Let, (a + ib)² = -11 - 60i

Now using, (a + b)² = a² + b² + 2ab

a² + (bi)² + 2abi = -11 - 60i

Since i² = -1

a² - b² + 2abi = -11 - 60i

now, separating real and complex parts, we get

a² - b² = -11…………..eq.1

2ab = -60…….. eq.2

a =

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

– b² = -11

900 – b⁴ = -11b²

b⁴- 11b² - 900= 0

Simplify and get the value of b², we get,

b² = 36 or b² = -25

as b is real no. so, b² = 36

b= 6 or b= -6

Therefore , a= -5 or a= 5

Hence the square root of the complex no. is -5 + 6i and 5 – 6i.