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Math Help - DeMoivre's Theorem

  1. #1
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    DeMoivre's Theorem

    My question is when we use this theorem

    z=4(cos60+isin60) * 4(cos 60 +i sin 60)
    z^2= 16 (cos 120 +isin 120) *4(cos 60 +i sin 60)
    z^3= 64 (cos 180 + i sin 180)

    Why do we just add the 60 degrees to each
    why do we not distribute? to get 16+ 8 cos 60 +8 i sin 60 + cos 60^2 + (i sin 60)^2

    if that is correct how does that become line 2? thanks

    Do we add the values because of the sum and difference formulas?
    Last edited by algebraisabeast; March 29th 2009 at 04:20 PM. Reason: Think figured out
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  2. #2
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    Quote Originally Posted by algebraisabeast View Post
    My question is when we use this theorem

    z=4(cos60+isin60) * 4(cos 60 +i sin 60)
    z^2= 16 (cos 120 +isin 120) *4(cos 60 +i sin 60)
    z^3= 64 (cos 180 + i sin 180)

    Why do we just add the 60 degrees to each
    why do we not distribute? to get 16+ 8 cos 60 +8 i sin 60 + cos 60^2 + (i sin 60)^2

    if that is correct how does that become line 2? thanks

    Do we add the values because of the sum and difference formulas?
    Please see here,
    http://en.wikipedia.org/wiki/De_Moivre%27s_formula

    Also,

    Z = 4 (\cos 60 + i \sin 60)

    Z^2 = 4^2 (\cos 60 + i \sin 60)^2

    Z^2 = 16 (\cos 60 + i \sin 60)(\cos 60 + i \sin 60)

    Z^2 = 16 (\cos^2 60 + i^2 \sin^2 60+i\sin 60. \cos 60 + i \cos 60.\sin 60)

    Z^2 = 16 [(\cos^2 60 - \sin^2 60)+i(\sin 60. \cos 60 + \cos 60.\sin 60)]

    Z^2 = 16 [\cos (60+60)+i\sin (60+ 60)]

    Z^2 = 16 (\cos 120+i\sin 120)


    Also,

    Z^3 = 4^3 (\cos 60 + i \sin 60)^3

    Z^3 = 4^2 (\cos 60 + i \sin 60)^2 \times 4(\cos 60 + i \sin 60)

    Z^3 = 64 (\cos 120 + i \sin 120)(\cos 60 + i \sin 60) [substitute the value of Z^2, proved above.]

    Z^3 = 64 (\cos 120\cos 60 + i^2 \sin 120. \sin 60+i\sin 120. \cos 60 + i \cos 120.\sin 60)

    Z^3 = 64 [(\cos 120\cos 60 - \sin 120. \sin 60)+i(\sin 120. \cos 60 + \cos 120.\sin 60)]

    Z^3 = 64 [\cos (120+60)+i\sin (120+ 60)]

    Z^3 = 64 (\cos 180+i\sin 180)

    Formula used:

    \cos A. \cos B -\sin A . \sin B = \cos (A+B)\;\;\;and\;\;\;<br />
\sin A. \cos B +\cos A . \sin B = \sin (A+B)
    Last edited by Shyam; March 29th 2009 at 06:02 PM.
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