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# Products and Quotients of Complex Numbers

## Strategies based on multiplying binomials and conjugates.

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Practice Products and Quotients of Complex Numbers
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Multiplying and Dividing Complex Numbers

Mr. Marchez draws a triangle on the board. He labels the height (2 + 3 i ) and the base (2 - 4 i ). "Find the area of the triangle," he says. (Recall that the area of a triangle is $A = \frac{1}{2}bh$ , b is the length of the base and h is the length of the height.)

### Watch This

First watch this video.

Then watch this video.

### Guidance

When multiplying complex numbers, FOIL the two numbers together (see Factoring when $a = 1$ concept) and then combine like terms. At the end, there will be an $i^2$ term. Recall that $i^2=-1$ and continue to simplify.

#### Example A

Simplify:

a) $6i(1-4i)$

b) $(5-2i)(3+8i)$

Solution:

a) Distribute the $6i$ to both parts inside the parenthesis.

$6i(1-4i)=6i-24i^2$

Substitute $i^2 = -1$ and simplify further.

$&=6i-24(-1)\\&=24+6i$

Remember to always put the real part first.

b) FOIL the two terms together.

$(5-2i)(3+8i) &= 15+40i-6i-16i^2\\&= 15+34i-16i^2$

Substitute $i^2 = -1$ and simplify further.

$&= 15+34i-16(-1)\\&= 15+34i+16\\&= 31+34i$

### More Guidance

Dividing complex numbers is a bit more complicated. Similar to irrational numbers, complex numbers cannot be in the denominator of a fraction. To get rid of the complex number in the denominator, we need to multiply by the complex conjugate . If a complex number has the form $a + bi$ , then its complex conjugate is $a-bi$ . For example, the complex conjugate of $-6 + 5i$ would be $-6-5i$ . Therefore, rather than dividing complex numbers, we multiply by the complex conjugate.

#### Example B

Simplify $\frac{8-3i}{6i}$ .

Solution: In the case of dividing by a pure imaginary number, you only need to multiply the top and bottom by that number. Then, use multiplication to simplify.

$\frac{8-3i}{6i}\cdot \frac{6i}{6i} &= \frac{48i-18i^2}{36i^2}\\&= \frac{18+48i}{-36}\\&= \frac{18}{-36}+\frac{48}{-36}i\\&= -\frac{1}{2}-\frac{4}{3}i$

When the complex number contains fractions, write the number in standard form, keeping the real and imaginary parts separate. Reduce both fractions separately.

#### Example C

Simplify $\frac{3-5i}{2+9i}$ .

Solution: Now we are dividing by $2 + 9i$ , so we will need to multiply the top and bottom by the complex conjugate, $2-9i$ .

$\frac{3-5i}{2+9i}\cdot \frac{2-9i}{2-9i} &= \frac{6-27i-10i+45i^2}{4-18i+18i-81i^2}\\&= \frac{6-37i-45}{4+81}\\&= \frac{-39-37i}{85}\\&= - \frac{39}{85}-\frac{37}{85}i$

Notice, by multiplying by the complex conjugate, the denominator becomes a real number and you can split the fraction into its real and imaginary parts.

In both Examples B and C, substitute $i^2 = -1$ to simplify the fraction further. Your final answer should never have any power of $i$ greater than 1.

Intro Problem Revisit The area of the triangle is $\frac{(2 + 3i)(2 - 4i)}{2}$ so FOIL the two terms together and divide by 2.

$(2 + 3i)(2 - 4i) = 4 - 8i + 6i -12i^2\\&= 4 - 2i - 12i^2$

Substitute $i^2 = -1$ and simplify further.

$&= 4 - 2i -12(-1)\\&= 4 - 2i + 12\\&= 16 - 2i$

Now divide this product by 2.

$\frac {16 - 2i}{2} = 8 - i$

Therefore the area of the triangle is $8 -i$ .

### Guided Practice

1. What is the complex conjugate of $7-5i$ ?

Simplify the following complex expressions.

2. $(7-4i)(6+2i)$

3. $\frac{10-i}{5i}$

4. $\frac{8+i}{6-4i}$

1. $7 + 5i$

2. FOIL the two expressions.

$(7-4i)(6+2i) &= 42+14i-24i-8i^2\\&= 42-10i+8\\&= 50-10i$

3. Multiply the numerator and denominator by $5i$ .

$\frac{10-i}{5i} \cdot \frac{5i}{5i} &= \frac{50i-5i^2}{25i^2}\\&= \frac{5+50i}{-25}\\&= \frac{5}{-25}+\frac{50}{-25}i\\&= -\frac{1}{5}-2i$

4. Multiply the numerator and denominator by the complex conjugate, $6 + 4i$ .

$\frac{8+i}{6-4i} \cdot \frac{6+4i}{6+4i} &= \frac{48+32i+6i+4i^2}{36+24i-24i-16i^2}\\&= \frac{48+38i-4}{36+16}\\&= \frac{44+38i}{52}\\&= \frac{44}{52} + \frac{38}{52}i\\&= \frac{11}{13}+\frac{19}{26}i$

### Vocabulary

Complex Conjugate

The “opposite” of a complex number. If a complex number has the form $a+bi$ , its complex conjugate is $a-bi$ . When multiplied, these two complex numbers will produce a real number.

### Practice

1. $i(2-7i)$
2. $8i(6+3i)$
3. $-2i(11-4i)$
4. $(9+i)(8-12i)$
5. $(4+5i)(3+16i)$
6. $(1-i)(2-4i)$
7. $4i(2-3i)(7+3i)$
8. $(8-5i)(8+5i)$
9. $\frac{4+9i}{3i}$
10. $\frac{6-i}{12i}$
11. $\frac{7+12i}{-5i}$
12. $\frac{4-2i}{6-6i}$
13. $\frac{2-i}{2+i}$
14. $\frac{10+8i}{2+4i}$
15. $\frac{14+9i}{7-20i}$

### Vocabulary Language: English

complex number

complex number

A complex number is the sum of a real number and an imaginary number, written in the form $a + bi$.