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Math Help - Complex Numbers ( Roots)

  1. #1
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    Complex Numbers ( Roots)

    Find the three cubic roots of -1

    possible answers
    a> cos π/3 + i sin π/3, cos 3π + i sin 3π , cos 2π/3 + i sin 2π/3
    b> cos π/3 + i sin π/3, cos π + i sin π, cos 5π/3 + i sin 5π/3
    c> cos π/2 + i sin π/2, cos π + i sin π, cos 5π/2 + i sin 5π/2
    d> cos π/4 + i sin π/4, cos π + i sin π, cos 5π/4 + i sin 5π/4

    Any help would be greatly appreciated
    please show step by step!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Daniels4691 View Post
    Find the three cubic roots of -1

    possible answers
    a> cos π/3 + i sin π/3, cos 3π + i sin 3π , cos 2π/3 + i sin 2π/3
    b> cos π/3 + i sin π/3, cos π + i sin π, cos 5π/3 + i sin 5π/3
    c> cos π/2 + i sin π/2, cos π + i sin π, cos 5π/2 + i sin 5π/2
    d> cos π/4 + i sin π/4, cos π + i sin π, cos 5π/4 + i sin 5π/4

    Any help would be greatly appreciated
    please show step by step!
    Hint:

    -1=e^{i(2n+1)\pi}, \ n\in \mathbb{Z}

    CB
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  3. #3
    MHF Contributor Amer's Avatar
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    let

    z^3=-1

    write -1 in exponential


    z^3=e^{i(\pi+2n\pi)} n take the values (1,2,3) since you want the cube square

    take the cube square for the both sides ......... the rest for you
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  4. #4
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    -1 = 1(\cos \pi + i \sin \pi)

    If z = r(\cos \theta + i \sin \theta), then the nth roots of a complex number would be in the form of

    \sqrt[n]{z} = \sqrt[n]{r}\left[\cos \left(\frac{\theta + 2\pi k}{n}\right) + i \sin \left(\frac{\theta + 2\pi k}{n}\right)\right] ,
    where k = 0, 1, 2, ... n - 1.

    In our case, n = 3, r = 1, and k will go from 0 to 2. The angle simplifies to
    \frac{\theta + 2\pi k}{n} = \frac{\pi + 2\pi k}{3} = \frac{\pi}{3} + \frac{2\pi k}{3}

    The three cube roots of -1 = 1(\cos \pi + i \sin \pi) are:
    \begin{aligned}<br />
z_1 &= \sqrt[3]{1}\left[\cos \left(\frac{\pi}{3} + 0\right) + i \sin \left(\frac{\pi}{3} + 0\right)\right] \\<br />
&= \cos\left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right) \\<br />
&= \frac{1}{2} + \frac{\sqrt{3}}{2}i<br />
\end{aligned}

    \begin{aligned}<br />
z_2 &= \sqrt[3]{1}\left[\cos \left(\frac{\pi}{3} + \frac{2\pi}{3}\right) + i \sin \left(\frac{\pi}{3} + \frac{2\pi}{3}\right)\right] \\<br />
&= \cos \pi + i \sin \pi \\<br />
&= -1<br />
\end{aligned}

    \begin{aligned}<br />
z_3 &= \sqrt[3]{1}\left[\cos \left(\frac{\pi}{3} + \frac{4\pi}{3}\right) + i \sin \left(\frac{\pi}{3} + \frac{4\pi}{3}\right)\right] \\<br />
&= \cos\left(\frac{5\pi}{3}\right) + i \sin \left(\frac{5\pi}{3}\right) \\<br />
&= \frac{1}{2} - \frac{\sqrt{3}}{2}i<br />
\end{aligned}

    ... or answer choice (B).


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