# The complex equation

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...

• 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... Jul 21 '14 at 15:46
• More context in the problem (particularly regarding what you've tried so far) greatly improves a problem statement such as this. Jul 21 '14 at 15:47
• Remember that $|z| \geq 0$. Thus the negative root (I assume) you found is therefore not valid. Jul 21 '14 at 15:50
• The polar form $z=r e^{i \phi}$ may also be useful. Jul 21 '14 at 15:51
• Observe that $1-2z$ is purely imaginary. Thus $z=\frac{1-bi}{2}$. Now solve for $b$. Bear in mind that $b>0$. Jul 21 '14 at 15:56

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