What are the solutions to x^3=−2−2i in polar form?
Select all correct answers.
8√6cis(5π/12)
8√6cis(13π/12)
8√6cis(7π/4)
8√6cis(2π/3)
22√3cis(2π/3)
22√3cis(7π/4)
22√3cis(13π/12)
22√3cis(5π/12)
8√6cis(5π/3)
+23246 -2 - 2i in r·cis(theta) form:
r = sqrt( (-2)2 + (-2)2 ) = sqrt(8)
theta = tan-1( -2/-2 ) = 5pi/4
x3 = sqrt(8) · cis ( 5pi/4 ) ---> x = [ sqrt(8) · cis ( 5pi/4 ) ]1/3
To find the principal cube root, take the cube root of sqrt(8) = (81/2)1/3 = (81/3)1/2 = sqrt(2)
and divide 5pi/4 by 3 = 5p/12.
So, the principal cube root is: sqrt(2)·cis( 5pi/12 )
Adding 2p/3, the second root is: sqrt(2)·cis( 13pi/12 )
and another 2p/3, the third root is: sqrt(2)·cis( 21pi/12 )
So, I don't get any of the answers ...
