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Thread: z^3 in Cartesian form please help

  1. #1
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    z^3 in Cartesian form please help

    z= 10 (cos(56π)+isin(56π))in polar form
    I'm not sure how to put z^3 in Cartesian form ..please help
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    Re: z^3 in Cartesian form please help

    56pi (or any even multiple of pi) is coterminal with 2pi.

    So, cos (56pi) = cos (2pi) = 1.

    Similarly sin(56pi) = sin(2pi) = 0

    So z = 10(1 + 0i) =10.

    Now z^3 is easy!!

    (unless you meant 5pi/6 instead of 56pi)
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    Re: z^3 in Cartesian form please help

    Quote Originally Posted by bee77 View Post
    z= 10 (cos(56 π)+isin(56 π))in polar form
    I'm not sure how to put z^3 in Cartesian form ..please help
    ${\large z=10\exp(56\pi\bf{i})}$ so $z^3=10^3\exp(168\pi\bf{i})$.

    BUT is it true that $z=10~?$
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    Re: z^3 in Cartesian form please help

    Oh sorry it should have been 5/6 PI i with both the cos and the sine ....
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    Re: z^3 in Cartesian form please help

    terribly sorry ,yes I meant 5pi/6
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    Re: z^3 in Cartesian form please help

    Quote Originally Posted by Plato View Post
    ${\large z=10\exp(56\pi\bf{i})}$ so $z^3=10^3\exp(168\pi\bf{i})$.

    BUT is it true that $z=10~?$
    Yes because exp(168pi i) = 1
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    Re: z^3 in Cartesian form please help

    Recall that if z = r(cos (A) + i sin (A)) then z^3 = (r^3)(cos(3A)+i sin(3A))
    Last edited by Debsta; Mar 20th 2017 at 09:30 PM.
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    Re: z^3 in Cartesian form please help

    Thanks a heap ,I'm pretty rusty on trig as it's been a very long time...really appreciate the help
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    Re: z^3 in Cartesian form please help

    Quote Originally Posted by Plato View Post
    ${\large z=10\exp(56\pi\bf{i})}$ so $z^3=10^3\exp(168\pi\bf{i})$.

    BUT is it true that $z=10~?$
    If it really were " 56\pi" then, yes, it would be 10.

    With 5\pi/6, z^2= 10^3(exp(3*(5/6)\pi i)= 1000 exp((5/2)\pi i)= 1000 exp((1/2)\pi)= 1000(cos(\pi/2)+ i sin(\pi/2))= 1000i.
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