Algebra Know-It-ALL

j−^6 = 1/(j^6 )

= 1/(−1)

=− 1

j−^8 = 1/(j^8 )

= 1/1

= 1

↓

and so on, forever

3. To solve this problem, we need a little intuition. First, let’s apply the difference of
   powers law. Note that

j−^1 =j^3 −^4

=j^3 /j^4

We have already determined that j^3 =−j and j^4 = 1. Therefore,

j^3 /j^4 = (−j)/1

=−j

We can conclude that j−^1 =−j. Now let’s use cross multiplication and see if we get the

same result. Note that j−^1 is the reciprocal of j, or 1/j. If we let this quantity equal an

unknown z, we can formulate this equation:

1/j=z/1

According to the law of cross multiplication, the above expression is equivalent to

1 × 1 =jz

which tells us that jz= 1. Let’s make an educated guess as to what z might be. It’s easy

enough to see that z can’t be equal to 1, −1, or j. How about −j? When we multiply j by −j,

we get

j× (−j)=j× (− 1 ×j)

=j×j× (−1)

=j^2 × (−1)

=− 1 × (−1)

= 1

It works! This tells us that z=−j, and therefore that 1/j=−j. We’ve now shown, in two

different ways, that the reciprocal of the unit imaginary number is the same as its nega-

tive. If you want to use technical language, the additive inverse of j is the same as its

multiplicative inverse. No real number behaves like that!

664 Worked-Out Solutions to Exercises: Chapters 21 to 29