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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]

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.

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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 ...

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