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Convert the complex number 2-2√3i into its polar representation.
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There is a more accurate way to solve these using arctangent and finding the magnitude of the complex value... but I like to do these a little weirder. \[\Large\rm 2-2\sqrt3 ~i\]If we pull a 4 out of each term, we can create these special trig values that we often see,
\[\Large\rm =4\left(\frac{1}{2}-\frac{\sqrt3}{2}~i\right)\]
And then think about which angle gives you:\[\Large\rm \cos \theta=\frac{1}{2}\]\[\Large\rm \sin \theta=-\frac{\sqrt3}{2}\]
If cosine is positive, and sine is negative, which quadrant are we in?
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4
Mm ya good :)
Can you figure out which angle it is? :U
@zepdrix is that choice b.
yay good job \c:/
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