Let the square root of 3 + 4i be x + iy.

So (x + iy) (x +iy) = 3 +4i

=> x^2 + xyi + xyi + i^2*y^2 = 3 + 4i

=> x^2 – y^2 + 2xyi = 3 + 4i

Equate the real and complex terms

=>...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

Let the square root of 3 + 4i be x + iy.

So (x + iy) (x +iy) = 3 +4i

=> x^2 + xyi + xyi + i^2*y^2 = 3 + 4i

=> x^2 – y^2 + 2xyi = 3 + 4i

Equate the real and complex terms

=> x^2 - y^2 = 3 and 2xy = 4

2xy = 4

=> xy = 2

=> x = 2/y

Substitute in

=> x^2 - y^2 = 3

=> 4/y^2 - y^2 = 3

=> 4 – y^4 = 3y^2

=> y^4 + 3y^2 – 4 = 0

=> y^4 + 4y^2 – y^2 – 4 =0

=> y^2(y^2 + 4) – 1(y^2 + 4) =0

=> (y^2 – 1) (y^2 + 4) =0

Therefore y^2 = 1, we ignore y^2 = -4 as it gives complex values of y.

Therefore y = 1 and x = 2/1 = 2

**The required square root of 3 + 4i is 2 + i.**