Problem: If x=1−i32x=\dfrac{1-i \sqrt{3}}{2}x=21−i3 where i=−1i=\sqrt{-1}i=−1, then 1x2−x\dfrac{1}{x^2-x}x2−x1 is equal to
Answer Choices:
A. −2-2−2
B. −1-1−1
C. 1+i31+i \sqrt{3}1+i3
D. 111
E. 222 Solution:
x2−x=14(1−i3)2−12(1−i3)=14(−2−2143)−12(1−i3)=−1x^{2}-x=\frac{1}{4}(1-i \sqrt{3})^{2}-\frac{1}{2}(1-i \sqrt{3})=\frac{1}{4} (-2-2 \frac{1}{4} \sqrt{3})-\frac{1}{2}(1-i \sqrt{3})=-1x2−x=41(1−i3)2−21(1−i3)=41(−2−2413)−21(1−i3)=−1 and the reciprocal requested is also −1-1−1.