Complex Numbers And Quadratic Equations question 253
Question: If $ \alpha ,\beta ,\gamma $ are the cube roots of $ p(p<0) $ , then for any $ x,y $ and $ z,\frac{x\alpha +y\beta +z\gamma }{x\beta +y\gamma +z\alpha }= $ [IIT 1989]
Options:
A) $ \frac{1}{2}(-1+i\sqrt{3}) $
B) $ \frac{1}{2}(1+i\sqrt{3}) $
C) $ \frac{1}{2}(1-i\sqrt{3}) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Since $ p<0 $ . Let $ p=-q $ , where $ q $ is positive. Therefore $ {p^{1/3}}=-{q^{1/3}}{{(1)}^{1/3}}. $ Hence $ \alpha =-{q^{1/3}} $ , $ \beta =-{q^{1/3}}\omega $ and $ \gamma =-{q^{1/3}}{{\omega }^{2}} $ The given expression $ \frac{x+y\omega +z{{\omega }^{2}}}{x\omega +y{{\omega }^{2}}+z}=\frac{1}{\omega }.\frac{z\omega +y{{\omega }^{2}}+z}{x\omega +y{{\omega }^{2}}+z} $ $ =\frac{1}{\omega }={{\omega }^{2}} $ .
BETA
