Question

If $|z|=maxof|z-1|,|z+1|$, then

• A. $|z+\overline{z}|=\frac{1}{2}$
• B. $z+\overline{z}=0$
• C. $|z+\overline{z}|=1$
• D. $zϵ\varphi$

Answer 1

The correct option is C $|z+\overline{z}|=1$

$\left|z\right|=maxof|z-1|,|z+1|$

Let $z=a+ib,$ where $a$ and $b$ are real number,

$\therefore \left|z\right|=\sqrt{{(a-1)}^2+b^2}$ or $\left|z\right|=\sqrt{{(a+1)}^2+b^2}$

In the first case,

$\Rightarrow \left|z\right|=\sqrt{{(a-1)}^2+b^2}=\sqrt{a^2+b^2}$

$\therefore a^2+b^2={(a-1)}^2+b^2$

Solving this, we get $a=\frac{1}{2}$

When, $z+\overline{z}=a+ib+a-ib=2a$

$\Rightarrow 2a=2\times \frac{1}{2}=1$

$\Rightarrow |z+\overline{z}|=\sqrt{1^2+0^2}=1$

$\Rightarrow |z+\overline{z}|=1$

Hence, the answer is $|z+\overline{z}|=1.$