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(-1/2) + i ( \(-\frac{\sqrt{3}}{2}\) )

I got : cos ( 4 pi/3 ) + sin ( 4 pi/3)

But the book says :
cos ( - 2pi/3 ) + i sin ( - 2pi/3). Which is also right...

isin

You're wrong.

Sorry typing mistake I meant i sin 4pi/3

^ cos ( 4 pi/3 ) + i sin(4pi/3)

Find the absolute value of this:
\[\left| -\frac{ 1 }{ 2 } -\frac{ i \sqrt{3} }{ 2 }\right|\]

So which should I prefer more ? -2pi/3 or 4 pi/3

So
\[=\sqrt{(-\frac{ 1 }{ 2 })^2+(-\frac{ \sqrt{3} }{ 2 })^2}\]

So r=1

Now draw it.

Drawing it gives you a way better picture of what the angle is.

\[\large{\sqrt{ (\frac{-1}{2} )^2 + ( \frac {- \sqrt{3}}{2})^2}}\]
right so it is : \(\1\)

|dw:1359153193735:dw|

You took the other angle

You go backwards meaning it will be negative.

By the diagram I think that the angle theta will surely lie in the III quadrant
i.e. 4 pi /3

Ok got it thanks.

If it's in the third of fourth quadrant go backwards.

third or fourth*