Answer:
Step-by-step explanation:
In the rectangular complex number -9√3 + 9i, which has a standard form a + bi, the a = -9√3 and the b = 9. We need this in polar form (r, θ) where
[tex]r=\sqrt{a^2+b^2}[/tex] and filling in:
[tex]r=\sqrt{(-9\sqrt{3})^2+(9)^2 }[/tex] (notice we do not put the i in there with the 9).
[tex]r=\sqrt{243+81}[/tex] so
r = 18. Now let's move on to the angle, which is a little more difficult. The angle is found in the inverse tangent ratio:
[tex]tan^{-1}(\frac{b}{a})[/tex] so filling that in, we have:
[tex]tan^{-1}(\frac{9}{-9\sqrt{3} })=\frac{1}{-\sqrt{3} }[/tex] Since tangent is the side opposite over the side adjacent, y is positive and x is negative in the second quadrant. This is a 30 degree angle in QII, which has a reference angle of 150 degrees. This angle in radians is [tex]\frac{5\pi}6}[/tex], so the polar form of that number is (18, [tex]\frac{5\pi}{6}[/tex])
