Question

The inverse of \(-i\) in the multiplicative group, \(1,-1, i,-i\) is

Answer 1

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 \(i\).