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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?

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  • $\begingroup$ Not sure what you mean by the remark about squaring. "$|z|=c$" is equivalent to "$|z|^2=c^2$" as long as $c\geq 0$ (in your equations, $c$ is $1$ or $2$, so this is certainly the case). $\endgroup$ – MPW May 3 '17 at 20:31
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    $\begingroup$ you are basicly intersecting two circles, can you see that? $\endgroup$ – Maffred May 3 '17 at 20:31
  • $\begingroup$ Simultaneous quadratics - eliminate $y^2$ and solve for $x$ $\endgroup$ – David Quinn May 3 '17 at 20:40
  • $\begingroup$ @MPW I'm not entirely sure what I meant either, just something vague I read haha! Thank you for clearing that up! $\endgroup$ – Emma Louise May 3 '17 at 20:41
  • $\begingroup$ @Maffred Ah, yeah I do see that now! Thanks :-) $\endgroup$ – Emma Louise May 3 '17 at 20:41
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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 $$ x= -\frac{1}{4}, \qquad y = \pm \frac{\sqrt{15}}{4} $$
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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 $\operatorname{Im}(z) = y = \pm \sqrt{1 - \operatorname{Re}^2(z)}\,$.

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