Complex Numbers 2 Question 31
32. If $\alpha, \beta$, $\gamma$ are the cube roots of $p, p<0$, then for any $x, y$ and $z$ then $\frac{x \alpha+y \beta+z \gamma}{x \beta+y \gamma+z \alpha}=\ldots$.
(1990, 2M)
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Answer:
Correct Answer: 32. Centre $=\frac{\alpha-k^{2} \beta}{1-k^{2}}$, Radius $=\frac{k(\alpha-\beta)}{1-k^{2}}$
Solution:
- $\frac{x \alpha+y \beta+z \gamma}{x \beta+y \gamma+z \alpha}=\frac{x(p)^{1 / 3}+y(p)^{1 / 3} \omega+z(p)^{1 / 3} \omega^{2}}{x(p)^{1 / 3} \omega^{2}+y(p)^{1 / 3} \omega^{3}+z(p)^{1 / 3} \omega}$
$$ \frac{\omega^{2}\left(x+y \omega+z \omega^{2}\right)}{\omega^{2}\left(x \omega+y \omega^{2}+z\right)} $$
$=\frac{\omega^{2}\left(x+y \omega+z \omega^{2}\right)}{x+y \omega+z \omega^{2}}=\omega^{2}$
