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Math Help - Polar form of complex numbers

  1. #1
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    Polar form of complex numbers

    hi guys , i really need help in these two questions :
    Express each of the following in the form
    a + ib and also in the polar form r\,
    where the angle is the principal value.
    (a)
    (-sqrt(3)- i )^7

    (b)(1 + i)^3(sqrt(3) + i)^3
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  2. #2
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    Start by using this, where z = x+yi then (x+yi)^n = z^n = r^n(\cos n\theta +i\sin n\theta)

    given r = \sqrt{x^2+y^2} and \theta = \tan^{-1}\frac{y}{x}
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  3. #3
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    hi

    i did start with that but im notgetting the right answer i dont know what im doing wrong
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  4. #4
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    Quote Originally Posted by sandy View Post
    i did start with that but im notgetting the right answer i dont know what im doing wrong
    We won't know either unless you show all your working.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    We won't know either unless you show all your working.
    Hi ,i tried to do it is this correct?

    (-sqrt(3) -i)^7=2^7(cos 7pi/6 + i sin 7pi/6)

    and how do we do it with the form a+ib
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  6. #6
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    Quote Originally Posted by sandy View Post

    and how do we do it with the form a+ib
    That's exactly what we are doing. you have -\sqrt{3}-i giving a = -\sqrt{3} and b = -1


    Quote Originally Posted by sandy View Post
    Hi ,i tried to do it is this correct?

    (-sqrt(3) -i)^7=2^7(cos 7pi/6 + i sin 7pi/6)
    -\pi < \theta < \pi
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  7. #7
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    Hi ,i tried to do it is this correct?

    (-sqrt(3) -i)^7=2^7(cos 7pi/6 + i sin 7pi/6)

    and how do we do it with the form a+ib




    Quote Originally Posted by pickslides View Post
    Start by using this, where z = x+yi then (x+yi)^n = z^n = r^n(\cos n\theta +i\sin n\theta)



    given r = \sqrt{x^2+y^2} and \theta = \tan^{-1}\frac{y}{x}
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  8. #8
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    So do we expand a+ib to the power of 7 by to the power of 2 multiplied by 3 and then multiply that by a+ib again?




    Quote Originally Posted by pickslides View Post
    That's exactly what we are doing. you have -\sqrt{3}-i giving a = -\sqrt{3} and b = -1




    -\pi < \theta < \pi
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  9. #9
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    Quote Originally Posted by sandy View Post
    So do we expand a+ib to the power of 7 by to the power of 2 multiplied by 3 and then multiply that by a+ib again?
    i got it !!
    z^7= 64*sqrt(3) - i64
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