What are the solutions to x^{3} = −4 + 4i in polar form?

Select **all** correct answers below.

^{6}√32*cis*(19π/12)

2 ^{3}√5*cis*(23π/12)

^{6}√32*cis*(2π/3)

^{6}√32*cis*(π/4)

2 ^{3}√5*cis*(7π/12)

^{6}√32*cis*(11π/12)

2 ^{3}√5*cis*(2π/3)

2 ^{3}√5*cis*(5π/4)

^{6}√32*cis*(4π/3)

**hey! I'm not asking to do the whole problem. if someone could help me solve/find one, I can do the rest. :) **

auxiarc May 17, 2020

#1**+1 **

x^{3 } = -4 + 4i

Step 1) write -4 + 4i in r·cs( theta ) form

r = sqrt( (-4)^{2} + (4^{2}) )

theta = tan^{-1}( 4 / -4)

that your angle is in the second quadrant>

Step 2) To get your first answer:

take the 3rd root of the r-value and divide the angle by 3

Step 3) To find the other roots:

-- add 2pi/3 to the angle of the first answer (keep r the same)

-- add 2pi/3 to the angle of the second answer (keep r the same)

geno3141 May 17, 2020

#3**+1 **

r is (32)^{1/2 } ---> r^{1/3 } = [ (32)^{1/2} ]^{1/3} = (32)^{1/6}

When you find theta, you use tan^{-1}( 4 / -4 ) = tan^{-1}( -1 )

One of the values of tan^{-1}( -1 ) is -45^{o} or -pi/4 -- however --

for this problem, since the number is

-4 + 4i -- which is in the second quadrant, you need to use either:

if you are using degrees: -45^{o} + 180^{o} = 135^{o}

or, if you are using radians: - pi/4 + pi = 3pi/4.

geno3141 May 20, 2020