Let, (a + ib)2 = 0 - 8i
Now using, (a + b)2 = a2 + b2 + 2ab
⇒ a2 + (bi)2 + 2abi = 0 - 8i
Since i2 = -1
⇒ a2 - b2 + 2abi = 0 - 8i
Now, separating real and complex parts, we get
⇒ a2 - b2 = 0 …………..eq.1
⇒ 2ab = -8…….. eq.2
⇒ a = -4/b
Now, using the value of a in eq.1, we get
⇒ \((-\frac{4}{b})^2\) – b2 = 0
⇒ 16 – b4 = 0
⇒ b4 = 16
Simplify and get the value of b2, we get,
⇒ b2 = - 4 or b2 = 4
As b is real no. so, b2 = 4
b = 2 or b = -2
Therefore , a = -2 or a = 2
Hence the square root of the complex no. is -2 + 2i and 2 - 2i.