Let, (a + ib)2 = - 4 - 3i
Now using, (a + b)2 = a2 + b2 + 2ab
⇒ a2 + (bi)2 + 2abi = -4 -3i
Since i2 = -1
⇒ a2 - b2 + 2abi = - 4 - 3i
Now, separating real and complex parts, we get
⇒ a2 - b2 = - 4…………..eq.1
⇒ 2ab = - 3…….. eq.2
⇒ a = - 3/2b
Now, using the value of a in eq.1, we get
⇒ \((-\frac{3}{2b})^2\) – b2 = - 4
⇒ 9 – 4b4 = -16b2
⇒ 4b4 - 16b2 - 9 = 0
Simplify and get the value of b2, we get,
⇒ b2 = 9/2 or b2 = - 2
As b is real no. so, b2 = 9/2
b = \(\frac{3}{\sqrt2}\) or b = -\(\frac{3}{\sqrt2}\)
Therefore, a = -\(\frac{1}{\sqrt2}\) or a = \(\frac{1}{\sqrt2}\)
Hence the square root of the complex no. is -\(\frac{1}{\sqrt2}\) + \(\frac{3}{\sqrt2}\)i and \(\frac{1}{\sqrt2}\) - \(\frac{3}{\sqrt2}\)i.