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I have managed to work out the answer in some dirty brute force manner and was wondering if there was an easier way.

The question is, find the locus of points of $z$ such that:

$$ (z-2+i)^2+(z-2-i)^2+2zz^*+6=0$$

I substituted $z=x+iy$ and just worked out a relation and got a parabola. Is there an easier way?

Thank you

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Expanding the squares and canceling out the $\,\pm 2 i (z-2)\,$ terms:

$$0=(z-2+i)^2+(z-2-i)^2+2zz^*+6 = 2(z-2)^2+2z z^*+4 \tag{1}$$

Taking the conjugate:

$$0 = 2(z^*-2)^2+2z^*z+4 \tag{2}$$

Subtracting $\frac{1}{2}(1)-\frac{1}{2}(2)\,$:

$$\require{cancel} 0 = (z-2)^2-(z^*-2)^2=(z-z^*)(z+z^*-4) \tag{3} $$

The solution set is therefore a subset of the union of:

  • $\,z-z^*=0\,$ $\iff$ $\,\operatorname{Im}(z)=0\,$ which is the real axis, but in this case $\,z=x \in \mathbb{R}\,$ and $(1)$ becomes $\,2(x-2)^2+2x^2+4\gt 0\,$, so there are no solutions in this set;

  • $\,z+z^*-4=0\,$ $\iff$ $\,\operatorname{Re}(z)=2\,$ which is the vertical line passing through $z=2$ on the real axis, but in this case $\,z=2+iy \mid y \in \mathbb{R}\,$ and $\,(1)\,$ becomes $\,2(\cancel{2}+iy-\cancel{2})^2+2(4+y^2)+4\,$ $\,=\cancel{-2y^2}+8+\cancel{2y^2}+4=12 \gt 0\,$, so there are no solutions in this set, either.

Therefore the equation has no solutions.



P.S.  In case the problem was mistyped, and the equation was, instead:

$$ (z-2+i)^2+(\color{red}{z^*}-2-i)^2+2zz^*+6=0$$

This can be rewritten as:

$$ \begin{align} 0 &= z^2-2z(2-i)+(2-i)^2 + {z^*}^2-2z^*(2+i)+(2+i)^2 + 2 z z^* + 6 \\ &= z^2 + 2 z z^* + {z^*}^2 - 4(z+z^*) + 2i(z - z^*) + 12 \\ &= (z+z^*)^2 - 4(z+z^*) +2i(z-z^*)+12 \end{align} $$

In cartesian coordinates $\,z=x+iy\,$, and using that $z+z^*=2x, z-z^*=2iy\,$, the above reduces to the equation of a parabola, indeed:

$$ 0 = 4x^2 - 8x - 4y + 12 = 4(x-1)^2 -4y + 8 \;\;\iff\;\; y = (x-1)^2 + 2 $$

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