# 5.3.4: Products and Quotients of Complex Numbers

- Page ID
- 4162

Strategies based on multiplying binomials and conjugates.

Multiplying and Dividing Complex Numbers

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

**Multiplying and Dividing Complex Numbers**

When multiplying complex numbers, FOIL the two numbers together and then combine like terms. At the end, there will be an \(i^2\) term. Recall that \(i^2=−1\) and continue to simplify.

Let's simplify the following expressions.

- Simplify \(6i(1−4i)\).

Distribute the 6i to both parts inside the parenthesis.

\(6i(1−4i)=6i−24i^2\)

Substitute \(i^2=−1\) and simplify further.

\(\begin{aligned} &=6i−24(−1) \\ &=24+6i \end{aligned}\)

Remember to always put the real part first.

\((5−2i)(3+8i)\)

FOIL the two terms together.

\(\begin{aligned} (5−2i)(3+8i)&=15+40i−6i−16i^2 \\ &=15+34i−16i^2 \end{aligned}\)

Substitute \(i^2=−1\) and simplify further.

\(\begin{aligned} &=15+34i−16(−1) \\ &=15+34i+16 \\ &=31+34i \end{aligned}\)

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.

- Simplify \(\dfrac{8−3i}{6i}\).

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.

\(\begin{aligned}

\dfrac{8-3 i}{6 i} \cdot \dfrac{6 i}{6 i} &=\dfrac{48 i-18 i^{2}}{36 i^{2}} \\

&=\dfrac{18+48 i}{-36} \\

&=\dfrac{18}{-36}+\dfrac{48}{-36} i \\

&=-\dfrac{1}{2}-\dfrac{4}{3} i

\end{aligned}\)

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

- Simplify \(\dfrac{3−5i}{2+9i}\).

Now we are dividing by \(2+9i\), so we will need to multiply the top and bottom by the complex conjugate, \(2−9i\).

\(\begin{aligned}

\dfrac{3-5 i}{2+9 i} \cdot \dfrac{2-9 i}{2-9 i} &=\dfrac{6-27 i-10 i+45 i^{2}}{4-18 i+18 i-81 i^{2}} \\

&=\dfrac{6-37 i-45}{4+81} \\

&=\dfrac{-39-37 i}{85} \\

&=-\dfrac{39}{85}-\dfrac{37}{85} i

\end{aligned}\)

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 the previous three problems above, we substituted \(i^2=−1\) to simplify the fraction further. ** Your final answer should never have any power of** \(i\)

**greater than 1.**Example \(\PageIndex{1}\)

Earlier, you were asked to find the area of the triangle.

**Solution**

The area of the triangle is \((2+3i)(2−4i)^2\) so FOIL the two terms together and divide by 2.

\(\begin{aligned} (2+3i)(2−4i)&=4−8i+6i−12i^2 \\&=4−2i−12i^2 \end{aligned}\)

Substitute \(i^2=−1\) and simplify further.

\(\begin{aligned} &=4−2i−12(−1) \\ &=4−2i+12 \\ &=16−2i\end{aligned}\)

Now divide this product by 2.

\(\dfrac{16−2i}{2}=8−i\)

Therefore the area of the triangle is \(8−i\).

Example \(\PageIndex{2}\)

What is the complex conjugate of \(7−5i\)?

**Solution**

\(7+5i\)

Example \(\PageIndex{3}\)

Simplify the following complex expression: \((7−4i)(6+2i)\).

**Solution**

FOIL the two expressions.

\(\begin{aligned} (7−4i)(6+2i)&=42+14i−24i−8i^2 \\ &=42−10i+8 \\ &=50−10i \end{aligned}\)

Example \(\PageIndex{4}\)

Simplify the following complex expression: \(\dfrac{10−i}{5i}\).

**Solution**

Multiply the numerator and denominator by \(5i\).

\(\begin{aligned}

\dfrac{10-i}{5 i} \cdot \dfrac{5 i}{5 i} &=\dfrac{50 i-5 i^{2}}{25 i^{2}} \\

&=\dfrac{5+50 i}{-25} \\

&=\dfrac{5}{-25}+\dfrac{50}{-25} i \\

&=-\dfrac{1}{5}-2 i

\end{aligned}\)

Example \(\PageIndex{5}\)

Simplify the following complex expression: \(\dfrac{8+i}{6−4i}\).

**Solution**

Multiply the numerator and denominator by the complex conjugate, \(6+4i\).

\(\begin{aligned}

\dfrac{8+i}{6-4 i} \cdot \dfrac{6+4 i}{6+4 i} &=\dfrac{48+32 i+6 i+4 i^{2}}{36+24 i-24 i-16 i^{2}} \\

&=\dfrac{48+38 i-4}{36+16} \\

&=\dfrac{44+38 i}{52} \\

&=\dfrac{44}{52}+\dfrac{38}{52} i \\

&=\dfrac{11}{13}+\dfrac{19}{26} i

\end{aligned}\)

**Review**

Simplify the following expressions. Write your answers in standard form.

- \(i(2−7i)\)
- \(8i(6+3i)\)
- \(−2i(11−4i)\)
- \((9+i)(8−12i)\)
- \((4+5i)(3+16i)\)
- \((1−i)(2−4i)\)
- \(4i(2−3i)(7+3i)\)
- \((8−5i)(8+5i)\)
- \(\dfrac{4+9i}{3i}\)
- \(\dfrac{6−i}{12i}\)
- \(\dfrac{7+12i}{−5i}\)
- \(\dfrac{4−2i}{6−6i}\)
- \(\dfrac{2−i}{2+i}\)
- \(\dfrac{10+8i}{2+4i}\)
- \(\dfrac{14+9i}{7−20i}\)

**Answers for Review Problems**

To see the Review answers, open this PDF file and look for section 5.9.

## Vocabulary

Term | Definition |
---|---|

complex number |
A complex number is the sum of a real number and an imaginary number, written in the form \(a+bi\). |