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