# System of equations with complex numbers and absolute value

I am trying to solve the following system of equations: $|z|=2 \ ; |z-2|=2$
If $\ \ z=a+bi$ is a complex number I have got that $a=1$, but I don't know how to calculate $b$ because it gives me a contradition. I'm not sure of the result of $a$, I would like to find a good method of resolution of the system.

$$\left| z \right| =2\\ \left| z-2 \right| =2\\ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } =2\Rightarrow { a }^{ 2 }+{ b }^{ 2 }=4\\ \sqrt { { \left( a-2 \right) }^{ 2 }+{ b }^{ 2 } } =2\Rightarrow { a }^{ 2 }-4a+4+{ b }^{ 2 }=4\Rightarrow a=1\\ { a }^{ 2 }+{ b }^{ 2 }=4\Rightarrow b=\pm \sqrt { 3 } \\ \\$$

If you write as $a+bi$ ($a,b\in\mathbb R$), then $|z|=2\iff a^2+b^2=4$ and $|z-2|=2\iff(a-2)^2+b^2=4\iff a^2+b^2=4a$. It follows from these two equalities that $a=1$. Therefore, $b=\pm\sqrt3$. So, the solutions are $1\pm\sqrt3i$.

$|z|=2$ then let $z=2e^{it}$ and $|z-2|=2$ so $|2e^{it}-2|=2$ that is $|e^{it}-1|=1$ or $$(\cos t-1)^2+\sin^2t=1$$ so $t=\pm\dfrac{\pi}{3}$ and finally $z=2e^{\pm i\frac{\pi}{3}}=\color{blue}{1\pm i\sqrt{3}}$.

$|z|=2\iff z\in\mathscr C(O,2)$

$|z-2|=2\iff z\in\mathscr C(A,2)$

The solutions are the intersections points $B$ and $B'$ of abscissa $a=1$ given by $I=(1,0)$ middle of $[OA]$.

The ordinate is $b=BI$ and Pythagoras gives $b^2=OB^2-OI^2=2^2-1^2=3$

Finally the solutions are $z=a+ib=1\pm i\sqrt{3}$

Alternately you can solve it algebraically

$\begin{cases} |z|=2\\ |z-2|=2\end{cases}\iff \begin{cases} a^2+b^2=4\\ (a-2)^2+b^2=4\end{cases}\iff \begin{cases} a^2-(a-2)^2=0\\ b^2=4-a^2\end{cases}\iff \begin{cases} a=1\\ b^2=3\end{cases}$

Detail:

$|z|^2=z\bar z=(a+ib)(a-ib)=a^2-i^2b^2=a^2+b^2$

$|z-2|^2=(z-2)\overline{(z-2)}=((a-2)+ib)((a-2)-ib)=(a-2)^2-i^2b^2=(a-2)^2+b^2$

A lil last for the road.

Let $u=\dfrac{z+(z-2)}2=z-1$

$\begin{cases} |z|=2 \\ |z-2|=2 \end{cases}\iff \begin{cases} |u+1|=2 \\ |u-1|=2 \end{cases}\iff \begin{cases} (u+1)(\bar u+1)=4 \\ (u-1)(\bar u-1)=4 \end{cases}\iff \begin{cases} u+\bar u=0\\ u\bar u=3\end{cases}$

So $\Re(u)=0$ and $\Im(u)^2=3$ and consequently $u=z-1=\pm i\sqrt{3}$