The electric field generated by a uniformly charged infinite rod, standing perpendicular to the $z$-plane at the point $z_0$ is given by $$E(z)=\frac{1}{\overline{z}-\overline{z_0}} $$ in appropriate units.
If three such rods are located at the points $1+i,-1+i$ and $0$, I want to find the points of equilibrium (that is, the points where the vector sum of the fields is zero). In other words, I want to solve the equation $$\frac{1}{\overline{z}}+\frac{1}{\overline{z}-(1-i)}+\frac{1}{\overline{z}-(-1-i)}=0.$$ Writing $z=x+yi$ and equating the real and imaginary parts, I got the simultaneous equations $$\begin{align} \frac{3 x^5+6 x^3 y^2-8 x^3 y+3 x y^4-8 x y^3+8 x y^2-8 x y+4 x}{\left(x^2+y^2\right) \left(x^2-2 x+y^2-2 y+2\right) \left(x^2+2 x+y^2-2 y+2\right)}&=0, \\ \frac{3 x^4 y-2 x^4+6 x^2 y^3-12 x^2 y^2+8 x^2 y-4 x^2+3 y^5-10 y^4+16 y^3-12 y^2+4 y}{\left(x^2+y^2\right) \left(x^2-2 x+y^2-2 y+2\right) \left(x^2+2 x+y^2-2 y+2\right)}&=0. \end{align} $$ Mathematica can solve them as $x+iy=(\pm\sqrt{2}+2i)/3 $. However, I don't know how to handle this system by hand. How does one go about solving these equations? Also, is there a shorter way to get to the answer?
Thank you!
We want to solve $$\frac{1}{\overline{z}}+\frac{1}{\overline{z}-(1-i)}+\frac{1}{\overline{z}-(-1-i)}=0 \tag 1$$ Multiply $(1)$ by $\overline{z}(\overline{z}-(1-i))(\overline{z}-(-1-i))$ and simplify to get $$3\overline{z}^2+4i\overline{z}-2=0.$$ Now use the quadratic formula to get $$\overline{z} = -\frac{2}{3} i \pm \frac{\sqrt{2}}{3}.$$