Find all couple $(x,y)$ for satisfy $\frac{x+iy}{x-iy}=(x-iy)$

Question

I have a problem to solve this exercise, I hope someone help me.

score 2 · Accepted Answer · answered Oct 08 '13 at 21:43

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 $(x,y)=(-\frac12,\pm\frac{\sqrt3}2)$.

score 2 · Answer 2 · answered Oct 08 '13 at 21:57

$z \equiv x + {\rm i}y$. $z/z^{*} = z^{*}$. Then, $\left\vert z\right\vert = 1$. $z = {\rm e}^{{\rm i}\theta}\,,\quad \theta \in {\mathbb R}$.

$$ {{\rm e}^{{\rm i}\theta} \over {\rm e}^{-{\rm i}\theta}} = {\rm e}^{-{\rm i}\theta}\,, \quad \theta = - 2\theta + 2n\pi\,, \quad \theta = {2n\pi \over 3}\,,\quad n \in {\mathbb Z} $$

$$ x_{n} = \cos\left(2n\pi \over 3\right)\,,\quad y_{n} = \sin\left(2n\pi \over 3\right) $$

$$\color{#ff0000}{\large% z_{1} = -\,{1 \over 2} + {\sqrt{3\,} \over 2}\,{\rm i}\,, \qquad z_{2} = -\,{1 \over 2} - {\sqrt{3\,} \over 2}\,{\rm i}\,, \qquad z_{3} = 1} $$

score 1 · Answer 3 · answered Oct 08 '13 at 21:55

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.

score 0 · Answer 4 · answered Oct 08 '13 at 21:36

0

Hint: multiply by $x-iy$ and equate real and imaginary parts. Two equations in two unknowns

answered Oct 08 '13 at 21:36

Ross Millikan

374,822

1

Sorry but using the modulus+argument route is much less tedious. – Did Oct 08 '13 at 21:37
I agree with you Did – Salvattore Oct 09 '13 at 01:08

score 0 · Answer 5 · answered Oct 08 '13 at 21:37

$$\frac{z}{\bar{z}} = \bar{z}$$ 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 $y = 0$

score 0 · Answer 6 · answered Oct 08 '13 at 21:37

0

$\frac{x+iy}{x-iy}=\frac{(x+iy)^2}{(x-iy)(x+iy)}=\frac{(x+iy)^2}{x^2+y^2}=\frac{x^2-y^2}{x^2+y^2}+i\frac{2xy}{x^2+y^2}$

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.

answered Oct 08 '13 at 21:37

Sabino Di Trani

5,101

Find all couple $(x,y)$ for satisfy $\frac{x+iy}{x-iy}=(x-iy)$

6 Answers6