Just as we can perform operations with real numbers, we can also add, subtract, multiply, and divide complex numbers. Recall the order of operations for real numbers here: Operations with Numbers

When performing operations with complex numbers, you may encounter three types of complex numbers:

*Real number*as complex number - imaginary part is equal to 0

__Examples__: \(2 , -6, 7.462437, \frac{3}{5}\)

**Tip:** Since all real numbers are also complex numbers, you can add \(0i\) to these numbers and they'll still be considered complex numbers!

*Imaginary number*as complex number - real part is equal to 0

__Examples__: \(i, 3i, \sqrt{2}i, \frac{1}{2}i\)

- Complex number composed of
*real and imaginary part*

__Examples__: \(3+4i, \frac{5}{2}-0.1i, -7.25-11i\)

To add or subtract complex numbers, just combine like terms as if the number \(i\) is a variable, then evaluate.

- For any two complex numbers, say \(x=a+bi\) and \(y=c+di\), we can
**add**them as follows:

\(x+y=(a+bi)+(c+di)=(a+c)+(b+d)i\)

- For any two complex numbers, say \(x=a+bi\) and \(y=c+di\), we can
**subtract**them as follows:

\(x-y=(a+bi)-(c+di)=(a-c)+(b-d)i\)

Example: Evaluate \((-4+5i)-(6-6i)+(2i+7)\).

**Solution:**

1. Combine like terms:

\((-4+5i)-(6-6i)+(2i+7)\)

\(=(-4-6+7)+(5i-(-6i)+2i)\)

2. Simplify and evaluate:

\(=-3+(5-(-6)+2)i\)

\(=-3+13i\)

To multiply complex numbers, you can first use one the FOIL process to expand the two binomials. The FOIL process is a quick method of apply the distributive law to expand two binomials:

\((a+b)(c+d)=ac+ad+bc+bd\)

where FOIL is an acronym for:

**F**irst**O**uter**I**nner**L**ast

since we multiply the "firsts" in the two binomials - \(a\) and \(c\) in the above case, then the "outer" terms - \(a\) and \(d\), then "inner" terms - \(b\) and \(c\), then "last" terms - \(b\) and \(d\), and add all these products up.

Once you've finished expanding, the final step is to combine like terms. To summarize, you can multiply any two complex numbers, say, \((a+bi)\) and \((c+di)\), as follows:

\((a+bi)(c+di)=(a)(c)+(a)(di)+(bi)(c)+(bi)(di)=ac+adi+bci+bd(-1)=(ac-bd)+(ad+bc)i\)

*__Note__: in the second last step, we were able to replace \(i\times i=i^2\) with -1 since, by definition, \(i^2=(\sqrt{-1})^2=-1\)

Example: Evaluate \((3+7i)(2-i)\).

**Solution:**

1. First, expand using the FOIL method:

\((3+7i)(2-i)\)

\(=(3)(2)+(3)(-i)+(7i)(2)+(7i)(-i)\)

2. Next, multiply out all the terms:

\(=6-3i+14i-7i^2\)

\(=6-3i+14i-7(-1)\)

\(=6-3i+14i+7\)

3. Evaluate by collecting like terms and simplifying:

\(=13+11i\)

Complex division is often represented using fractions, in the general form of \(\frac{a+bi}{c+di}\) for any two complex numbers. Notice that the denominator is the sum of two terms, so we cannot find the quotient directly.

Recall that the **conjugate** of a complex number \(z=a+bi\) is: \(\bar{z}=a-bi\). This will be helpful to us when dividing complex numbers since multiplying a complex number by its conjugate gives us a real number product.

For any two complex numbers, say \(x=a+bi\) and \(y=c+di\), we can divide \(x\) by \(y\) (i.e. evaluate \(\frac{a+bi}{c+di}\)) by following these steps:

1. Determine the conjugate of the **denominator **(which is \(c-di\) here). Then multiply the numerator __and__ denominator by this conjugate:

\(\frac{a+bi}{c+di} \cdot \frac{c-di}{c-di}\)

*__Note__: this step does __not__ change the original expression since we're actually just multiplying it by \(\frac{c-di}{c-di}=1\)!

2. Expand the expression by multiplying:

\(=\frac{(a+bi)(c-di)}{(c+di)(c-di)}\)

\(=\frac{ac-adi+bci+bd}{c^2-d^2i^2}\)

3. Simplify by collecting like terms and reducing fractions (if applicable):

\(=\frac{(ac-bd)+(bc-ad)i}{c^2+d^2}\)

Example: Evaluate \(\frac{3+5i}{2+4i}\).

See the below video for the solution:

Time to do some practice!

Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License.

- Last Updated: Apr 20, 2023 12:36 PM
- URL: https://libraryguides.centennialcollege.ca/c.php?g=717490
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