I'm having troubles, finding how solution would look like for complex equation of the form $\bar{z} = |z|$. Taking $z = x + iy$, we get the following: $$x - iy = \sqrt{x^2 + y^2},$$ then raising it to the 2nd power we get $$x^2 - 2ixy - y^2 = x^2 + y^2,$$ $$2y^2 = -2ixy.$$ Now, taking complex terms as one equation and real as second, we construct the system of equations: $$-2xy = 0$$ $$2y^2 = 0$$ Which again yields equation: $$y^2 + xy = 0.$$ Here we get that $\Im{(z)} = 0$. But what is $\Re{(z)}$ equal to?

Is my solution correct?

  • $\begingroup$ Looks ok. Wouldn't the real part just be the modulus at that point? At which point it would have to be positive. $\endgroup$ Jul 26, 2016 at 21:18
  • $\begingroup$ @AndréNicolas: It should be $\bar z = \sqrt{z\bar z}$, not $\bar z = z\bar z$. $\endgroup$
    – Jack Lee
    Jul 26, 2016 at 21:20

2 Answers 2


$|z|$ is real and nonnegative. Therefore, so is $\overline{z}$. I.e., $\overline{z}$ is a number on the nonnegative real semiaxis, and, conversely, every such number equals its own modulus.

  • $\begingroup$ May I ask You to write this reasoning more "rigorously"? :-) I got the $|z|$ point, because it gives us a real number, which is always positive. But why is $\bar{z}$ then always positive? $\endgroup$ Jul 26, 2016 at 21:29
  • 2
    $\begingroup$ @AcceleratetotheInfinity $\lvert z \rvert$ is real and nonnegative, therefore so is $\bar{z}$. $\endgroup$
    – user4894
    Jul 26, 2016 at 21:32
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    $\begingroup$ @AcceleratetotheInfinity keep in mind that your solutions are $z \in \mathbb C$ so that $|z| = \bar{z}$. So any property of $|z|$ is a property of $\bar{z}$ for these $z$. $\endgroup$ Jul 26, 2016 at 21:41
  • $\begingroup$ "But why is z¯ then always positive?" Nonnegative, as user4894 pointed out. And, $\overline{z}$ is then always nonnegative because, in the conditions of your problem, it is equal to $|z|$. $\endgroup$
    – avs
    Jul 26, 2016 at 22:15
  • $\begingroup$ @avs: There are countries (France, fir instance) where ‘positive’ means ‘non-negative’, and positive is said ‘strictly positive’. $\endgroup$
    – Bernard
    Jul 26, 2016 at 22:44

From $2y^2 = 0$, you have $y=0$ or $\Im(z) = 0$. But then you have the problem that $x$ can be anything. This is because by squaring the first equation, you "threw away" the information that the solutions require $x \geq 0$. (I.e., $x^2 = |x|^2$ for all $x \in \Bbb{R}$.)

You would be better served working with $z = r \mathrm{e}^{\mathrm{i}\theta}$ with $r \in \Bbb{R}_{\geq 0}$ and $\theta \in \Bbb{R}$. Then $$ r = |z| = \overline{z} = r \mathrm{e}^{-\mathrm{i}\theta} \text{.} $$ If $r=0$, we have the solution $z=0$. If $r \neq 0$, we can divide through, getting $\mathrm{e}^{-\mathrm{i}\theta} = 1$, so $\theta = 2 \pi k$ for $k \in \Bbb{Z}$. That is, $z \in \Bbb{R}_{>0}$.

Gluing the two solution sets together, we have $z \in \Bbb{R}_{\geq 0}$.


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