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.
 A: 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 $\lvert\operatorname{Re} z\rvert=1.$
A: Let me establish that no $z\in\mathbb{C}$ exists such that $|z|=\max\big\{|z-2|,|z+2|\big\}$.  Suppose such a $z$ exists, then $|z|\geq|z+2|$ and $|z|\geq|z-2|$.  By the Triangle Inequality, $$2|z| \geq |z+2|+|z-2| \geq \big|(z+2)+(z-2)\big|=|2z|=2|z|\,,$$ 
whence we have an equality, which holds if and only if $z \in (-\infty,-2]\cup[+2,+\infty)$.  However, if $z\in(-\infty,-2]$, we have $|z-2|>|z|$; on the other hand, if $z \in [+2,+\infty)$, $|z+2|>|z|$.
Any inference with a false premise is always valid.
As other people have noted, if you indeed wanted to find $z$ such that $|z|=\min\big\{|z-2|,|z+2|\big\}$, then all solutions are $z\in\mathbb{C}$ with $\text{Re}(z)\in\{-1,+1\}$.
A: A requirement for $|z| = \max{|z-2|, |z+2|}$ is that $|z| = |z-2|$ or $|z| = |z+2|$. (It's a requirement, it's not sufficient!).
In $\mathbb{C}$, $|x| = |y|$ if they lie on the same circle centered at zero. If $|z|$ and $|z-2|$ are on the same circle, then $\Re (z) > 0$ and thus $|z-2| > |z|$, so the statement is false.
If $|z|$ and $|z+2|$ are on the same circle, then $|Re(z) < 0$ and thus $|z+2| > |z|$ so the statement is false.
Therefore, no such $z$ exists and the inference is (vacuously) true.
A: 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
