If $| z_{1} | = | z_{2} | = | z_{3} | = ...... = | z_{n} | = 1,$ then $| z_{1} + z_{2} + z_{3} + ....... + z_{n} | =$

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$| \frac{1}{z_{1}^{2}} - \frac{1}{z_{2}^{2}} + \frac{1}{z_{3}^{3}} + ....... + \frac{1}{z_{n}^{2}} |$

$| \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} + ....... + \frac{1}{z_{n}} |$

$| \frac{1}{z_{1}^{2}} + \frac{1}{z_{2}^{2}} + \frac{1}{z_{3}^{3}} + ....... + \frac{1}{z_{n}^{2}} |$

$| \frac{1}{z_{1}} - \frac{1}{z_{2}} + \frac{1}{z_{3}} + ....... + \frac{1}{z_{n}} |$

answer is A.

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Detailed Solution

Z is unimodular $\Rightarrow \bar{z} = \frac{1}{z}$ and $| z | = | \bar{z} |$

$\begin{aligned} |z_{1}| = |z_{2}| = |z_{3}| = \dots = |z_{n}| = 1 \\ \Rightarrow {|z_{1}|}^{2} = {|z_{2}|}^{2} = {|z_{3}|}^{2} = \dots = {|z_{n}|}^{2} = 1 \\ \Rightarrow z_{1} {\bar{z}}_{1} = 1, z_{2} {\bar{z}}_{2} = 1, z_{3} {\bar{z}}_{3} = 1, \dots, z_{n} {\bar{z}}_{n} = 1 \\ \Rightarrow \frac{1}{z_{1}} = {\bar{z}}_{1}, \frac{1}{z_{2}} = {\bar{z}}_{2}, \frac{1}{z_{3}} = {\bar{z}}_{3}, \dots, \frac{1}{z_{n}} = {\bar{z}}_{n} \end{aligned}$