Question

If z=x-iy and w=y-ix, find all values of z which satisfy z^2=w. (note:a+bi=c+di and only if a=b and c=d.)

Answer 1

we have:
z = x-iy
w = y-ix
z^2 = w
thus
(x-iy)^2 = y-ix
x^2 - 2xiy - y^2 = y-ix
obviously this will have quite a few solutions, so it's k to leave it in this form. can you solve it?

Answer 2

Okay. I think I may be able to solve it. I am not sure where exactly to go from here.

Answer 3

just play around with it, you might get somewhere

Answer 4

Okay. I will do that. Thank you.

Answer 5

\[
x^2 - 2xiy - y^2 = y-ix \implies \color{red}{x^2-y^2} + \color{blue}{-2xy}i = \color{red}y+\color{blue}{-x}i
\]

Answer 6

\[
-2xy = -x\\
x^2-y^2=y
\]

Answer 7

When \(x=0\), we get \[
0=0\\
-y^2 = y
\]When \(y=0\) then both work, so \((x,y) = (0,0)\) is one possible solution.
We if \(x=0,y\neq 0\) then: \[
-y^2=y\implies -y = 1 \implies y=-1
\]So \((x,y) = (0,-1)\) is another solution.

Then consider \(x\neq 0\).\[
-2xy=-x \implies-2y = -1 \implies y=\frac 12\\
x^2 - \left(\frac 12 \right)^2 = \frac{1}{2} \implies x^2 = \frac 34\implies x = \pm \frac{\sqrt 3}{2}
\]

Answer 8

It helps to set up two cases where a variable is zero and where a variable is not 0, so that you can divide by that variable.

Answer 9

The solutions I am getting are \[
(x,y) \in \{(0,0),(0,-1), (\sqrt 3/2,1/2), (-\sqrt 3/2,1/2)\}
\]

Answer 10

Oh yeah, \((x,y) \to x-yi =z\).

Answer 11

For example \(z=0\), \(z=i\), and \(z=(\sqrt 3 -i)/2\).

Answer 12

Okay. Awesome. You have truly been a lifesaver. Thank you so much! I appreciate it! The answer would be the solution set you gave correct? Thank you so much again!

Answer 13

I would double check that the solutions work

Answer 14

Okay. I will. Thank you again! (: