Download 6-10 Simplifying Expressions Containing Complex Numbers

6-10 Simplifying Expressions Containing Complex Numbers
As we saw in the last example from 6-9, when we multiply two imaginary numbers
together, we get a real number.
This being the case we can use the concept of conjugates that we have had so much
experience with in the past few sections.
a+bi and a-bi are conjugates.
Let’s see it in action.
Simplify 6  3i 6  3i  .
Use FOIL.
6  3i 6  3i 
36  18i  18i  9i 2
36  9(1)
45
What if we interject some variables?
Simplify x  2 yi x  2 yi  .
x  2 yi x  2 yi 
x 2  2 xyi  2 xyi  4 y 2 i 2
x 2  4 y 2 (1)
x2  4y2
Since i is technically a radical, and since we must eliminate radicals from the
denominator to fully simplify a fraction, we must also eliminate imaginary numbers.
Simplify
2  8i
3i
2  8i 2  8i i


3i
3i i
2i  8i 2
3i 2
2i  8 8  2i
or
3
3
If we find we have a complex number in the denominator, we can use its conjugate to
clear the radical.
Simplify
3  9i
.
4  2i
3  9i 3  9i 4  2i


4  2i 4  2i 4  2i
12  6i  36i  18i 2
16  4i 2
12  42i  18
20
 6  42i  3  21i

20
10
Last, but not least, our old friend the multiplicative inverse, rears its head.
Find the multiplicative inverse of 7  2i . Simple, right? It is
1
. All done, right?
7  2i
Wrong!! You can’t have radicals in the denominator. Simplify.
1
1
7  2i


7  2i 7  2i 7  2i
7  2i
49  4i 2
7  2i
53