still can't seem to get the answer right, so to find the root of w, i divided the argument by 6, but i manage to get -1*Pi/9, and that is not what the answer states
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    Quote Originally Posted by Rine198 View Post
    yeah, i got that part, but i still can't seem to get the answer right, so to find the root of w, i divided the argument by 6, but i manage to get -1*Pi/9, and that is not what the answer states
    But the question states that the answer is in the first quadrant.
    \dfrac{2\pi}{6}=\dfrac{3\pi}{9} so add.
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  5. #5
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    i don't get how u got 2*Pi/6 as well as 3*Pi/9
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  6. #6
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    Quote Originally Posted by Rine198 View Post
    i don't get how u got 2*Pi/6 as well as 3*Pi/9
    Those six sixth roots are located on a circle centered at 0.
    They are equally spaced, separated by \dfrac{2\pi}{6}.
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  7. #7
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    Quote Originally Posted by Rine198 View Post
    Find the complex number w, lying in the first quadrant, and having the largest possible real part, which satisfies the equation:

    w^6=-6-6*I*sqrt(3)

    write your answer for w in polar form

    Can u please help me with this question? the answer is 12^(1/6)*exp(2/9*I*Pi)

    Thanks!

    Did you mean to write w^6=6-6\sqrt{3}i above? If so you want the 6-th root

    of 6-6\sqrt{3}i=12\,cis(-\pi/3+2k\pi)=12e^{-\pi i/3+2k\pi i}=12e^{5\pi i/3+2k\pi i} with the

    largest real part , so using de Moivre's formula we get:

    \displaystyle{w_k=\left(12e^{5\pi i/3+2k\pi i}\right)^{1/6}=12^{1/6}e^{\frac{5\pi i}{18}+\frac{2k\pi i}{6}}=12^{1/6}e^{\frac{\pi i}{18}(5+6k)}\,,\,\,k=0,1,2,...,5 .

    Now just write down each root above and check which one has the largest real part. For example,

    Re(w_0)=Re\left(12^{1/6}e^{5\pi i/18}\right)=12^{1/6}\cos\left(5\pi/18)\cong 0.9726 ,

    Re(w_1)=12^{1/6}\cos\left(11\pi/18)\cong -0.5175 , etc.

    Hint: Of course, you should check only those roots with angle in \left(-\pi/2\,,\,\pi/2\right) radians (why?)

    Tonio
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