Q50 / 125 · BITSAT 2020 · Mathematics

If $\left|z_{1}\right|=\left|z_{2}\right|=\ldots \ldots \ldots . .\left|z_{n}\right|=1,$ then the value of $\left|z_{1}+z_{2}+\ldots \ldots . z_{n}\right|-$ $\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\ldots \ldots .+\frac{1}{z_{n}}\right|$ is,

  • Correct answer
    A
    0
  • B
    1
  • C
    -1
  • D
    None

Solution

$z_{1} \bar{z}_{1}=z_{2} \bar{z}_{2}=\ldots .=z_{n} \bar{z}_{n}=1$ $\Rightarrow \overline{\mathrm{z}}_{1}=\frac{1}{\mathrm{z}_{1}}, \overline{\mathrm{z}}_{2}=\frac{1}{\mathrm{z}_{2}}, \overline{\mathrm{z}}_{3}=\frac{1}{\mathrm{z}_{3}}, \ldots \ldots, \overline{\mathrm{z}}_{\mathrm{n}}=\frac{1}{\mathrm{z}_{\mathrm{n}}}$ $\therefore\left|z_{1}+z_{2}+\ldots .+z_{n}\right|-\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\ldots .+\frac{1}{z_{n}}\right|$ $=\left|z_{1}+z_{2}+\ldots+z_{n}\right|-\left|\bar{z}_{1}+\bar{z}_{2}+\ldots .+\bar{z}_{n}\right|=0$

Hence, the correct answer is option: 'A'.

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