If $|{z}|=maxbig{|{z}+2|,|{z}-2|big}$, then which is true: $|zpm bar{z}|=2$ or $|zpm bar{z}|=1/2$?

Question

If $|{z}|=maxbig{|{z}+2|,|{z}-2|big}$, then
(a) $|{z}-bar{z}|=1 / 2$.
(b) $|{z}+overline{{z}}|={2}$.
(c) $|{z}+overline{{z}}|=1 / 2$.
$({d})|{z}-overline{{z}}|={2}$.

My approach
$|z|=|z+2|$
$Rightarrow {zoverline{z}}=({z}+2)(overline{{z}}+2)$
$Rightarrow {z}+{overline{z}}=-2 Rightarrow|{z}+{overline{z}}|=2$
$|z|=|z-2| Rightarrow z bar{z}=(z-2)(bar{z}-2)$
$Rightarrow {z}+overline{{z}}=2 Rightarrow|{z}+overline{{z}}|=2$
I guess it is right.
My question is how can I  come to solution using any graphical approach!
for ref:- https://www.desmos.com/calculator/nnnnairelh

user376343 · Answer

For a geometric approach:
$$|z|=|z-2|$$ is the equation of the bissector of the segment with ends $0$ and $2.$
Its equation can be written $Re(z)=1.$
It is not excluded that the author of the MCQ was thinking of such approach.
BUT:

If the real part satisfies $Re(z)<0$ then the distance from $z$ to $2$ is greater than the distance to $-2.$
In other words, $|z-2|$ is the maximum. However, $|z|=|z-2|$ is the above mentionned bissector and lies in the half-plane with positive real parts.

If $Re(z)>0,$ we arrive to a similar contradiction.

Cesareo · Answer

If
$$
|z|=maxleft(|z+2|,|z-2|right)Rightarrow 1 = maxleft(|1+2/z|,|1-2/z|right)
$$
but
$$
maxleft(|1+2/z|,|1-2/z|right) gt 1
$$

If $|{z}|=maxbig{|{z}+2|,|{z}-2|big}$, then which is true: $|zpm bar{z}|=2$ or $|zpm bar{z}|=1/2$?

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