What are the two square roots of -18 + 18√3i?

Question

a) 6√3 + 6i, -6√3 - 6i 
b) 6 + 6√3i, -6 - 6√3i 
c) 3√3 + 3i, -3√3 - 3i 
d) 3 + 3√3i, -3 - 3√3i

Answer 1

a) The roots are (6√3 + 6i) and (-6√3 - 6i)

Therefore (6√3 + 6i) (-6√3 - 6i)

= 6√3(-6√3) + 6√3( - 6i) + 6i(-6√3) + 6i( - 6i)

= -36(3) - 36√3i - 36√3i - 36(i)²

= -108 - 72√3i - 36(-1)

= -72 - 72√3i  ≠  -18 + 18√3i

b) The roots are (6 + 6√3i) and (-6 - 6√3i)

Therefore (6 + 6√3i) (-6 - 6√3i)

= 6(-6) + 6( - 6√3i) + 6√3i(-6) + 6√3i( - 6√3i)

= -36 -36√3i - 36√3i - 36(3)(i)² 

= -36 -72√3i -108(-1)

= -36 - 72√3i + 108

= 72 - 72√3i ≠  -18 + 18√3i

Answer 2

c) The roots are (3√3 + 3i) and (-3√3 - 3i)

Therefore (3√3 + 3i) (-3√3 - 3i)

= 3√3(-3√3) + 3√3( - 3i) + 3i(-3√3) + 3i( - 3i)

= -9(3) - 9i√3 - 9i√3 - 9(i)²

= -27 - 18i√3 - 9(-1)

= -27 - 18i√3 + 9

= -18 - 18i√3 ≠ -18 + 18√3i

Answer 3

d) The roots are (3 + 3√3i) and (-3 - 3√3i)

Therefore (3 + 3√3i) (-3 - 3√3i)

= 3(-3) + 3( - 3√3i) + 3√3i(-3) + 3√3i( - 3√3i)

= -9 - 9√3i - 9√3i - 9(3)(i)²

= -9 - 18√3i - 27(-1)

= -9 - 18√3i + 27

= 18 - 18√3i ≠ -18 + 18√3i