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Math Help - Polar form of complex numbers

  1. #1
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    Polar form of complex numbers

    Hi

    here





    Last edited by mr fantastic; April 9th 2010 at 12:15 AM. Reason: Changed post title
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  2. #2
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    -4.88 - 3.17i.

    Notice that this is in the third quadrant.


    Its length is \sqrt{(-4.88)^2 + (-3.17)^2}

     = \sqrt{23.8144 + 10.0489}

     = \sqrt{33.8633}

     \approx 5.81922.


    The angle is in the third quadrant.

    So it's calculated as

    \theta = -180^{\circ} + \arctan{\frac{3.17}{4.88}}

    \theta \approx -146.993^{\circ}.


    Therefore:

    -4.88 - 3.17i \approx 5.81922 \,\textrm{cis}\,(-146.993^{\circ}).
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  3. #3
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    First you need to convert to polars.

    |1 - i\sqrt{3}| = \sqrt{1^2 + (-\sqrt{3})^2}

     = \sqrt{1 + 3}

     = \sqrt{4}

     = 2.


    Notice that the angle is in the fourth quadrant.

    So it's calculated by

    \theta = 2\pi - \arctan{\frac{\sqrt{3}}{1}}

     = 2\pi - \arctan{\sqrt{3}}

     = 2\pi - \frac{\pi}{3}

     = \frac{5\pi}{3}.



    So 1 - i\sqrt{3} = 2\,\textrm{cis}\,\left(\frac{5\pi}{3}\right)


    Therefore (1 - i\sqrt{3})^6 = \left[2\,\textrm{cis}\,\left(\frac{5\pi}{3}\right)\right]^6

     = 2^6\,\textrm{cis}\,\left(\frac{6\cdot 5\pi}{3}\right)

     = 64 \,\textrm{cis}\,\left(10\pi\right)

     = 64(\cos{10\pi} + i\sin{10\pi})

     = 64(1 + 0i)

     = 64.
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  4. #4
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    thanks but why you wrote ( cis ) no e
    Last edited by r-soy; April 8th 2010 at 12:23 PM.
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  5. #5
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    \textrm{cis}\,\theta is just a shorthand way of writing \cos{\theta} + i\sin{\theta}.

    As we know, \cos{\theta} + i\sin{\theta} = e^{i\theta}.


    So I could have written e, I just prefer using the \textrm{cis} notation, as I find it easier to convert between polars and cartesians.
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  6. #6
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    Hi prove it

    I want ask you if your answers are completed
    I mean if there any answer after your solving
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  7. #7
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    Yes, the solutions are complete with answers.
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