Search any question & find its solution
Question:
Answered & Verified by Expert
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,
Options:
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,
Solution:
2399 Upvotes
Verified Answer
The correct answer is:
0
$z_{1} \bar{z}_{1}=z_{2} \bar{z}_{2}=\ldots .=z_{n} \bar{z}_{n}=1$
$\Rightarrow \overline{\mathrm{z}}_{1}=\frac{1}{\mathrm{z}_{1}}, \overline{\mathrm{z}}_{2}=\frac{1}{\mathrm{z}_{2}}, \overline{\mathrm{z}}_{3}=\frac{1}{\mathrm{z}_{3}}, \ldots \ldots, \overline{\mathrm{z}}_{\mathrm{n}}=\frac{1}{\mathrm{z}_{\mathrm{n}}}$
$\therefore\left|z_{1}+z_{2}+\ldots .+z_{n}\right|-\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\ldots .+\frac{1}{z_{n}}\right|$
$=\left|z_{1}+z_{2}+\ldots+z_{n}\right|-\left|\bar{z}_{1}+\bar{z}_{2}+\ldots .+\bar{z}_{n}\right|=0$
$\Rightarrow \overline{\mathrm{z}}_{1}=\frac{1}{\mathrm{z}_{1}}, \overline{\mathrm{z}}_{2}=\frac{1}{\mathrm{z}_{2}}, \overline{\mathrm{z}}_{3}=\frac{1}{\mathrm{z}_{3}}, \ldots \ldots, \overline{\mathrm{z}}_{\mathrm{n}}=\frac{1}{\mathrm{z}_{\mathrm{n}}}$
$\therefore\left|z_{1}+z_{2}+\ldots .+z_{n}\right|-\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\ldots .+\frac{1}{z_{n}}\right|$
$=\left|z_{1}+z_{2}+\ldots+z_{n}\right|-\left|\bar{z}_{1}+\bar{z}_{2}+\ldots .+\bar{z}_{n}\right|=0$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.