The square root of 3 - 4i is
±(2-i).
Let √3−4i=x+iy ⇒ 3-4i= x2-y2+2ixy
⇒ x2-y2=3, 2xy=-4 ..........(i)
⇒ (x2+y2)2=(x2−y2)2+4x2y2=(3)2+(−4)2=25
⇒ (x2+y2)2=5 ............(ii)
From equation (i) and (ii) ⇒ x2=4 ⇒x=±2,
y2=1 ⇒y=±1.Hence the square root of (3-4i)
is ±(2-i).