Test: Square Root Of A Complex Number - JEE MCQ

(x+iy)(x−iy) = (a+ib)(a−ib)/(c+id)(c−id) 
⇒x²−iy² = √[(a²−i²b²)/(c²−i²d²)]
⇒x²+y² = √[(a²+b²)/(c²+d²)]  [1i² = -1]
(x_²+y²)² = (a²+b²)/(c²+d²)

(a + ib)^1/2 = (x + iy)
Squaring both sides,
a + ib = (x + iy)²
a + ib = x² - y² + 2ixy
Equating real and imaginary 
a = x² - y²    b = 2xy............(1)
Using (x² + y²)² = (x² - y²)² + 4xy 
(x² + y²)² = a² + b²
(x² + y²) = (a² + b²)^1/2.......(2)
Adding (1) and (2)
2x² = (a² + b²)^1/2 + a
x = +-{1/2(a² + b²)^1/2 + a}^1/2
Subtract (2) from (1)
2y² = (a² + b²)^1/2 - a
y = x = +-{1/2(a² + b²)^1/2 - a}^1/2
Therefore, (a+ib)^1/2 = x+iy
=> +-{1/2(a² + b²)1/2 + a}^1/2 + i+-{1/2(a² + b²)^1/2 - a}^1/2

(a + ib)² = 5 + 12i
use (a + b)² = a² + b² + 2*a*b
=> a² + b²*i² + 2*a*b*i = 5 + 12i
i² = -1
⇒ a² - b² + 2*a*b*i = 5 + 12i
equate the real and complex coefficients
⇒ a² - b² = 5 and ab = 6
a = 6/b
substitute in a² - b² = 5
⇒ 36/b² - b² = 5
⇒ 36 - b⁴ = 5b²
⇒ b⁴ + 5b² - 36 = 0
⇒ b⁴ + 9b² - 4b² - 36 = 0
⇒ b²(b² + 9) - 4(b² + 9) = 0
⇒ (b² - 4)(b² + 9) = 0
⇒ b² = 4 and b² = -9
b is a real number, so we eliminate b² = -9
b² = 4
⇒ b = 2 and b = -2
a = 3 and a = -3
The required number can be 3 + 2i and -3 -2i

Let √(5 – 12i) = x + iy
Squaring both sides, we get
5 – 12i = x² + 2ixy +(iy)² = x² – y² + 2xyi.
Comparing real and imaginary parts , we get
5 = x² – y² ———– (1) and xy = – 6 ———— (2)
Squaring (1), we get
25 = (x² – y²)2 = (x² + y²)² – 4x²y²
⇒ 25 = (x² + y²)² – 4(– 6)²
⇒ (x² + y²)² = 169
⇒ x² + y² = 13 ———- (3)
Adding (1) and (3) we get
2x² = 18
⇒ x = ± 3.
Subtracting (1) from (3) we get
2y² = 8
⇒ y = ± 2.
Hence, square root of √(5 – 12i) is (3 – 2i)
Similarly, √(5 + 12i) is (3 + 2i)
√(5 + 12i) + √(5 – 12i)
⇒ (3 + 2i) + (3 - 2i)
⇒ 6