Question

How do you convert `-6 - 6i sqrt 3` to polar form?

Answer 1

`12(sin ((-5pi)/6)) - i cos ((-5pi)/6))`

Explanation:

In polar form,`(rsin theta + r i cos theta)` ... `I`
Comparing it with `-6-6isqrt3`
we get `rarr -6 = r sin theta` ... 1
`rarr -6sqrt3 = r cos theta` ... 2

Squaring and adding 1 and 2,
`r^2 sin^2 theta + r^2 cos^theta = (-6)^2 + (-6sqrt3)^2`
`r^2(sin^2 theta + cos^2 theta) = 36 + 108`
`r^2 = 144`
`r = 12`

Dividing equation 1 by 2
`tan theta =( -6)/(-6sqrt3)`
`tan theta = 1/sqrt3`
`theta = pi/6`

Real part of complex no = `-6 (-x)`
Imaginary part of complex no = `-6sqrt3 (-y)`
`:.`the point is in 3rd quadrant.

At 3rd quadrant `alpha = theta - pi`
`alpha = (-5pi)/6`

Substituting the value in equation `I` we get,
`12(sin ((-5pi)/6)) - i cos ((-5pi)/6))`