Find all couple $(x,y)$ for satisfy $\frac{x+iy}{x-iy}=(x-iy)$ I have a problem to solve this exercise, I hope someone help me.
Find all couple $(x,y)$ for satisfy $\frac{x+iy}{x-iy}=(x-iy)$
 A: One looks for the complex numbers $z=x+\mathrm iy$, $z\ne0$, such that $z=\bar z^2$. 
In particular, $|z|=1$. Plugging $z=\mathrm e^{\mathrm it}$ in $z=\bar z^2$ yields $\mathrm e^{3\mathrm it}=1$ hence $t$ is a multiple of $\frac23\pi$. 
The arguments $t=0$, $t=\frac23\pi$ and $t=\frac43\pi$ yield $z=1$ and $z=-\frac12\pm\mathrm i\frac{\sqrt3}2$ respectively, thus the solutions are $(x,y)=(1,0)$ and $(x,y)=(-\frac12,\pm\frac{\sqrt3}2)$.
A: $z \equiv x + {\rm i}y$. $z/z^{*} = z^{*}$. Then, $\left\vert z\right\vert = 1$.
$z = {\rm e}^{{\rm i}\theta}\,,\quad \theta \in {\mathbb R}$.
$$
{{\rm e}^{{\rm i}\theta} \over {\rm e}^{-{\rm i}\theta}}
=
{\rm e}^{-{\rm i}\theta}\,,
\quad
\theta =  - 2\theta + 2n\pi\,,
\quad
\theta = {2n\pi \over 3}\,,\quad n \in {\mathbb Z}
$$
$$
x_{n} = \cos\left(2n\pi \over 3\right)\,,\quad
y_{n} = \sin\left(2n\pi \over 3\right)
$$
$$\color{#ff0000}{\large%
z_{1} = -\,{1 \over 2} + {\sqrt{3\,} \over 2}\,{\rm i}\,,
\qquad
z_{2} = -\,{1 \over 2} - {\sqrt{3\,} \over 2}\,{\rm i}\,,
\qquad
z_{3} = 1}
$$
A: Let $z=x+iy,\overline z=x-iy$. Since $\dfrac z{\overline z}=\overline z$, we have$$|z|=|\overline z|=|\dfrac z{\overline z}|=\dfrac{|z|}{|\overline z|}=\dfrac{|z|}{|z|}=1,$$i.e., $|z|=1$, and so$$\dfrac1z=\dfrac{\overline z}{z\overline z}=\dfrac{\overline z}{|z|^2}=\overline z,$$so your equation can be rewritten as $$\dfrac z{1/z}=\dfrac1z,$$that is,$$z^3=1.$$The solutions are the three cube roots of unity, which you can find by factoring $$z^3-1=(z-1)(z^2+z+1)$$and solving the quadratic equation.
A: Hint: multiply by $x-iy$ and equate real and imaginary parts. Two equations in two unknowns
A: $$\frac{z}{\bar{z}} = \bar{z}$$
Take a look at $|\cdot|$ of the two sides and you get $|z| = 1 (=x^2+y^2)$ for free.
Now expand with $z$ (since $z \neq 0$):
$$z^2 = \bar{z}$$
Now consider $\Re z^2 = x^2 - y^2$ and $\Re \bar z = x$ to get
$$x^2 - x =y^2$$
so
$$2x^2 - x = 1$$
which has solutions $x = \pm 1$ implying $y = 0$
A: $\frac{x+iy}{x-iy}=\frac{(x+iy)^2}{(x-iy)(x+iy)}=\frac{(x+iy)^2}{x^2+y^2}=\frac{x^2-y^2}{x^2+y^2}+i\frac{2xy}{x^2+y^2}$
Then you have to impose 
$x=\frac{x^2-y^2}{x^2+y^2}$  
$y=\frac{2xy}{x^2+y^2}$
and solve the system.
