How do I solve $z+z^{-1}=2\bar z$ using $z=x+iy$ and $\bar z =x-iy$ substitution?

Question

How do I solve $z+z^{-1}=2\bar z$ using $z=x+iy$ and $\bar z =x-iy$ substitution?

zconjugate is also $\bar z$ in some countries.

Answer 1

With

$z = x + iy, \tag 1$

we have

$z^{-1} = \dfrac{x - iy}{x^2 + y^2}; \tag 2$

thus the equation

$z + z^{-1} = 2 \bar z \tag 3$

becomes

$x + iy + \dfrac{x - iy}{x^2 + y^2} = 2x -2iy, \tag 4$

or

$x + \dfrac{x}{x^2 + y^2} + i \left (y - \dfrac{y}{x^2 + y^2} \right ) = 2x - 2iy; \tag 5$

if we equate the real and imagiary parts from the two sides of this equation, we find

$x + \dfrac{x}{x^2 + y^2} = 2x, \tag 6$

$y - \dfrac{y}{x^2 + y^2} = -2y; \tag 7$

whence

$\dfrac{x}{x^2 + y^2} = x, \tag 8$

$\dfrac{y}{x^2 + y^2} = 3y; \tag 9$

if $x \ne 0$, (8) implies

$x^2 + y^2 = 1; \tag{10}$

this makes $y \ne 0$ impossible, since

$y \ne 0 \Longrightarrow x^2 + y^2 = \dfrac{1}{3}; \tag{11}$

so one set of solutions is

$x^2 = 1, \; y = 0, \tag{12}$

or

$x = \pm 1, \; y = 0 \Longrightarrow z = \pm 1, \tag{13}$

which is easily checked. A similar argument shows that we may also have

$x = 0, \; y = \pm \dfrac{\sqrt 3}{3}, \tag{15}$

or

$z = \pm \dfrac{\sqrt 3}{3} i, \tag{16}$

also easily checked; thus the complete list of solutions is

$z = -1, z = 1, z = \dfrac{\sqrt 3}{3} i, z = -\dfrac{\sqrt 3}{3} i. \tag{17}$

Answer 2

if $z\neq0$ then $$z+\dfrac{1}{z}=2\bar z$$ $$z^2+1=2z\bar{z}=2|z|^2$$ $$x^2-y^2+2ixy+1=2(x^2+y^2)$$ \begin{cases} x^2-y^2+1&=2(x^2+y^2)\\ 2xy&=0. \end{cases} Only $x=0$ or $y=0$ should occur.

Can you continue?

Answer 3

Alt. hint: $\;z+\frac{1}{z}=2\bar z \;\;\iff\;\; z^2 - 2 z \bar z + 1 = 0 \;\;\iff\;\; z^2 = 2 |z|^2 - 1 \in \mathbb{R}\,$.

The square of a complex number is real iff the number is either real, or purely imaginary. This leaves two cases to consider for $\,z=x \in \mathbb{R}\,$ and $\,z = i y \in i\mathbb{R}\,$.