Complex Numbers And Quadratic Equations question 626
Question: If $ \alpha ,\beta ,\gamma $ and a, b, c are complex numbers such that $ \frac{\alpha }{a}+\frac{\beta }{b}+\frac{\gamma }{c}=1+i $ and $ \frac{a}{\alpha }+\frac{b}{\beta }+\frac{c}{\gamma }=0, $ then the value of $ \frac{{{\alpha }^{2}}}{a^{2}}+\frac{{{\beta }^{2}}}{b^{2}}+\frac{{{\gamma }^{2}}}{c^{2}} $ is equal to
Options:
A) $ 0 $
B) $ -1 $
C) $ 2i $
D) $ -2i $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{\alpha }{a}+\frac{\beta }{b}+\frac{\gamma }{c}=1+i $ squaring $ \frac{{{\alpha }^{2}}}{a^{2}}+\frac{{{\beta }^{2}}}{b^{2}}+\frac{{{\gamma }^{2}}}{c^{2}}+2( \frac{\alpha \beta }{ab}+\frac{\beta \gamma }{bc}+\frac{\gamma \alpha }{ac} )=2i $ or $ \frac{{{\alpha }^{2}}}{a^{2}}+\frac{{{\beta }^{2}}}{b^{2}}+\frac{{{\gamma }^{2}}}{c^{2}}+\frac{2\alpha \beta \gamma }{abc}( \frac{c}{\gamma }+\frac{a}{\alpha }+\frac{b}{\beta } )=2i $
$ \therefore \frac{{{\alpha }^{2}}}{a^{2}}+\frac{{{\beta }^{2}}}{b^{2}}+\frac{{{\gamma }^{2}}}{c^{2}}=2i $
BETA
