# 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

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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 **

We have z = x - iy and w = y - ix

We need the values of z that satisfy z^2 = w

z^2 = (x - iy)^2 = x^2 - y^2 - 2xyi = w

So we have x^2 - y^2 - 2xyi = y - ix

equate the real and complex coefficients

x^2 - y^2 = y and -2xy = -x

2xy = x

=> y = 1/2

x^2 - y^2 = y

=> x^2 = 1/2 + (1/2)^2

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

=> x^2 = 3/4

=> x = -(sqrt 3)/2 and (sqrt 3)/2

**This gives z = -(sqrt 3)/2 - (1/2)*i and (sqrt 3)/2 - (1/2)*i**