Find the four fourth roots of -1 - sqrt3i ??

Question

I'm dying here. Help? please explain your answer.

Answer 1

First, convert the number to polar form:

|-1-i√3| = √(1+3) = 2
tan φ = -√3/(-1) = √3
φ = -2π/3 (remembering that this is a third quadrant angle)

-1-√3i = 2e^(-2πi/3)

Now, to find one of the fourth roots of this, we take the fourth root of the magnitude and divide the argument by 4:

∜(2e^(-2πi/3)) = ∜2 e^(-πi/6)

Rewriting this in rectangular form:

∜2 * (√3/2 - i/2)
2^(-3/4)√3 - 2^(-3/4)i

To find the other fourth roots, we multiply this number by the fourth roots of unity, which are 1, i, -1, and -i. So we have:

2^(-3/4)√3 - 2^(-3/4)i
2^(-3/4) + 2^(-3/4)√3 i
-2^(-3/4)√3 + 2^(-3/4)i
-2^(-3/4) - 2^(-3/4)√3 i

And we are done.

Answer 2

1. Convert your number into (r, θ) form

2. Use DeMoivre's theorem

3. Substitute n = 0,1,2,3


1.

-1 - √3 i = 2[cos (240 + 360n) + i sin (240 + 360n)]


2.

fourth root of -1 - √3 i

= 2^ 1/4 .[cos (240 + 360n) + i sin (240 + 360n)]^(1/4)

= 2^ 1/4 .[cos (60 + 90n) + i sin (60 + 90n)]

3.

n = 0 ... z1 = 2^ 1/4 [½ + i√3 /2] = 2^(-3/4) [1 + i√3]

n = 1 ... z2 = 2^ 1/4 [-½ + i√3 /2] = 2^(-3/4) [-1 + i√3]

n = 2 ... z3 = 2^ 1/4 [-½ - i√3 /2] = 2^(-3/4) [-1 - i√3]

n = 3 ... z4 = 2^ 1/4 [½ - i√3 /2] = 2^(-3/4) [1 - i√3]

Answer 3

Worry not, this problem's cake. Euler's formula:

exp(itheta)=cos(theta) + isin(theta)

Think about it . . .

Ask yourself what exp(i(pi/2)) is (use above) . . .

The cube root of exp(i(pi/2)) is exp(i(pi/6))

recall that (-1)*(-1)=1=1*1 and that (-i)(-i)=-1=i*i for all the permuations for the roots