Question

If $|z_1|=|z_2|=...=|z_n|=1$, then the value of $|z_1+z_2+z_3+....+z_n|$ is :

• A. $1$
• B. $|z_1|+|z_2|+..|z_n|$
• C. $|\frac{1}{z_1}+\frac{1}{z_2}+..+\frac{1}{z_n}|$
• D. $Noneofthese$

Answer 1

The correct option is C $|\frac{1}{z_1}+\frac{1}{z_2}+..+\frac{1}{z_n}|$

Given that $\left|z_1\right|=\left|z_2\right|=\left|z_1\right|=...=\left|z_n\right|=1$

We need to prove that

$|z_1+z_2+z_3+...+z_n|=|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}+...+\frac{1}{z_n}|$

Consider $|z_1+z_2+z_3+...+z_n|=|\frac{z_1\overline{z_1}}{z_1}+\frac{z_2\overline{z_2}}{z_2}+\frac{z_3\overline{z_3}}{z_3}+...+\frac{z_n\overline{z_n}}{z_n}|$

$=|\frac{{\left|z_1\right|}^2}{\overline{z_1}}+\frac{{\left|z_2\right|}^2}{\overline{z_2}}+\frac{{\left|z_3\right|}^2}{\overline{z_3}}+...+\frac{{\left|z_n\right|}^2}{\overline{z_n}}|$

$=\frac{|\frac{1}{\overline{z_1}}+\frac{1}{\overline{z_2}}+\frac{1}{\overline{z_3}}+...+\frac{1}{\overline{z_n}}|}{|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}+...+\frac{1}{z_n}|}$ since $\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=\left|z_4\right|=...=\left|z_{n1}\right|=1$

$=|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}+...+\frac{1}{z_n}|$