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I've tried solving this, but I'm stuck at one point.

Here's what I did:

Let $ z = x + yi $, where $x, y \in \mathbf R$

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

Or, $x^2 + {(yi)}^2 + 2xyi + \sqrt{x^2 + y^2} = 0 $

Or, $ x^2 - y^2 + 2xyi + \sqrt{x^2 + y^2} = 0 + 0i$

Thus, $ x^2 - y^2 + \sqrt{x^2 + y^2} = 0\qquad\qquad\qquad\qquad\qquad\qquad\ (i)$
and $2xy = 0 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(ii)$

If $2xy = 0$, then either $x = 0 $ or $y = 0$

Now, if I take $ x = 0$, and subsitute in $(i)$, I get either $y = 0$ or $y = 1$.

So far, so good, but if I take $y = 0$, and substitute in $(ii)$:

We have $x^2 + \sqrt{x^2} = 0$

so $x^2 = -\sqrt{x^2} $

or $x^2 = -x$

or $\frac{x^2}{x} = -1 $

or $x = -1$

However, this solution doesn't satistfy the equation $x^2 + \sqrt{x^2}$ or the original equation.

What am I doing wrong here ?

Nathuram
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  • Why do you have $+y^2$ in (i)? Shouldn't it be $-$? – Yuriy S Sep 17 '17 at 11:05
  • Good catch, thanks. – Nathuram Sep 17 '17 at 11:06
  • Let $z=re^{i\theta}$, then $r^2e^{2i\theta}+r=r(re^{2i\theta}+1)=0\iff r=0\text{ or }re^{2i\theta}+1=0$. Solve and reconstruct your complex numbers from $r$ and $\theta$. The trivial solution is $z=0$ (corresponding to $r=0$)... – pshmath0 Sep 17 '17 at 11:07
  • Apart from the obvious solution $z=0$, $|z|$ is always a positive real. That maske $z^2$ a negative real and $z=iy$ purely imaginary. The equation becomes $-y^2+|y|=0$ and as $y^2=|y|^2$, this leads to $|y|=1$. – Hagen von Eitzen Sep 17 '17 at 11:35
  • @HagenvonEitzen I don't understand the bit about $z^2$ being a negative real. How does $\left\vert z \right\vert$ being a positive real make $z^2$ a negative real ? – Nathuram Sep 17 '17 at 22:11

2 Answers2

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In $x^2 = -\sqrt{x^2}$, note that $x$ is real, so you have a non-negative on the left, and a non-positive on the right. Therefore they must both be $0$.

Alternative solution: $$ z^2=-|z|\\ |z|^2=|z|\\ |z|=0\quad \text{or}\quad|z|=1 $$ then for $|z|=1$, solve $z^2+1=0$.

Arthur
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Let $x,y\in\mathbb{R};\;z=x+iy$ the given equation $z^2+|z|=0$ becomes

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

$x^2-y^2+\sqrt{x^2+y^2}+ixy=0$

which translates in the system

$\left\{ \begin{array}{l} x^2-y^2+\sqrt{x^2+y^2}=0 \\ y-2 x y=0 \\ \end{array} \right. $

$\left\{ \begin{array}{l} x^2-y^2+\sqrt{x^2+y^2}=0 \\ 2xy=0 \\ \end{array} \right. $

Which splits into two systems

$\left\{ \begin{array}{l} x^2-y^2+\sqrt{x^2+y^2}=0 \\ x=0 \\ \end{array} \right. $ $\left\{ \begin{array}{l} x^2-y^2+\sqrt{x^2+y^2}=0 \\ y=0 \\ \end{array} \right. $

$\left\{ \begin{array}{l} -y^2+\sqrt{y^2}=0 \\ x=0 \\ \end{array} \right. $ $\left\{ \begin{array}{l} x^2+\sqrt{x^2}=0 \\ y=0 \\ \end{array} \right. $

$\left\{ \begin{array}{l} y^2=|\,y\,| \\ x=0 \\ \end{array} \right. $ $\left\{ \begin{array}{l} x^2=-|\,x\,| \\ y=0 \\ \end{array} \right. $

$\left\{ \begin{array}{l} y=0\lor y= \pm 1 \\ x=0 \\ \end{array} \right. $ $\left\{ \begin{array}{l} x=0\\ y=0 \\ \end{array} \right. $

So we have the solutions

$z=0\lor z=\pm i$

Hope this helps

Raffaele
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