Answer:
the answer is 2 + i.
Step-by-step explanation:
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.
