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Convert into polar form : (-1/2) + i ( \(\frac{-\sqrt{3}}{2}\) )

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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...

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Other answers:

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 : \(\1\)
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*

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