Let, (a + ib)2 = 0 + i
Now using, (a + b)2 = a2 + b2 + 2ab
⇒ a2 + (bi)2 + 2abi = 0 + i
Since i2 = -1
⇒ a2 - b2 + 2abi = 0 + i
Now, separating real and complex parts, we get
⇒ a2 - b2 = 0 …………..eq.1
⇒ 2ab = 1…….. eq.2
⇒ a = 1/2b
Now, using the value of a in eq.1, we get
⇒ \((\frac{1}{2b})^2\) – b2 = 0
⇒ 1 – 4b4 = 0
⇒ 4b2 = 1
Simplify and get the value of b2 , we get,
⇒ b2 = -\(\frac{1}{2}\) or b2 = \(\frac{1}{2}\)
As b is real no. so, b2 = 3
b = \(\frac{1}{\sqrt2}\) or b = -\(\frac{1}{\sqrt2}\)
Therefore , a = \(\frac{1}{\sqrt2}\) or a = -\(\frac{1}{\sqrt2}\)
Hence the square root of the complex no. is \(\frac{1}{\sqrt2}\) + \(\frac{1}{\sqrt2}\)i and - \(\frac{1}{\sqrt2}\) - \(\frac{1}{\sqrt2}\)i.