24.Find the square root of 5+12 i

1.	24.Find the square root of 5+12 i
Answer» Let √(5 – 12i) = x + iy Squaring both sides, we get 5 – 12i = x^2+ 2ixy +(iy)^2= x^2– y^2+ 2xyi. Comparing real and imaginary parts , we get 5 = x^2– y^2———– (1) and xy = – 6 ———— (2) Squaring (1), we get 25 = (x^2– y^2)^2= (x^2+ y^2)^2– 4x^2y^2 ⇒25 = (x^2+ y^2)^2– 4(– 6)^2 ⇒(x^2+ y^2)^2= 169 ⇒ x^2+ y^2= 13 ———- (3) Adding (1) and (3) we get 2x^2= 18=> x = ± 3. Subtracting (1) from (3) we get 2y^2= 8=> y = ± 2. Therefore, √(5 – 12i) = 3 + 2i or, 3 – 2i or, – 3 + 2i or, – 3 – 2i. As imaginary part of 5 – 12i is negative, the square root is ±(3 – 2i)

Answer»

Let √(5 – 12i) = x + iy

Squaring both sides, we get

5 – 12i = x^2+ 2ixy +(iy)^2= x^2– y^2+ 2xyi.

Comparing real and imaginary parts , we get

5 = x^2– y^2———– (1) and xy = – 6 ———— (2)

Squaring (1), we get

25 = (x^2– y^2)^2= (x^2+ y^2)^2– 4x^2y^2

⇒25 = (x^2+ y^2)^2– 4(– 6)^2

⇒(x^2+ y^2)^2= 169

⇒ x^2+ y^2= 13 ———- (3)

Adding (1) and (3) we get

2x^2= 18=> x = ± 3.

Subtracting (1) from (3) we get

2y^2= 8=> y = ± 2.

Therefore, √(5 – 12i) = 3 + 2i or, 3 – 2i or, – 3 + 2i or, – 3 – 2i.

As imaginary part of 5 – 12i is negative, the square root is ±(3 – 2i)

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