Question

What is the polar form of Negative 9 StartRoot 3 EndRoot + 9 i?

Answer 1

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

$r=\sqrt{a^2+b^2}$ and filling in:

$r=\sqrt{(-9\sqrt{3})^2+(9)^2 }$ (notice we do not put the i in there with the 9).

$r=\sqrt{243+81}$ 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:

$tan^{-1}(\frac{b}{a})$ so filling that in, we have:

$tan^{-1}(\frac{9}{-9\sqrt{3} })=\frac{1}{-\sqrt{3} }$ 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 $\frac{5\pi}6}$, so the polar form of that number is (18, $\frac{5\pi}{6}$)