Saturday, November 17, 2018

Solve the equation $ z^2 + leftvert z rightvert = 0 $, where $z$ is a complex number.

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 ?

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