Question

convert 2sqrt3+2i into its polar form

Answer 1

@mathstudent55 can you help???

Answer 2

Hey there Jules c:
\[\Large\rm a+b \mathcal i\]

Converting to polar form:\[\Large\rm r(\cos \theta+\mathcal i \sin \theta)\]Where r is given by,\[\Large\rm r=\sqrt{a^2+b^2}\]and the angle theta is found by using,\[\Large\rm \tan \theta=\frac{b}{a}\]

Answer 3

So for our problem here,\[\Large\rm 2\sqrt3+2\mathcal i\]Our r value,\[\Large\rm r=\sqrt{(2\sqrt3)^2+(2)^2}\]Simplify to get your radius.

And our theta,\[\Large\rm \tan \theta=\frac{2}{2\sqrt3}=\frac{1}{\sqrt3}\]This corresponds to one of your special angles.
Since both the `real` and `imaginary` components of our point were `positive`, we're in the first quadrant.