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Math Help - find the cube roots

  1. #1
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    find the cube roots

    I am learning how to find cube roots of a complex number and I wanted to see if I am doing this right. could someone let me know?

    find the cube roots of -8i:

    <br />
8(cos (3\pi/2)+ i sin(3\pi/2))<br />

    <br />
(8)^1/3[cos((3\pi/2)+2\pi(k)/3) + i sin((3\pi/2)+2\pi(k)/3)]<br />

    <br />
k= 0, 1, 2<br />

    <br />
2(cos(\pi/2) + i sin(\pi/2))<br />

    Is this good so far? I just done the first one to see if I done it right.
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  2. #2
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    Dear robasc,



    It's not trully correct. Check it cubing for k=1 or 2.
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  3. #3
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    Hello, robasc!

    I'm pretty sure you've got it right . . . I'll tidy-up a bit just to make sure.


    Find the cube roots of -8i

    z\;=\;8\left(\cos\frac{3\pi}{2}+ i\sin\frac{3\pi}{2}\right)


    z^{\frac{1}{3}} \;=\;8^{\frac{1}{3}}\bigg[\cos\left(\frac{\frac{3\pi}{2}+2\pi k}{3}\right) + i\sin\left(\frac{\frac{3\pi}{2}+2\pi k}{3}\right)\bigg] \;= . 2\bigg[\cos\left(\frac{\pi}{2} + \frac{2\pi}{3}k\right) + i\sin\left(\frac{\pi}{2} + \frac{2\pi}{3}k\right) \bigg]

    . . for k\:=\: 0, 1, 2




    2\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)

    Is this good so far? I just done the first one to see if I done it right.

    Looks good to me!

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  4. #4
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    do you mean check for the rest of them?


    <br />
2(cos(7\pi/6)+ i sin(7\pi/6))<br />

    <br />
2(cos(11\pi/6)+ i sin(11\pi/6))<br />
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  5. #5
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    Okey, than you only misspeled the parenthesis but though right.
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  6. #6
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    Thanks you friends!
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