0
$\begingroup$

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

The task is:

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Matti P. Sep 18 at 9:13
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you !! :) $\endgroup$ – Srdjan Pesevic Sep 18 at 9:47
2
$\begingroup$

$z=0$ 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 $i.$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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 $\{0,-i,i\}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ You've still your two cases mixed up, or your signs wrong. $\endgroup$ – Jaap Scherphuis Sep 18 at 9:27
  • $\begingroup$ @JaapScherphuis Thanks... corrected hopefully now! $\endgroup$ – mathcounterexamples.net Sep 18 at 9:28
  • $\begingroup$ Check again. The given equation is $z^2=-|z|$. $\endgroup$ – Jaap Scherphuis Sep 18 at 9:33
  • $\begingroup$ Houuuuups...... $\endgroup$ – mathcounterexamples.net Sep 18 at 9:36
  • $\begingroup$ Thank you ! :)) $\endgroup$ – Srdjan Pesevic Sep 18 at 9:47
0
$\begingroup$

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 $$\mathrm e^{i\tfrac\pi 2}=i,\quad\text{ and }\quad\mathrm e^{-i\tfrac\pi 2}=-i.$$

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.