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 =

Now, using the value of a in eq.1, we get

– b² = 0

1 – 4b⁴ = 0

4b² = 1

Simplify and get the value of b² , we get,

b² = - or b² =

As b is real no. so, b² = 3

b= or b=

Therefore , a= or a= -

Hence the square root of the complex no. is + i and - - i.