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Convert the complex number 2 - 2 (sqrt 3) i into its polar representation.
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2 sqrt 2(cos(240) + isin(240)) -4(cos(120) + isin(120)) 4(cos(300) + isin(300)) 4 sqrt 3(cos(60) + isin(60))
I don't even know where to start...
\[\Large\rm z=2-2\sqrt3~\mathcal i\] We can write our point in polar form:\[\Large\rm z=r(\cos \theta+\mathcal i \sin \theta)\]By figuring out the \(\Large\rm r\) and \(\Large\rm \theta\).
When we have a point of the form: \[\Large\rm z=a+b\mathcal i\]The radius is given by:\[\Large\rm r=\sqrt{a^2+b^2}\]And theta is given by:\[\Large\rm \tan \theta=\frac{b}{a}\qquad\to\qquad \theta=\arctan \frac{b}{a}\]
So what is our radius in this case?\[\Large\rm r=\sqrt{2^2+\left(-2\sqrt3\right)^2}\]
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