Please explain to me why: $$ Z^2 = \bar{Z} $$

Has a 3rd and 4th solutions: $1, 0$ ?

I found the first two complex numbers, but why also $1$ and $0$ ?

I replaced $Z^2$ with $(x+yi)^2$ and $\bar{Z}$ with $x-yi$ . Why did I only find the first two solutions? What am I missing?

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    $\begingroup$ Your $Z^2$ replacement is wrong... It should be $Z^2=(x+iy)^2$. $\endgroup$ – abiessu Aug 13 '15 at 21:21
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    $\begingroup$ It's impossible to say from the question itself what you missed, because you left out some of the steps you took to find the solutions you found. $\endgroup$ – David K Aug 13 '15 at 22:21

$$(x+yi)^2=x-yi$$ $$x^2-y^2+2yxi=x-yi$$ From the imaginary part: $2yx=-y$. So $y=0$ or $x=-\frac{1}{2}$

Then, from the real part:

For case $y=0$, we get $x^2=x$. So $x=0$ or $x=1$.

For case $x=-\frac{1}{2}$, we have $\frac{1}{4}-y^2-yi= -\frac{1}{2}-yi$. That is $\frac{1}{4}-y^2= -\frac{1}{2}$. So $y=\pm\frac{i\sqrt{3}}{2}$

So there are four solutions are $z=0$, $z=1$, $z=-\frac{1}{2}+\frac{i\sqrt{3}}{2}$ and $z=-\frac{1}{2}-\frac{i\sqrt{3}}{2}$.

  • $\begingroup$ I see. But if I got $x = -\frac {1}{2}$ why should I also check for $y = 0$? I have never done it before so I don't understand the reason $\endgroup$ – Sahar Avr Aug 13 '15 at 21:54
  • $\begingroup$ You solve the imaginary part ($2yx=-y$) and we get two cases ($y=0$ and $x=-\frac{1}{2}$). Then for each case we must solve the real part. Note that when solving the the real part for $x=-\frac{1}{2}$, I did NOT assume $y=0$. $\endgroup$ – user261922 Aug 13 '15 at 22:04
  • $\begingroup$ Why two cases? Why can't I just solve for x and return to the real part? $\endgroup$ – Sahar Avr Aug 13 '15 at 22:09
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    $\begingroup$ Because when we solve the imaginary parte ($2yx=-y$) we find two possibilities: either $y=0$ (and then we know nothing about $x$) or $x=-\frac{1}{2}$ (and then we know nothing about $y$). We can NOT simply throw away any of those two possibilities. $\endgroup$ – user261922 Aug 13 '15 at 22:17

Taking the modulus of both sides, $|z|^2 = |z|$. Hence $|z| = 0$ or $|z| = 1$.

If $|z| = 0$ then $z = 0$.

If $|z| = 1$ then note we also have that $z^3 = |z|^2$ and thus $z$ is one of the cube roots of unity.

Hence the four solutions are $\displaystyle 0, 1, -\frac 12 \pm i\frac{\sqrt 3} {2}$.

  • $\begingroup$ Multiplication by $z$ gives you a solution $z=0$ (which we were lucky to have anyway, but formally it should be treated carefully). Do you really need to multiply first? Taking abs gives $|z|^2=|z|$ etc. $\endgroup$ – A.Γ. Aug 13 '15 at 21:35
  • $\begingroup$ Yes $z^2 = \bar z$ implies $|z|^2 = |z|$ and thus $|z| = 0, 1$. What that doesn't give is $z^3 = 1$, which we need. But I take your point: answer edited. $\endgroup$ – Simon S Aug 13 '15 at 21:36
  • $\begingroup$ I mean $z^3=1$ we get later when we start looking for $|z|=1$. Then we can equivalently multiply. $\endgroup$ – A.Γ. Aug 13 '15 at 21:38

Let $z=x+iy$ then $$(x+iy)^2=x^2-y^2+2ixy=x-iy \ \ \ \Rightarrow \ \ \ \ x=x^2-y^2 \ \ \ \text{and} \ \ \ -y=2xy \ (*)$$

If $y=0$ then $x=x^2 \ \Rightarrow \ x=0,1$ therefore two solutions are $z_1=0$ and $z_2=1$

If $y \neq 0$ then we can cancel out the y's in (*) and we'll have $2x=-1 \ \Rightarrow x=\frac{-1}{2}$ , plug it into the other equation, you will get $y^2=\frac{3}{4}$ so the other two solutions are $z_3=\frac{-1}{2} + i\frac{\sqrt{3}}{2}$ and $z_4=\frac{-1}{2} - i\frac{\sqrt{3}}{2}$.


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