- Last updated
- Save as PDF
- Page ID
- 4162
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)
\( \newcommand{\vectorC}[1]{\textbf{#1}}\)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}}\)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}\)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)
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 \(a+bi\). |
