Compute $\bar z - iz^2 = 0$

Question

$\bar z - iz^2 = 0, i = $ complex unit.

I've found 2 solutions to this, like this:

$x - iy - i(x+iy)^2 = 0$

$i(-x^2+y^2-y)+2xy+x=0$

$2xy + x = 0$ --> $y=-\frac{1}{2}$

$x^2 - y^2 + y = 0$ --> $x_1=\sqrt\frac34$ $x_2=-\sqrt\frac{3}{4}$

Solution 1: $z = \sqrt\frac34 - \frac12i $

Solution 2: $z = -\sqrt\frac34 - \frac12i $

That's great and everything, but Wolfram gives me another solution, which is $z = i$. How do I get that? LINK to wolfram.

Answer 1

Starting from the step $0 = i(-x^2 + y^2 - y) + (2xy + x)$, we see that both the real and imaginary components must be equal to zero for the equation to hold. Equating the real component to zero gives $$x(2y+1) = 0,$$ which implies either $x = 0$ or $y = -1/2$, or both. Now look at the imaginary component for each of these cases: if $x = 0$, then $y^2 - y = y(y-1) = 0$, which gives $y = 0$ or $y = 1$. We note that the simultaneous condition $(x,y) = (0,-1/2)$ is therefore not a solution, and we have two solutions $(x,y) = (0,0)$ and $(x,y) = (0,1)$. Now if $y = -1/2$, we get $x^2 = 3/4$, or $x = \pm \sqrt{3}/2$. This gives the remaining two solutions $(x,y) = (\sqrt{3}/2, -1/2)$ and $(x,y) = (-\sqrt{3}/2, -1/2)$. Having exhausted all possibilities, we summarize the solution set as $$z \in \left\{ 0, i, e^{-\pi i/6}, e^{-5 \pi i/6} \right\}.$$

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Note. I had previously made a rather embarrassing error when converting from rectangular to polar representations of the solution set. There are in fact four distinct solutions, which are now correctly described above.

Answer 2

Did's suggestion is excellent: setting $z = re^{i\theta}$ we have $\bar z = re^{-i\theta}$ and $z^2 = r^2e^{2i\theta}$ whence

$re^{-i\theta} = i r^2 e^{2i\theta}; \tag{1}$

using

$i = e^{i \pi/2} \tag{2}$

yields

$r e^{-i\theta} = r^2 e^{i(2\theta + \pi/2)}; \tag{3}$

taking moduli,

$r = r^2 \Rightarrow r = 0, 1; \tag{4}$

when $r = 0$, $z = 0$; for $r = 1$,

$e^{-\theta} = e^{2\theta+ \pi/2}͵ \tag{5}$

whence

$e^{i(3\theta + \pi/2)} = e^{\pi i/2}(e^{i\theta})^3 = i(e^{i\theta})^3 = 1, \tag{6}$

or

$(e^{i\theta})^3 = -i. \tag{7}$

When $r = 1$, $z = re^{i\theta} = e^{i\theta}$, so (7) tells us that $z$ must be a cube root of $-i$; furthermore, we can work backwards from (7): if

$z^3 = -i, \tag{8}$

then, taking moduli,

$\vert z \vert^3 = 1, \tag{9}$

so that

$\vert z \vert = 1; \tag{10}$

then (8) implies

$i z^2 = i \vert z \vert^2 z^2 = i \bar z z z ^2 = i \bar z z^3 = i (-i) \bar z = \bar z; \tag{11}$

we see the non-zero solutions of

$\bar z = i z^2. \tag{12}$

are precisely the three cube roots of $-i = e^{3\pi i/2}$, viz. $e^{3 \pi i/6} = e^{\pi i / 2} = e^{-3 \pi i / 2} = i$, $e^{(1/2 + 2/3) \pi i} = e^{7 \pi i / 6} = e^{-5 \pi i / 6}$, $e^{(1/2 + 4/3) \pi i} = e^{11 \pi i / 6} = e^{- \pi i / 6}$.

We check this result for $z = e^{7 \pi i / 6}$:

$z = e^{7 \pi i / 6} = -\dfrac{\sqrt{3}}{2} - \dfrac{1}{2} i \tag{13}$

$z^2 = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i; \tag{14}$

$iz^2 = -\dfrac{\sqrt{3}}{2} + \dfrac{1}{2} i = \bar z; \tag{15}$

a similar calculation validates $z = e^{11 \pi i / 6}$.

It is easy to see that $i(i)^2 = -i$ and the solution $z = 0$ "checks itself", as it were. The complete solution set is thus

$\{ 0, i, e^{7 \pi i / 6}, e^{11 \pi i / 6} \}. \tag{16}$

Note: Not to put too fine a point on it, but for the sake of accuracy it should be observed that $2$ of heropup's proposed solutions, namely $e^{2 \pi i / 3}$ and $e^{-2 \pi i / 3} = e^{4 \pi i / 3}$, do not satisfy the given equation (12):

$e^{2 \pi i / 3} = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2} i; \tag{17}$

$(e^{2 \pi i / 3})^2 = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2} i; \tag{18}$

$\overline{e^{2 \pi i / 3}} = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2} i \ne \dfrac{\sqrt{3}}{2} - \dfrac{1}{2} i = i (e^{2 \pi i / 3})^2, \tag{19}$

with a similar calculation for $e^{4 \pi i / 3}$. One can also simply write

$(e^{2 \pi / 3})^3 = e^{2 \pi i} = 1 \ne -i, \tag{20}$

etc. I do not know exactly where heropup's error lies, but we can see from the above his proposed solutions do not all satisfy $\bar z = i z^2$. End of Note.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

Answer 3

$$\bar z=iz^2$$ Equate absolute values: $$|z|=|\bar z|=|iz^2|=|z|^2\Longrightarrow|z|=0,1.$$ One solution is $z=0$; other solutions satisfy $|z|=1$, so $\bar z=\frac{|z|^2}z=\frac1z$ and the equation simplifies to $$\frac1z=iz^2,$$ that is, $$z^3=\frac1i=i^3.$$ One solution obviously is$$z=i.$$ The other cube roots of $i^3$ are $i$ times a cube root of unity, that is, $$i\omega=i(\cos120^\circ+i\sin120^\circ)=i\left(-\frac12+\frac{\sqrt3}2i\right)=-\frac{\sqrt3}2-\frac12i$$ and $$i\omega^2=i\bar\omega=i(\cos240^\circ+i\sin240^\circ)=i\left(-\frac12-\frac{\sqrt3}2i\right)=\frac{\sqrt3}2-\frac12i.$$ Alternatively, $$z^3-i^3=(z-i)(z^2+iz+i^2)=(z-i)(z^2+iz-1)$$ so the last two solutions can be found by solving the quadratic equation $$z^2+iz-1=0.$$