# 5.3.4: Products and Quotients of Complex Numbers

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.

1. 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.

1. 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.

1. 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

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. $$\dfrac{4+9i}{3i}$$
10. $$\dfrac{6−i}{12i}$$
11. $$\dfrac{7+12i}{−5i}$$
12. $$\dfrac{4−2i}{6−6i}$$
13. $$\dfrac{2−i}{2+i}$$
14. $$\dfrac{10+8i}{2+4i}$$
15. $$\dfrac{14+9i}{7−20i}$$

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