I am trying to solve the following problem:
Let : $$ \bbox[,5px] { (\forall z\in \mathbb{C}): A(z)=z^2+2z+2 } $$
Prove that : $$ \bbox[pink,5px] { (\forall z\in \mathbb{C}) ; (A(z)=A(\bar{z})) \Leftrightarrow (z=\bar{z} \text{ $$ or $$ } \Re(z)=-1) } $$
I tried to solve the problem by considering $$ \bbox[,px] (A(z)=A(\bar{z})) $$; letting z=x+iy but I am stuck, can someone please provide a hint ? am I even on the right track ?
I couldn't find any error in my calculation. Any help would be highly appreciated. Thanks in advance.
Note that $$A(z)=z^2+2z+2 \quad A(\bar z)=\bar z^2+2\bar z+2$$
then
$$A(z)=A(\bar z) \iff z^2+2z+2= \bar z^2+2\bar z+2 \iff z^2+2z= \bar z^2+2\bar z\\\iff z^2-\bar z^2 = 2\bar z -2z \iff (z-\bar z)(z+\bar z) = -2(z-\bar z)\iff\\ z=\bar z\quad \lor\quad z+\bar z=2\cdot\Re(z)=-2$$