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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]

https://brainmass.com/math/complex-analysis/trigonometric-form-complex-number-4014

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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 trigonometric forms.

Even though you will get theta as pi/3 you must remember the fact that theta must be in the 4th quadrant. Remember the ASTC rule, in the first quadrant All trigonometric functions are +ve ,in second quadrant Sin and its inverse are positive(all others are negative), in the third quadrant Tan and its inverse are postive (all others are negative), in the fourth quadrant Cos and its inverse (sec) are positive while all others are negative.
You see, in our question, theta must be in the fourth quadrant, so that Sin(theta) is negative and Cos(theta) is positive.
Now, how will you select theta so that Sin is negative and Cos is positive ?? Simply 60 degrees (= pi/3) ??
No you have to select theta in the 4th quadrant.
and it is 5pi/3
How will you get this ?
You want theta to be in the 4th quadrant, and remember the following thing:

Sin(- theta) = - sin(theta)
Cos (- theta) = Cos (theta)

Yes your theta must be negative..

it's -pi/3 or it is (2pi - pi/3)= 5pi/3

You can replace 0 by a 2pi because 2pi = 360 degrees and no change in trigonometric quantities.

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