The real part of a complex number such that $|z|=\max\{|z-2|,|z+2|\}$ If |z|=max{|z-2|,|z+2|} then -
INFERRENCE - |Re(z)|=1

Is the inferrence incorrect?

My approach is -
|z|=|z-2| when |z-2| {i.e. distance of z from 2 is greater}is greater OR |z+2| when |z+2| is greater.

|z-2|>|z+2| means Re(z)>0....and so on.
If $\lvert z\rvert=\lvert z-2\rvert,$ then one can show that $\operatorname{Re} z=1,$ and so $\lvert z+2\rvert\gt\lvert z\rvert.$ But then $\lvert z+2\rvert\gt\lvert z-2\rvert$ by assumption, whence $$\lvert z\rvert\lt\max\bigl(\lvert z-2\rvert,\lvert z+2\rvert\bigr).$$

On the other hand, if $\lvert z\rvert=\lvert z+2\rvert,$ then one can show that $\operatorname{Re} z=-1,$ and so $\lvert z-2\rvert\gt\lvert z\rvert.$ But then $\lvert z-2\rvert\gt\lvert z+2\rvert$ by assumption, whence $$\lvert z\rvert\lt\max\bigl(\lvert z-2\rvert,\lvert z+2\rvert\bigr).$$

Hence, there is no $z$ such that $$\lvert z\rvert=\max\bigl(\lvert z-2\rvert,\lvert z+2\rvert\bigr),$$ so (vacuously), if $z$ satisfies said impossible condition, then $\lvert\operatorname{Re} z\rvert=1.$ More interestingly, we can see by the above reasoning that $$\lvert z\rvert=\min\bigl(\lvert z-2\rvert,\lvert z+2\rvert\bigr)$$ if and only if $\lvert\operatorname{Re} z\rvert=1.$
So , the conclusion is the given inference is right (though vacuously) and if max is replaced with min , the INFERENCE is more meaningful and correct