Question

solve for x? if x^2+j16=0 where j=sqrt(-1)

Answer 1

unusual to have complex coefficients, but

x^2 = -16j = 16cis(3π/2)
x = ±4 cis(3π/4)
= ±4 (-1/√2 + 1/√2 j)
= ±2√2 (-1+j)