Solving simultaneous equations with complex numbers and moduli Would anyone be able to give me a starting point as to how to approach this? A friend suggested squaring both sides, but I read something that said you can't square both sides of an equation(?).
Complex numbers and simultaneous equations question
If $x$ and $y$ are both real numbers, find all the solutions $(x,y)$ of the simultaneous equations
$$
\begin{cases}
\lvert x + i y \rvert &= 1\\[3pt]
\big\lvert x + i y - \frac{3}{2} \big\rvert &= 2
\end{cases}
$$
Thank you in advance!
Edit: So, would you square the real and imaginary parts separately? Giving: $x^2+y^2=1$ for the first one? Then, $(x-3/2)^2+y^2=4$ for the second one?
 A: From the first equation,
$$
 \color{blue}{\lvert x + i y \rvert = \sqrt{x^{2}+y^{2}} = 1} \qquad \Rightarrow \qquad \color{blue}{y_{1}=\pm \sqrt{1-x^2}}
$$
From the second equation,
$$
 \color{red}{\lvert x -\frac{3}{2} + i y  \rvert =\sqrt{\left(x-\frac{3}{2}\right)^2+y^2}}
\qquad \Rightarrow \qquad 
\color{red}{y_{2} =\frac{1}{2} \sqrt{-4 x^2+12 x+7}}
$$

Solve
$$
 \color{blue}{y_{1}(x)} = \color{red}{y_{2}(x)}
$$
to see that
$$
 x= -\frac{1}{4}, \qquad y = \pm \frac{\sqrt{15}}{4}
$$


A: Hint: let $z=x+iy$ and write the system as:
$$
\begin{cases}
\begin{align}
\lvert z \rvert &= 1 \\
\left| z - \frac{3}{2} \right| &= 2
\end{align}
\end{cases}
$$
Since $z \bar z = |z|^2=1$ the second equation is equivalent to:
$$
2^2 = \left| z - \frac{3}{2} \right|^2 = \left( z - \frac{3}{2} \right)  \left( \bar z - \frac{3}{2} \right) = z \bar z - \frac{3}{2}(z+\bar z)+\frac{9}{4} = -3 \operatorname{Re}(z)+\frac{13}{4}
$$
Therefore $\operatorname{Re}(z)=x=-\cfrac{1}{4}\,$, then $\operatorname{Im}(z) = y = \pm \sqrt{1 - \operatorname{Re}^2(z)}\,$.
