Let, (a + ib)2 = 3 + 4√7 i
Now using, (a + b)2 = a2 + b2 + 2ab
⇒ a2 + (bi)2 + 2abi = 3 + 4√7 i
Since i2 = -1
⇒ a2 - b2 + 2abi = 3 + 4√7 i
Now, separating real and complex parts, we get
⇒ a2 - b2 = 3 …………..eq.1
⇒ 2ab = 4√7 …….. eq.2
⇒ a = 2√7/b
Now, using the value of a in eq.1, we get
⇒ \((\frac{2\sqrt7}{b})^2\) – 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 = - 2
Therefore, a = √7 or a = -√7
Hence the square root of the complex no. is √7 + 2i and -√7 - 2i.