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

Question

If $|{z}|=\max\big\{|{z}+2|,|{z}-2|\big\}$, then

(a) $|{z}-\bar{z}|=1 / 2$.

(b) $|{z}+\overline{{z}}|={2}$.

(c) $|{z}+\overline{{z}}|=1 / 2$.

$({d})|{z}-\overline{{z}}|={2}$.

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My approach

$|z|=|z+2|$ $\Rightarrow {z\overline{z}}=({z}+2)(\overline{{z}}+2)$

$\Rightarrow {z}+{\overline{z}}=-2 \Rightarrow|{z}+{\overline{z}}|=2$

$|z|=|z-2| \Rightarrow z \bar{z}=(z-2)(\bar{z}-2)$

$\Rightarrow {z}+\overline{{z}}=2 \Rightarrow|{z}+\overline{{z}}|=2$

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

Answer 1

If

$$ |z|=\max\left(|z+2|,|z-2|\right)\Rightarrow 1 = \max\left(|1+2/z|,|1-2/z|\right) $$

but

$$ \max\left(|1+2/z|,|1-2/z|\right) \gt 1 $$

Answer 2

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.