I have the following set $A$ be $A$={$z \in \mathbb{C}:z*\bar{z}=2^2$}. Which is the value of $\sup(Re(A))+\inf(Im(A))$ ?
We know that $z*\bar{z}$=$|z|^{2}$. Therefore $|z|^{2}$=$2^{2}$ and we have z=2 or z=-2. so $A$={$2$,$-2$}. And therefore the Supremum would be the $2$ but regarding the $\inf (Im(A))$ wouldn't that be $0$ since $A$ doesn't have an imaginary part, but only has $2$ and $-2$? According to our book the answer should be $0$, so surely I am getting something wrong. I would be really thankful for your help.
Annalisa
Actually,$$|z|^2=2^2\iff|z|=2\iff z=2\bigl(\cos(\theta)+i\sin(\theta)\bigr),$$for some $\theta\in\Bbb R$. And therefore, if $z\in A$, $\operatorname{Re}(z)$ can take any value in $[-2,2]$, and so does $\operatorname{Im}(z)$. So$$\sup\bigl(\operatorname{Re}(A)\bigr)+\inf\bigl(\operatorname{Im}(A)\bigr)=2+(-2)=0.$$