Complex number doubt

Question

If $|z| = \max \{|z-1|,|z+1|\}$, then:

1. $\left| z + \overline{z} \right| =1/2$
2. $z + \overline{z} =1$
3. $\left| z + \overline{z} \right| =1$
4. $z - \overline{z} = 5$

I am totally confused that how to calculate maximum of $|z-1|$ and $|z+1|$, I think they represent line $x=1$ and $x=-1$ and if so then how to calculate maximum.

Please! Guide me how to proceed.

Answer 1

Welcome to MSE. note that $|Z|,|z-1|,|z+1|$ are real numbers, so we can solve the equation, here split in two cases $$ \max \{|z-1|,|z+1|\}=|z-1| or |z+1|\\ (1):|z| =|z-1|\to put \space z=x+iy\\|x+iy|=|x+iy-1|\\\sqrt{x^2+y^2}=\sqrt{(x-1)^2+y^2}\\x^2+y^2=x^2+y^2-2x+1\to x=\frac12\\ (2):|z| =|z+1|\to put \space z=x+iy\\|x+iy|=|x+iy+1|\\\sqrt{x^2+y^2}=\sqrt{(x+1)^2+y^2}\\x^2+y^2=x^2+y^2+2x+1\to x=-\frac12\\$$you can see $Re(z)=\pm \frac 12 \to z=\pm \frac 12 +iy$ now find $z+\bar{z}=(\pm \frac 12+iy )+(\pm \frac 12 -iy) =\pm1$ it seems $Z+\bar{Z}=1$ but when you put $x=\pm 0.5$ in the equation you will find, the question is wrong

let's see $$|\pm 0.5 +iy| = \max \{|\pm 0.5 +iy-1|,|\pm 0.5 +iy+1|\}\\ \sqrt{(\pm0.5)^2+y^2}=\max \{\sqrt{(\pm 0.5+1)^2+y^2},\sqrt{(\pm0.5-1)^2+y^2} \}$$ but it is impossible $$\sqrt{(0.5)^2+y^2}<\sqrt{(1.5)^2+y^2}$$

Answer 2

There is no complex number $z$ such that $|z|$ is the maximum of $|z-1|$ and $|z+1|$.

One can say that all four options are vacuously true!

Proof: write $z$ as $\frac {(z-1)+(z+1)} 2$. This gives $|z| \leq \frac {|z-1|+|z+1|} 2 \leq \max \{|z-1|,|z+1|\}=|z|$. Hence equality holds throughout. In particular this implies $|z-1|=|z+1|$ so $z$ must lies in the imaginary axis. But then $z=iy$ for some real number $y$ and the given equation fails.