Problem: If i2=−1i^{2}=-1i2=−1, then (i−i−1)−1=\left(i-i^{-1}\right)^{-1}=(i−i−1)−1=
Answer Choices:
A. 000
B. −2i-2 i−2i
C. 2i2 i2i
D. −i/2-i / 2−i/2
E. i/2i / 2i/2
Solution:
(i−i−1)−1=(i−1i)−1=(i2−1i)−1=(−2i)−1=−i2\left(i-i^{-1}\right)^{-1}=\left(i-\dfrac{1}{i}\right)^{-1}=\left(\dfrac{i^{2}-1}{i}\right)^{-1}=\left(\dfrac{-2}{i}\right)^{-1}=-\dfrac{i}{2} (i−i−1)−1=(i−i1)−1=(ii2−1)−1=(i−2)−1=−2i
OR\textbf{OR} OR
Since i−1=−ii^{-1}=-ii−1=−i, it follows that
(i−i−1)−1=(i−(−i))−1=(2i)−1=12i=−i2\left(i-i^{-1}\right)^{-1}=(i-(-i))^{-1}=(2 i)^{-1}=\dfrac{1}{2 i}=-\dfrac{i}{2} (i−i−1)−1=(i−(−i))−1=(2i)−1=2i1=−2i