Download a . 49 = 300 = i i - Dorman High School

Notes on complex numbers:
Complex numbers are of the form a +bi or a – bi, where “a” is a real number and “i” is an imaginary unit that represents
1 . In the Real number system, all squares are positive, therefore in the Real number system the square root of a
negative number is undefined.
We use
1 to help simplify negative roots and we use i 2 = 1 when rationalizing the denominator using F.O.I.L.
We have learned to add, subtract, multiply and rationalize fractions using “i.”
We have also learned how to simplify square roots using “i.”
Hopefully, the examples below will assist you as you try to understand the concept of “I” as well as complex numbers.
A. To add or subtract complex numbers, combine the Real with the Real and the Imaginary with the Imaginary.
Example: (5 4i ) ( 3 8i )
5 + -3 and 4i + 8i = 2 + 12i
Example: (8 3i ) ( 6 4i )
8 - -6 and -3i – 4i = 14 – 7i
B. To multiply: either distribute or use FOIL.
Example: 2i ( 7 – 4i) = 2i(7) – 2i(4i)
14i - 8 i 2
14i – 8(-1) because i 2 = -1
14i +8, then 8 + 14i (Real ,imaginary)
Example 2. ( 4 – 6i) ( 5 + 3i) Using FOIL
F
4 * 5 = 20
O
4 * 3i = 12i
I
-6i *5 = -30i
L
-6i * 3i = -18 i 2 = -18(-1)= 18
Now combine F & L to get the Real parts of your complex number
20 + 18 = 26
O & I to get the Imaginary parts of your complex number
12i + -30i = -18i
So your final answer is 38 – 18i
C. To simplify negative square roots- think of breaking down the square root into the product of
Example:
49 =
1 *
1 times
a.
49 = 7i
300 =
1 * 100 * 3 = 10i 3 Notice the order. The perfect square root is first, followed by I,
Example 2:
then any left over radicals last. Never put i behind the radical.
D. To simplify I in the denominator: Just like -1 and
, a fraction ( division problem) cannot be left with any form of
“I” in the denominator. Therefore, we either multiply by
i
or by the “conjugate” of the complex number in the
i
denominator. A conjugate is the same complex number, with an opposite sign between the Real and the Imaginary
parts. (3 + 5i) is the conjugate of (3 – 5i).
Simplify
4 5i
i
First multiply the entire fraction by . Distribute in the numerator, i(4 + 5i) = 4i + 5 i 2 = 4i + -5 since i 2
3i
i
=-1. Then make sure to report the complex number with the real before the imaginary, thus -5 + 4i. Now for the
denominator: 3i * I = 3 i 2 = -3. Assemble your fraction back together as
distinct sections, and simplify:
5 4i
. Separate the numerator into two
3
5
4i 5 4i
+
=
.
3
3 3 3
Last example: to simplify a fraction with a complex number in the denominator, multiply both the numerator and the
denominator by the conjugate of the denominator.
Example:
5 3i
3 4i
First, determine the conjugate of the denominator- in this case 3 – 4i
Next, multiply both the numerator and the denominator by 3 – 4i
5 3i 3 4i
*
3 4i 3 4i
Now using FOIL for both the numerator as well as the denominator, you will
eliminate the “i” terms in the denominator completely.
Numerator: F = 15 0= -20i I = 9i L = -12 i 2 = 12 then add F & L for your Reals, 15 + 12 = 27 and -20i + 9i = -11i for the
imaginary and you will have your numerator of
27 – 11i.
Denominator: F = 9 O = -12i I = 12i L = -16 i 2 = 16 Notice that the outer and inner terms cancel out completely.
Thus the new denominator is a result of combining the F & L terms only = 25.
The simplified problem would look like:
27 11i
27 11i
=
.
25
25 25
I hope this helps! You would also benefit from getting a classmate’s notes and examples.