to find the square root of a cormplex number

to find sqrt(3-4i)

let sqrt(3-4i)= x+iy

squaring 3-4i = x² -y² +2ixy

equating real parts , x² -y² =3
equating imaginary parts 2xy = -4

(x² +y²)² =(x² - y²)² + (2xy)² = 9 +16 = 25
therefore
(x² +y²) = 5 ----------(1)
also
x² -y² =3-----------(2)

(1) +(2) implies

2 x² = 8 or x² = 4 or x = ± 2
using 2xy = -4
if x=2 , y = -1
if x = -2 , y =1

so sqrt(3-4i) = 2-1 i or-2 + 1i


sqrt(3-4i) = ± (2- i )

some notes on complex numbers
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