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Math Help - Expressing complex numbers in polar form.

  1. #1
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    Expressing complex numbers in polar form.

    Hi, I'm not too sure with this one, can I get some help?

    If Z= 3 - 3i;what are Z, 1/Z and Z^{4} in polar form?

    I'm not sure if my answers are correct...

    Thanks, any help would be greatly appreciated!

    Cheers.
    Last edited by mr fantastic; March 25th 2011 at 03:28 AM. Reason: Title.
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  2. #2
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    Well, what are your answers?
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  3. #3
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    Mmmm... I got:

    z = 3\sqrt{2}\left ( cos(-\frac{\pi }{4}) + i sin(-\frac{\pi }{4})\right )

    z^{4} = 324\left ( cos(\pi) + i sin(\pi)\right )

    and

    \frac{1}{z} = \frac{\sqrt{2}}{6} ( cos(\frac{\pi }{4}) + i sin(\frac{\pi }{4})\right )
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  4. #4
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    Quote Originally Posted by rorosingsong View Post
    Mmmm... I got:

    z = 3\sqrt{2}\left ( cos(-\frac{\pi }{4}) + i sin(-\frac{\pi }{4})\right ) Mr F says: Correct.

    z^{4} = 324\left ( cos(\pi) + i sin(\pi)\right ) Mr F says: Argument is wrong. How did you get it?

    and

    \frac{1}{z} = \frac{\sqrt{2}}{6} ( cos(\frac{\pi }{4}) + i sin(\frac{\pi }{4})\right ) Mr F says: Correct.
    ..
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  5. #5
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    Oh whoops... silly mistake, methinks.

    z^{4} = 324 (cos(-\pi )+ i sin(-\pi ))

    Is that right?
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    Yes
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    Quote Originally Posted by rorosingsong View Post
    Oh whoops... silly mistake, methinks.

    z^{4} = 324 (cos(-\pi )+ i sin(-\pi ))

    Is that right?
    Of course, adding 2\pi to the argument does't change the value but gives 324(cos(\pi)+ i sin(\pi)), exactly what you had before.
    Last edited by mr fantastic; March 26th 2011 at 03:00 PM. Reason: Fixed math tag.
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  8. #8
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    Quote Originally Posted by HallsofIvy View Post
    Of course, adding 2\pi to the argument does't change the value but gives 324(cos(\pi)+ i sin(\pi)), exactly what you had before.
    True, but it is not the correct application of DeMoivre's Theorem.
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