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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Complex Number

BITSAT 2020 · Mathematics · Complex Number

If |z_ 1 |=|z_ 2 |= . .|z_ n |=1, then the value of |z_ 1 +z_ 2 + ....

If ∣z1∣=∣z2∣=………..∣zn∣=1,\left|z_{1}\right|=\left|z_{2}\right|=\ldots \ldots \ldots . .\left|z_{n}\right|=1, then the value of ∣z1+z2+…….zn∣−∣1z1+1z2+…….+1zn∣\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,

Options

  1. A0
  2. B1
  3. C-1
  4. DNone

Correct answer

A. 0

Step-by-step 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
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