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I'm trying to solve this complex number equation: $$z^5=\bar z$$ As far as I understand, every complex number $z$ should be written in trigonometric form: $$r^5(\cos(5\phi) + i\sin(5\phi)) = r(\cos(-\phi) + i\sin(-\phi))$$

Unfortunately, at this moment, I'm stacked. Could you give some ideas that would help me to solve it?

Martin Sleziak
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Artem
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    $|z|^5=|z|$ says what about the absolute value of $z$? Then, we can compute $\bar z$ in terms of $z$. – robjohn Dec 06 '19 at 10:33
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    Think about what this means geometrically first. Multiplying your angle by $5$ is the same thing as flipping your angle across the axis. Also, your radius to the fifth power must be be same as your initial radius ($|z| = |\bar z|$) – Brevan Ellefsen Dec 06 '19 at 10:33
  • Yes, thinking in $e^{i\theta}$ form will help more - the geometric interpretation is more obvious – Dhanvi Sreenivasan Dec 06 '19 at 10:43
  • You were too hasty in expanding the trigonometric form. $r^5e^{i5\theta}=re^{-i\theta}$ is easier to handle. –  Dec 06 '19 at 11:11
  • Compare https://math.stackexchange.com/q/2309331, https://math.stackexchange.com/q/1283252, https://math.stackexchange.com/q/766599, https://math.stackexchange.com/q/1067784 for some quite similar questions. – Martin R Dec 06 '19 at 13:38

4 Answers4

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Note that:

$$z = re^{i\phi}, z^5 = r^5 e^{i5\phi}~\text{and}~\bar{z} = re^{-i\phi},$$

where $r \geq 0$. Therefore:

$$z^5 = \bar{z} \Rightarrow \begin{cases} r^5 = r\\ 5\phi + 2k\pi= -\phi + 2h\pi \end{cases},$$

where $k, h \in \mathbb{Z}.$

The previous system can be rewritten as: $$ \begin{cases} r(r^4-1) = 0\\ 6\phi = 2s\pi \end{cases},$$

where $s = k-h \in \mathbb{Z}.$

The first equation has $3$ distinct roots: $r=-1$, $r=0$ and $r=1.$ Of course, $r=-1$ should be discarded. This means that for $r=0$, $z = 0$ is a solution, which obviously does not depend on the phase $\phi.$ Moreover, for $r=1$, the phase is important. Solving the second equation, we get:

$$\phi = \frac{s\pi}{3},$$

and hence $z = e^{\frac{is\pi}{3}}$ for $s \in \mathbb{Z}$, together with $z=0$ represent the solution of the equation.

the_candyman
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We have that

$$z^5=\bar z \implies z^5z=\bar zz \implies z^6=|z|^2$$

which requires $|z|=0$ or $|z|=1$ and $z^6=1$.

user
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$z=0$ is an obvious solution. Then the equation can be written

$$z^6=|z|^2$$ where the RHS is real. Hence $z$ is proportional to a sixth root of unity, let $\omega\ne1$.

Now $$z^6=|z|^6=|z|^2$$ requires $|z|=1$.

Finally

$$z=0\lor z=\omega^k$$ for integer $k\in[0,5]$.

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The hard way:

$$(x+iy)^5=x^5+5ix^4y -10x^3y^2 -10ix^2y^3+ 5xy^4 +iy^5=x-iy$$

or

$$\begin{cases}x^5-10x^3y^2+5xy^4=x,\\5x^4y-10x^2y^3+y^5=-y.\end{cases}$$

We have the obvious solutions $x=0,y=0$ and $y=0,x=\pm1$.

Then

$$\begin{cases}x^4-10x^2y^2+5y^4-1=0,\\5x^4-10x^2y^2+y^4+1=0.\end{cases}$$

By elimination,

$$-24x^4+40x^2y^2-6=0$$

and

$$9-128x^4-256x^8=0.$$

The only real solutions are $x=\pm\dfrac12$, implying $y=\pm\dfrac{\sqrt 3}2$.

In total, $7$ solutions:

$$(0,0),(1,0),(-1,0),\tfrac12(1,\sqrt3),\tfrac12(1,-\sqrt3),\tfrac12(-1,\sqrt3),\tfrac12(-1,-\sqrt3).$$