1994 AHSME Problems/Problem 12

Problem

If $i^2=-1$ , then $(i-i^{-1})^{-1}=$

$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ -2i \qquad\textbf{(C)}\ 2i \qquad\textbf{(D)}\ -\frac{i}{2} \qquad\textbf{(E)}\ \frac{i}{2}$

Solution

We simplify step by step as follows: $\begin{align*}(i-i^{-1})^{-1}&=\frac{1}{i-i^{-1}}\\&=\frac{1}{i-\frac{1}{i}}\\&=\frac{1}{\left(\frac{i^2-1}{i}\right)}\\&=\frac{i}{i^2-1}\\&=\boxed{\textbf{(D) }-\frac{i}{2}.}\end{align*}$

--Solution by TheMaskedMagician

1994 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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1994 AHSME Problems/Problem 12

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