# 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]

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##### 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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