Difference between revisions of "1994 AHSME Problems/Problem 12"
(Created page with "==Problem== If <math>i^2=-1</math>, then <math>(i-i^{-1})^{-1}=</math> <math> \textbf{(A)}\ 0 \qquad\textbf{(B)}\ -2i \qquad\textbf{(C)}\ 2i \qquad\textbf{(D)}\ -\frac{i}{2} \qq...") |
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<math> \textbf{(A)}\ 0 \qquad\textbf{(B)}\ -2i \qquad\textbf{(C)}\ 2i \qquad\textbf{(D)}\ -\frac{i}{2} \qquad\textbf{(E)}\ \frac{i}{2}</math> | <math> \textbf{(A)}\ 0 \qquad\textbf{(B)}\ -2i \qquad\textbf{(C)}\ 2i \qquad\textbf{(D)}\ -\frac{i}{2} \qquad\textbf{(E)}\ \frac{i}{2}</math> | ||
==Solution== | ==Solution== | ||
| + | We simplify step by step as follows: <cmath>\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*}</cmath> | ||
Revision as of 19:20, 20 July 2014
Problem
If 
, then 
Solution
We simplify step by step as follows: 
