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# hard imaginary arithmetic

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What is $$(i-i^{-1})^{-1}$$?

I honestly just don't know how to do this because it just looks so confusing. Please help! All help is greatly appreciated! :D

Sep 16, 2020

#1
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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

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

Sep 16, 2020
#2
+734
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thank you so much! i understand it a lot better now!!!