find the square root of z=3-4j

Find the square root of z=3-4j. Hint: let z=(x+yj)^2

Let z=(x+yj)^2 where x and y real
z=x^2+2xyj+(yj)^2
=x^2+2xyj-y^2
=(x^2-y^2)+2xyj

z=3-4j

x^2-y^2=3 . . . . . (1)
2xy=-4 . . . . . . . . (2)

From (2) :
y=-2/x

Substitute y=-2/x into (1) :
x^2-(-2/x)^2=3
x^2-4=3x^2
x^4-3x^2-4=0
(x^2-4)(x^2+1)=0
x^2-4=0 or x^2+1=0
x^2=4 or x^2=-1 (rejected, since x is real)
x=2 or x=-2
y=-1 or y=1

z=(2-j)^2 or z=(-2+j)^2
√z=2-j or √z=-2+j

∴√z=+/-(2-j)


2010-10-30 01:36:10 補充：
Alternative solution :

Let z=(x+yj)^2

z=3-4j
=(4-1)-4j
=4-4j-1
=2^2-2(2)j+j^2
=(2-j)^2

i.e. x=2 and y=-1

∴ √z=+/-(2-j)