hard imaginary arithmetic

Remember that raising a number to the power of  -1  "flips" it so the numerator becomes the denominator and vice versa.

Like this:          ( a / b )^-1   =   b / a          and          ( a )^-1   =   ( a / 1 )^-1   =   1 / a

 

 

\((i-i^{-1})^{-1}\\~\\ =\quad(i-\frac1i)^{-1}\)

                                     Get a common denominator between  \(i\)  and  \(\frac1i\)  by multiplying the first term by  \(\frac{i}{i}\)

\(=\quad(i\cdot\frac{i}{i}-\frac1i)^{-1}\\~\\ =\quad(\frac{i^2}{i}-\frac1i)^{-1}\\~\\ =\quad(\frac{i^2-1}{i})^{-1}\)

                                     Replace  \(i^2\)  with  -1

\(=\quad(\frac{-1-1}{i})^{-1}\\~\\ =\quad(\frac{-2}{i})^{-1}\\~\\ =\quad\frac{i}{-2}\\~\\ =\quad-\frac{1}{2}i\)

 

 

Check:  https://www.wolframalpha.com/input/?i=%28i-i%5E-1%29%5E-1