How find this $a$ such $(z_{1}+a)^2+a\overline{z_{1}}\neq (z_{2}+a)^2+a\overline{z_{2}}$

Question

Question:

find all the complex $a$,such for every complex $z_{1},z_{2}(|z_{1}|,|z_{2}|<1,z_{1}\neq z_{2})$,such $$(z_{1}+a)^2+a\overline{z_{1}}\neq (z_{2}+a)^2+a\overline{z_{2}}$$

My idea: let $$a=x+yi$$ then $$(z_{1}+z_{2}+2a)(z_{1}-z_{2})\neq a(\overline{z_{2}}-\overline{z_{1}})$$ so $$a=\dfrac{z_{1}+z_{2}}{\dfrac{\overline{z_{1}}-\overline{z_{2}}}{z_{1}-z_{2}}+2}$$ then I can't

Answer 1

This is not a solution. Just some thoughts.

$$(z_{1}+z_{2}+2a)(z_{1}-z_{2})\neq a(\overline{z_{2}}-\overline{z_{1}})\tag{1}$$

From (1), we have:

$$(z_{1}+z_{2})(z_{1}-z_{2})\neq a(\overline{z_{2}}-\overline{z_{1}})-(2a)(z_{1}-z_{2})\tag{2}$$

or

$$(z_{1}+z_{2})(z_{1}-z_{2})\neq a(\overline{z_{2}}-\overline{z_{1}}-2z_{1}+2z_{2})\tag{3}$$

or (if the denominator in (4) is not zero)

$$a\neq \frac{(z_{1}+z_{2})(z_{1}-z_{2})}{(\overline{z_{2}}-\overline{z_{1}}-2z_{1}+2z_{2})}\tag{4}$$

Answer 2

We want that $$(z_1+z_2+2a)(z_1-z_2)+a(\bar z_1-\bar z_2)=0,\quad z_1,z_2\in D:=\bigl\{z\>\bigm|\>|z|<1\bigr\}\tag{1}$$ together imply $z_1=z_2$.

Write $$z_1+z_2=:2z, \quad z_1-z_2=\rho\>e^{i\phi}\ .$$ Then $z\in D$, and the pair $(z, \rho e^{i\phi})$ determines $z_1$ and $z_2$. Rewriting $(1)$ in terms of the new variables we obtain the equation $$2(z+a)\rho e^{i\phi}+a\rho e^{-i\phi}=0\ .\tag{2}$$ When $\rho\ne0$ this is equivalent to $$z=-a-{a\over2}e^{-2i\phi}\ .\tag{3}$$ From $(3)$ we draw the following conclusions:

(I)$\quad$ If the circle with center $-a$ and radius ${|a|\over2}$ intersects $D$ in a point $z$ then $(2)$ has a solution $(z,\rho e^{i\phi})$ with $z\in D$ and arbitrary $\rho>0$. If $\rho$ is choosen sufficiently small the two corresponding points $z_1$, $z_2$ both lie in $D$, are different, and satisfy $(1)$. It follows that such an $a$ is forbidden.

(II)$\quad$ If the circle with center $-a$ and radius ${|a|\over2}$ does not intersect $D$ then $(2)$ has no solutions $(z,\rho e^{i\phi})$ with $z\in D$ and $\rho>0$. It follows that such an $a$ is allowed.

On account of (I) and (II) the set of allowed values for $a$ is characterized by $|a|\geq2$.