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    Trigonometric form of the complex number

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    Find the trigonometric form of the complex number where 0 <= theta < 2pi on the equation:
    r= 1/2 - [(sqrt(3)/2)i]

    © BrainMass Inc. brainmass.com March 4, 2021, 5:38 pm ad1c9bdddf
    https://brainmass.com/math/complex-analysis/trigonometric-form-complex-number-4014

    Solution Preview

    Any complex number z is in the form z=x+iy where

    x=rcos(theta)
    y=rsin(theta)

    r=|1/2 - [(sqrt(3)/2)i]|
    r=[(1/2)^2] + [(sqrt(3)/2)^2]
    r=sqrt[(1/4) + (3/4)]
    r=sqrt(1)
    r=1
    Now cos(theta)=1/2
    sin(theta)=-sqrt(3/2)
    so theta is in IVth quadrant and equal to 5pi/3
    So 1/2-[(sqrt(3)/2)i]=cos(5pi/3)+isin(5pi/3)=
    =e^(i*5pi/3) Finding ...

    Solution Summary

    Step by step to the final answer. Explanation about the ASTC rule are provided. The trigonometric form of the complex numbers are analyzed.

    $2.19

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