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Math Help - Express the eq in a+ib form

  1. #1
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    Express the eq in a+ib form

    Express z = \frac{(cos \theta + i sin \theta)^3(sin  \theta + i cos \theta)}{(cos2 \theta - i sin2 \theta)} in the form a+ib

    Solution :

    z = \frac{(cos \theta + i sin \theta)^3(sin  \theta + i cos \theta)}{(cos2 \theta - i sin2 \theta)}

     = \frac{(cos \theta + i sin \theta)^3(sin  \theta + i cos \theta)}{(cos \theta + i sin \theta)^{-2}}


     = (cos \theta + i sin \theta)^5(sin  \theta + i cos \theta)

    Stuck !!!!!
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  2. #2
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    Quote Originally Posted by TomJerry View Post
    Express z = \frac{(cos \theta + i sin \theta)^3(sin \theta + i cos \theta)}{(cos2 \theta - i sin2 \theta)} in the form a+ib

    Solution :

    z = \frac{(cos \theta + i sin \theta)^3(sin \theta + i cos \theta)}{(cos2 \theta - i sin2 \theta)}

     = \frac{(cos \theta + i sin \theta)^3(sin \theta + i cos \theta)}{(cos \theta + i sin \theta)^{-2}}


     = (cos \theta + i sin \theta)^5(sin \theta + i cos \theta)

    Stuck !!!!!
    Assuming your last line is correct (and I haven't bothered to check) then you can use deMoivre's Theorem to write ( \cos \theta + i \sin \theta )^5 = \cos (5 \theta) + i \sin  (5 \theta). Now expand  (\cos (5 \theta) + i \sin  (5 \theta)) (\cos ( \theta) + i \sin  (\theta)) and group the real and imaginary parts.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Assuming your last line is correct (and I haven't bothered to check) then you can use deMoivre's Theorem to write ( \cos \theta + i \sin \theta )^5 = \cos (5 \theta) + i \sin  (5 \theta). Now expand  (\cos (5 \theta) + i \sin  (5 \theta)) (\cos ( \theta) + i \sin  (\theta)) and group the real and imaginary parts.
    U are missing the question if u look closely the last line is

    (cos 5\theta + i sin 5\theta) (sin \theta + i cos \theta)

    and not as u have mentioned it

    (cos 5 \theta + i sin 5\theta) (cos \theta + i sin \theta)
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  4. #4
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    Quote Originally Posted by TomJerry View Post
    U are missing the question if u look closely the last line is

    (cos 5\theta + i sin 5\theta) (sin \theta + i cos \theta)

    and not as u have mentioned it

    (cos 5 \theta + i sin 5\theta) (cos \theta + i sin \theta)
    You should realise that my small error has absolutely no affect on the advice I have given you.
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    sin (theta)=cos((pi)/2-(theta))
    cos(theta)=sin((pi)/2-(theta))
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    Quote Originally Posted by mr fantastic View Post
    You should realise that my small error has absolutely no affect on the advice I have given you.
    how will u group the real parts when the one is a cos and the other is a sin ..........
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  7. #7
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    Quote Originally Posted by TomJerry View Post
    how will u group the real parts when the one is a cos and the other is a sin ..........
    Can you not expand (\cos (5\theta) + i \sin (5\theta)) (\sin \theta + i \cos \theta) and then group together the real and imaginary bits??
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  8. #8
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    Quote Originally Posted by mr fantastic View Post
    Can you not expand (\cos (5\theta) + i \sin (5\theta)) (\sin \theta + i \cos \theta) and then group together the real and imaginary bits??
    could u please show me a example or even a link would be more help.......i just need to know how to carry out this .......different signs
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  9. #9
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    Quote Originally Posted by TomJerry View Post
    could u please show me a example or even a link would be more help.......i just need to know how to carry out this .......different signs
    (\cos{5\theta} + i\sin{5\theta})(\sin{\theta} + i\cos{\theta}) = \cos{5\theta}\sin{\theta} + i\cos{5\theta}\cos{\theta} + i\sin{5\theta}\sin{\theta} + i^2\sin{5\theta}\cos{\theta}

     = \cos{5\theta}\sin{\theta} - \sin{5\theta}\sin{\theta} + i(\cos{5\theta}\cos{\theta} + \sin{5\theta}\sin{\theta}).


    What do you think a and b are?
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  10. #10
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    Here are some short cuts.
    \frac{{\left[ {\cos (\varphi ) + i\sin (\varphi )} \right]^3 \left( {\sin (\varphi ) + i\cos (\varphi )} \right)}}{{\cos (2\varphi ) - i\sin (2\varphi )}}  = \left[ {\cos (3\varphi ) + i\sin (3\varphi )} \right]\left( {\sin (\varphi ) + i\cos (\varphi )} \right)\left( {\cos (2\varphi ) + i\sin (2\varphi )} \right)
    \left( {\sin (\varphi ) + i\cos (\varphi )} \right) = \left( {\cos (\varphi ) - i\sin (\varphi )} \right)i = \left( {\cos ( - \varphi ) + i\sin ( - \varphi )} \right)i
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  11. #11
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    thanks
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