In solving $|z|i +2z =1$, I seem to be constantly getting two solutions while both answer key and Wolfram claim to be only one. What am I doing wrong?

Let's share the fun:

$(\sqrt{x^2 +y^2}) i +2x +2iy =1$

leading to the system of : $$ 2y+ \sqrt{x^2 +y^2}= 0 $$ $$2x=1$$

upon solving we get $(0.5, \sqrt{1/12})$, or $(0.5, -\sqrt{1/12})$ as possible real coordinates of the complex number which may solve said equation...

  • 2
    $\begingroup$ We can't say what you're doing wrong unless we know what it is you're doing. Hint though: write $z=x+iy$ and group real and imaginary parts... $\endgroup$
    – lemon
    Jul 21 '14 at 15:46
  • $\begingroup$ More context in the problem (particularly regarding what you've tried so far) greatly improves a problem statement such as this. $\endgroup$ Jul 21 '14 at 15:47
  • $\begingroup$ Remember that $|z| \geq 0$. Thus the negative root (I assume) you found is therefore not valid. $\endgroup$
    – Winther
    Jul 21 '14 at 15:50
  • $\begingroup$ The polar form $z=r e^{i \phi}$ may also be useful. $\endgroup$ Jul 21 '14 at 15:51
  • $\begingroup$ Observe that $1-2z$ is purely imaginary. Thus $z=\frac{1-bi}{2}$. Now solve for $b$. Bear in mind that $b>0$. $\endgroup$
    – Anurag A
    Jul 21 '14 at 15:56



But $\displaystyle\sqrt{x^2+y^2}\ge0\implies y\le0$


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