Let, (a + ib)2 = 5 + 12i
Now using, (a + b)2 = a2 + b2 + 2ab
⇒ a2 + (bi)2 + 2abi = 5 + 12i
Since i2 = -1
a2 - b2 + 2abi = 5 + 12i
Now, separating real and complex parts, we get
⇒ a2 - b2 = 5…………..eq.1
⇒ 2ab = 12……..eq.2
⇒ a = 6/b
Now, using the value of a in eq.1, we get
⇒ (6/b)2 – b2 = 5
⇒ 36 – b4 = 5b2
⇒ b4 + 5b2 - 36 = 0
Simplify and get the value of b2, we get,
b2 = - 9 or b2 = 4
As b is real no. so, b2 = 4
b = 2 or b = - 2
Therefore, a = 3 or a = - 3
Hence the square root of the complex no. is 3 + 2i and - 3 -2i.