Let, (a + ib)² = 1 - i

Now using, (a + b)² = a² + b² + 2ab

a² + (bi)² + 2abi = 1 – i

Since i² = -1

a² - b² + 2abi = 1 - i

Now, separating real and complex parts, we get

a² - b² = 1…………..eq.1

2ab = -1…….. eq.2

a =

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

– b² = 1

1 – 4b⁴ = 4b²

4b⁴ + 4b² -1= 0

Simplify and get the value of b², we get,

b² =

As b is real no. so, b² =

b² =

b= or b= -

Therefore , a= - or a=

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