# How do I find all complex numbers $z$ such that $z^2+|z|=0$? [duplicate]

I need help solving this task, if anyone had a similar problem, it would help me.

Determine the complex numbers $$z$$ from the condition. $$z^2+|z|=0$$

By my logic, the solutions are : $$-i$$ and $$i$$. But I don’t know how to come up with those solutions.

• I would start by writing $z=a+bi$ and then calculating the equivalent equations for $a$ and $b$. Please try that and come back with what you get from that. – Matti P. Sep 18 at 9:13

Hint : Assume that $$z = x + i(y)$$, where $$x,y \in \mathbb{R}.$$

Then express both $$(z^2)$$ and $$|z|$$ in terms of $$x$$ and $$y$$.

Then use the expressions to create an equation between $$x$$ and $$y$$.

Then, (if possible), simplify the equation as much as possible.

Hint (Alternate):

First check if $$z = 0$$ fits the constraint.

Then, (separately) assume that $$|z| = r$$, where $$r \in \mathbb{R^+}.$$

Then, assume that $$z = r(\cos \theta + i\sin \theta),$$ where $$\theta \in (-\pi, \pi].$$

Then, set up an equation between $$r$$ and $$\theta$$ (if possible). Then, try to simplify this equation.

• Thank you !! :) – Srdjan Pesevic Sep 18 at 9:47

$$z=0$$ is also a solution.

Indeed let $$z=x+iy$$ then $$z^{2}+|z|=(x+iy)^{2}+|x+iy|=x^2-y^2+\sqrt{x^2+y^2}+2ixy=0.$$

Now comparing the real and imaginary parts we must have $$x^2-y^2+\sqrt{x^2+y^2}=0\space\space\space\space\space\space\space\space \text{and} \space\space\space\space\space\space\space xy=0$$

From $$xy=0$$ we have either $$x=0$$ or $$y=0$$ or both $$x=y=0$$.

Clearly $$x=y=0$$ satisfies both equations, thus $$z=0$$ is a solution.

For $$x=0$$ the other equation gives $$y^{2}=|y|$$ and considering the definition of $$|y|$$ the possible values are $$y=1,-1$$ or $$0.$$ These give $$z=x+iy$$ to be $$z=i,-i$$ or $$z=0.$$

Similarly when $$y=0$$ you'll find that $$x=0,-i$$ or $$i.$$

As $$\vert z \vert$$ is a real, $$z^2$$ has to be a real too. Hence $$z=a$$ or $$z=ia$$ with $$a \in \mathbb R$$.

In the first case $$a^2 = -\vert a \vert$$ and $$a=0$$ is a unique solution as $$\vert a \vert \ge 0$$.

In the second case, you get $$a^2 = \vert a \vert$$. Which is equivalent to $$a \in \{0,-1,1\}$$.

Finally the solution set is $$\{0,-i,i\}$$.

• You've still your two cases mixed up, or your signs wrong. – Jaap Scherphuis Sep 18 at 9:27
• @JaapScherphuis Thanks... corrected hopefully now! – mathcounterexamples.net Sep 18 at 9:28
• Check again. The given equation is $z^2=-|z|$. – Jaap Scherphuis Sep 18 at 9:33
• Houuuuups...... – mathcounterexamples.net Sep 18 at 9:36
• Thank you ! :)) – Srdjan Pesevic Sep 18 at 9:47

There's naturally the obvious solution $$z=0$$.

For the other solutions, it is very short using the exponential form of complex numbers. So set $$z= r\mathrm e^{i\theta}\qquad(r>0,\; -\pi <\theta\le \pi).$$ The equation becomes $$r=-r^2\mathrm e^{2i\theta}\iff -1=re^{2i\theta}\iff \begin{cases}r=1\\2\theta\equiv \pi\bmod 2\pi\end{cases}\iff \begin{cases}r=1\\\theta\equiv \frac\pi 2\bmod \pi\end{cases}$$ so the nonzero solutions are $$\mathrm e^{i\tfrac\pi 2}=i,\quad\text{ and }\quad\mathrm e^{-i\tfrac\pi 2}=-i.$$