Difference between revisions of "1994 AHSME Problems/Problem 12"
(→Solution) |
m (→Solution) |
||
| Line 5: | Line 5: | ||
==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> | 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> | ||
| + | |||
| + | --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] | ||
, then 
