Find the square-root of 3 + 4i

Let √(3 + 4i) = x + yi

⇒ (x + yi)² = 3 + 4i

⇒ x² - y² + 2xyi = 3 + 4i

Equating real and imaginary parts:

x² - y² = 3 ………………………………(1)

2xy = 4 ⇒ xy = 2 ………………………(2)

Now, (x² + y²)² = (x² - y²)² + (2xy)²

= 3² + 4² = 25

⇒ x² + y² = 5 ………………………….(3)

From (1) and (3):

x² = 4, y² = 1 ⇒ y = ±1

Since xy = 2 is positive, when x = 2 ⇒ y = 1, and when x = -2 ⇒ y = -1

Therefore, the two square roots of 3 + 4i are:

2 + i and -2 - i