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5.9: Multiplying and Dividing Complex Numbers

Difficulty Level: At Grade Created by: CK-12
Atoms Practice
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Practice Products and Quotients of Complex Numbers
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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 \begin{align*}A = \frac{1}{2}bh\end{align*}A=12bh, b is the length of the base and h is the length of the height.)

Watch This

First watch this video.

Khan Academy: Multiplying Complex Numbers

Then watch this video.

Khan Academy: Dividing Complex Numbers


When multiplying complex numbers, FOIL the two numbers together (see Factoring when \begin{align*}a = 1\end{align*}a=1 concept) and then combine like terms. At the end, there will be an \begin{align*}i^2\end{align*}i2 term. Recall that \begin{align*}i^2=-1\end{align*}i2=1 and continue to simplify.

Example A


a) \begin{align*}6i(1-4i)\end{align*}6i(14i)

b) \begin{align*}(5-2i)(3+8i)\end{align*}(52i)(3+8i)


a) Distribute the \begin{align*}6i\end{align*}6i to both parts inside the parenthesis.


Substitute \begin{align*}i^2 = -1\end{align*}i2=1 and simplify further.

\begin{align*}&=6i-24(-1)\\ &=24+6i\end{align*}=6i24(1)=24+6i

Remember to always put the real part first.

b) FOIL the two terms together.

\begin{align*}(5-2i)(3+8i) &= 15+40i-6i-16i^2\\ &= 15+34i-16i^2\end{align*}(52i)(3+8i)=15+40i6i16i2=15+34i16i2

Substitute \begin{align*}i^2 = -1\end{align*} and simplify further.

\begin{align*}&= 15+34i-16(-1)\\ &= 15+34i+16\\ &= 31+34i\end{align*}

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 \begin{align*}a + bi\end{align*}, then its complex conjugate is \begin{align*}a-bi\end{align*}. For example, the complex conjugate of \begin{align*}-6 + 5i\end{align*} would be \begin{align*}-6-5i\end{align*}. Therefore, rather than dividing complex numbers, we multiply by the complex conjugate.

Example B

Simplify \begin{align*}\frac{8-3i}{6i}\end{align*}.

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.

\begin{align*}\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\end{align*}

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 \begin{align*}\frac{3-5i}{2+9i}\end{align*}.

Solution: Now we are dividing by \begin{align*}2 + 9i\end{align*}, so we will need to multiply the top and bottom by the complex conjugate, \begin{align*}2-9i\end{align*}.

\begin{align*}\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\end{align*}

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 \begin{align*}i^2 = -1\end{align*} to simplify the fraction further. Your final answer should never have any power of \begin{align*}i\end{align*} greater than 1.

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

\begin{align*}(2 + 3i)(2 - 4i) = 4 - 8i + 6i -12i^2\\ &= 4 - 2i - 12i^2\end{align*}

Substitute \begin{align*}i^2 = -1\end{align*} and simplify further.

\begin{align*}&= 4 - 2i -12(-1)\\ &= 4 - 2i + 12\\ &= 16 - 2i\end{align*}

Now divide this product by 2.

\begin{align*} \frac {16 - 2i}{2} = 8 - i\end{align*}

Therefore the area of the triangle is \begin{align*}8 -i\end{align*}.

Guided Practice

1. What is the complex conjugate of \begin{align*}7-5i\end{align*}?

Simplify the following complex expressions.

2. \begin{align*}(7-4i)(6+2i)\end{align*}

3. \begin{align*}\frac{10-i}{5i}\end{align*}

4. \begin{align*}\frac{8+i}{6-4i}\end{align*}


1. \begin{align*}7 + 5i\end{align*}

2. FOIL the two expressions.

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

3. Multiply the numerator and denominator by \begin{align*}5i\end{align*}.

\begin{align*}\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\end{align*}

4. Multiply the numerator and denominator by the complex conjugate, \begin{align*}6 + 4i\end{align*}.

\begin{align*}\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\end{align*}


Complex Conjugate

The “opposite” of a complex number. If a complex number has the form \begin{align*}a+bi\end{align*}, its complex conjugate is \begin{align*}a-bi\end{align*}. When multiplied, these two complex numbers will produce a real number.


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

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


complex number

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

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Difficulty Level:
At Grade
Date Created:
Mar 12, 2013
Last Modified:
Jun 07, 2016
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