Solve equation $|z|^2+z|z|+\bar{z} = 0$

Question

I literally have no idea on how to solve equations like this: $$|z|^2+z|z|+\bar{z} = 0$$ Even if I try to rewrite it as $$(x^2+y^2)+(x+iy)\sqrt{x^2+y^2}+(x-iy) = 0$$ how should I proceed?

Answer 1

Hint: Use polar coordinates: $z=re^{i\theta}$:

$$r^2 +re^{i\theta}\cdot r + re^{-i\theta} = 0 \\ r^2 +r^2 e^{i\theta}+ re^{-i\theta} = 0$$

Answer 2

Looking at imaginary parts: $$y\sqrt{x^2+y^2}-y=0,$$ so $y=0$ or $\sqrt{x^2+y^2}-1=0$.

In the case $y=0$, you then just have the equation $2x^2+x=0$, so $x=0$ or $-\frac{1}{2}$, so $z=0$ or $z=-\frac{1}{2}$. But we find if $z=-\frac{1}{2}$, the real part is not zero, so we reject this solution.

In the case $\sqrt{x^2+y^2}=1$, we have $|z|=1$, so the equation becomes $1+z+\overline{z}=0$, which implies $x=-\frac{1}{2}$, so $y=\pm\sqrt{1-x^2}=\pm\frac{\sqrt{3}}{2}$ and $z=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$.

Therefore, the solutions are $z=0,-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$.

Answer 3

Equate the real & the imaginary parts

$$y\sqrt{x^2+y^2}-y=0\iff y(\sqrt{x^2+y^2}-1)=0\ \ \ \ (1)$$

$$x^2+y^2+x\sqrt{x^2+y^2}+x=0\ \ \ \ (2)$$

Case $\#1:$

By $(1)$ if $y=0$ by $(2),$

$$x^2+|x|+x=0, x=?$$

Check when $x\ge0$ and when $x<0?$

Case $\#2:$

By $(1)$ if $\sqrt{x^2+y^2}=1$ by $(2),1+x+x=0$

Answer 4

Write $z=x+iy$ and take the imaginary part: $$|z|y-y=0$$ so either (i) $y=0$ or (ii) $|z|=1$. In (ii) you get $1+z+\overline z=0$. Then $z=\cos t+i\sin t$, so that $1+2\cos t=0$. Then $\cos t=-1/2$ and $\sin t=\pm\sqrt 3/2$. In (i), $z$ is real, and I'll leave that to you.

Answer 5

Given $|z|^2+z|z|+\bar{z} = 0$, $z |z|+\bar{z} $ is real, i.e.

$$z |z|+\bar{z} = \bar{z} |z|+z$$

Factorize to get

$$(|z|-1)(z-\bar{z}) =0$$

So, two cases to consider:

Case 1) $z-\bar{z} =0$ yields the solution $z=0$.

Case 2) $|z|-1=0$ leads to

$$\frac{z^3-1}{z-1} = z^2+z+1= \frac1{\bar z}[(|z|-1) z|z| +(|z|^2+z|z|+\bar z)]=0$$ or $z^3=1$ and $z\ne1$, which yields the solutions $z=e^{i2\pi/3}$ and $z=e^{i4\pi/3}$.