Let, (a + ib)² = 3 + 4i

Now using, (a + b)² = a² + b² + 2ab

a² + (bi)² + 2abi = 3 + 4i

Since i² = -1

a² - b² + 2abi = 3 + 4i

now, separating real and complex parts, we get

a² - b² = 3 …………..eq.1

2ab =4…….. eq.2

a =

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

– b² = 3

12 – b⁴ = 3b²

b⁴ + 3b² - 28= 0

Simplify and get the value of b², we get,

b² = -7 or b² = 4

as b is real no. so, b² = 4

b= 2 or b=

Therefore , a= or a= -

Hence the square root of the complex no. is + 2i and - -2i.