Complex Numbers And Quadratic Equations question 643
Question: If $ | z_1 |=| z_2 |=…….| z_{n} |=1, $ then the value of $ | z_1+z_2+….z_{n} |-| \frac{1}{z_1}+\frac{1}{z_2}+……+\frac{1}{z_{n}} | $ is,
Options:
A) 0
B) 1
C) $ -1 $
D) None
Show Answer
Answer:
Correct Answer: A
Solution:
$ z_1{{\bar{z}}1}=z_2{{\bar{z}}2}=……..=z{n}{{\bar{z}}{n}}=1 $
$ \Rightarrow {{\bar{z}}1}=\frac{1}{z_1},{{\bar{z}}2}=\frac{1}{z_2},{{\bar{z}}3}=\frac{1}{z_3},…..,{{\bar{z}}{n}}=\frac{1}{z{n}} $
$ \therefore |z_1+z_2+…….+z{n}|-| \frac{1}{z_1}+\frac{1}{z_2}+……+\frac{1}{z_{n}} | $
$ \therefore |z_1+z_2+….+z_{n}|-|{{\bar{z}}_1}+{{\bar{z}}2}+….+{{\bar{z}}{n}}|=0 $
BETA
