If z=x-iy and w=y-ix, how could i find all values of z which satisfy z^2=w?i stands for imaginary

z = x- iy

w = y-
ix

We need to find the value of z such that z^2 =
w

First we will square z= x-
iy

==> z^2 = (x-iy)^2

==>
z^2 = (x^2 - 2xy*i + y^2*i^2)

But we know that i^2 =
-1

==> z^2 = (x^2 - 2xy*i -
y^2)

==> z^2 = (x^2-y^2) -
2xy*i

Now we will compare z^2 =
w

==> (x^2-y^2) - 2xy*i = y-
i*x

==> x^2 - y^2 =
y...........(1)

==> 2xy =
x

Divide by x.

==> 2y =
1

==> y= 1/2

Now we will
substitute into (1).

==> x^2 - y^2 =
y

==> x^2 - 1/4 = 1/2

==>
x^2 = 1/2 +1/4 = 3/4

==> x^2 =
3/4

==> x = +-sqrt3 / 2

Then
possible z values are:

z = x-
iy

z1 = (sqrt3)/2 -
(1/2)*i

z2= (-sqrt3) /2 - (1/2)*i