Complex Numbers And Quadratic Equations question 243
Question: If $ x=a,y=b\omega ,z=c{{\omega }^{2}} $ , where $ \omega $ is a complex cube root of unity, then $ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}= $ [AMU 1983]
Options:
A) 3
B) 1
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Given that $ x=a,y=b\omega ,z=c{{\omega }^{2}} $ Then $ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=\frac{a}{a}+\frac{b\omega }{b}+\frac{c{{\omega }^{2}}}{c}=1+\omega +{{\omega }^{2}}=0 $
BETA
