A complex number has to be found that has a square of 3i.

Let the number be (a + ib)

(a + ib)^2 = 3i

=> a^2 + 2iab + i^2*b^2

=> a^2 - b^2 + 2ab*i

equate the real and imaginary coefficients

2ab = 3

=> a = 3/2*b

substitute in a^2 - b^2 = 0

=> 9/4b^2 = b^2

=> 9 = 4b^4

=> 4b^4 - 9 = 0

=> (2b^2 - 3)(2b^2 + 3) = 0

2b^2 = 3

=> b = sqrt(3/2) and b = -sqrt(3/2)

The other root gives complex values for b and can be ignored.

a = 3/2*b = 3/2*sqrt (3/2) = sqrt (3/2) and a = -sqrt (3/2)

The required complex number is sqrt(3/2) + i*sqrt (3/2) and sqrt -(3/2) - i*sqrt(3/2)