Algebra Know-It-ALL

The power of product rule allows us to rewrite the left side of this equation, so it becomes

[(−1)j]^2 =− 1

But (−1)j is the same thing as −j, so we can simplify further to get

(−j)^2 =− 1

This is the same result as we obtain by squaring j.

Question 21-3

How is an imaginary number “put together”?

Answer 21-3

An imaginary number is the product of j and a real number. Suppose we call the real
numberb. If b is positive, then we write the product of j and b as jb. If b is negative, we write
the product as −jb. (We always put the minus sign first.) If b= 0, then we can write the product
asj0. But because j 0 = 0, we would more likely write j0 simply as 0.

Question 21-4

How can we show that j 3 +j7 is the same as j 7 +j3, assuming that the familiar distributive
law of arithmetic works with the unit imaginary number?

Answer 21-4

When we apply the distributive law for multiplication over addition “backward” to the first
expression, we get

j 3 +j 7 =j(3+ 7)

The commutative law for addition tells us that 3 + 7 = 7 + 3. By substitution on the right
side, we get

j 3 +j 7 =j(7+ 3)

Applying the distributive law “forward” on the right side, we get

j 3 +j 7 =j 7 +j 3

Question 21-5

How can we show, again assuming that the distributive law works with the unit imaginary
number, that if a and b are any two real numbers, then ja+jb is the same as jb+ja?

Answer 21-5

We can use the same proof procedure as we did in Answer 21-4, using letter constants instead
of numbers! The distributive law tells us that

ja+jb=j(a+b)

Part Three 499