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Math Help - Questions in polar and rectangular i want help

  1. #1
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    Questions in polar and rectangular i want help

    Hi

    I want check my answer if correct or not

    write in rectangular form
    1) Z2 = 3e^(-60)i
    2) z3 = 7.19e^-2.13i
    3) z1 = root2 e^(-pi/2)i
    4)z2 = 3e^120i
    5 ) z3 = 6.49e^-2.08i
    ------



    my answer :



    1)
    = 3{cos (-60 ) + i sin ( -60) }
    = 3(cos(1/2) + i sin (-root 3/2 ) }
    = 3(1/2 + - root3/2)
    = 1/2 + - root 3 /6

    2 )
    = 7.19 { cos ( -2.13 ) - i sin (-2.13 ) }
    = 7.19{ 0.999 - -(0.037 ) }
    = 7.182 - 0.266

    3 )

    = root2 { cos (-pi/2) - i sin (-pi / 2 ) }
    = root 2 ( 0- 1 )
    = 0 - root2

    4 )

    = 3{ cos (120) - i sin (120) }
    3(1/2) - ( root 3 /2 )
    = 1/2 - root3/6

    5 )
    6.49 {(cos ( -2.08 ) - i sin ( -2.08 ) i )
    = 6.49 (0.999 ) - ( -0.036 )
    = 6.18 + -236
    =6.18 - 236
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  2. #2
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    Quote Originally Posted by r-soy View Post
    Hi

    I want check my answer if correct or not

    write in rectangular form
    1) Z2 = 3e^(-60)i
    2) z3 = 7.19e^-2.13i
    3) z1 = root2 e^(-pi/2)i
    4)z2 = 3e^120i
    5 ) z3 = 6.49e^-2.08i
    ------



    my answer :



    1)
    = 3{cos (-60 ) + i sin ( -60) }
    = 3(cos(1/2) + i sin (-root 3/2 ) }
    = 3(1/2 + - root3/2)
    = 1/2 + - root 3 /6


    It's almost correct but the +-sign is confusing and you made a mess with the square roots: it must be \frac{1}{2}-\frac{3\sqrt{3}}{2} . Check the other ones likewise.

    Tonio


    2 )
    = 7.19 { cos ( -2.13 ) - i sin (-2.13 ) }
    = 7.19{ 0.999 - -(0.037 ) }
    = 7.182 - 0.266

    3 )

    = root2 { cos (-pi/2) - i sin (-pi / 2 ) }
    = root 2 ( 0- 1 )
    = 0 - root2

    4 )

    = 3{ cos (120) - i sin (120) }
    3(1/2) - ( root 3 /2 )
    = 1/2 - root3/6

    5 )
    6.49 {(cos ( -2.08 ) - i sin ( -2.08 ) i )
    = 6.49 (0.999 ) - ( -0.036 )
    = 6.18 + -236
    =6.18 - 236
    .
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