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The inverse of $-i$ in the multiplicative group, $1,-1, i,-i$ is
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To find the multiplicative inverse of a complex number, we need to find a number that, when multiplied with the given complex number, results in 1 . The given multiplicative group is $1,-1, i,-i$.
Let's find the inverse of $-i$ :
$-i \cdot a=1$
Where $a$ is the multiplicative inverse we're looking for. We can test the elements of the given group to find the one that satisfies this equation:
1. $-i \cdot 1=-i$
2. $-i \cdot(-1)=i$
3. $-i \cdot i=-i^2=-(-1)=1$
4. $-i \cdot(-i)=-i^2=1$
As we can see, the third case gives us the desired result:
$-i \cdot i=1$
So, the multiplicative inverse of $-i$ in the given group is $i$.

$(-i) \cdot a = 1 \\ (-i) \cdot i = 1$

Therefore, the inverse of $-i$ is $i$.
by Diamond (89,043 points)

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