Complex numbers are used in many math, science and engineering applications. In this section, we will learn the basics of complex number operations.
In this subsection, we will learn how to add, subtract and multiply complex numbers.
A complex number is a number that can be expressed in the form \(a + bi\text{,}\) where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit, satisfying the equation \(i^2 = -1\text{.}\) In this expression, \(a\) is the real part and \(b\) is the imaginary part of the complex number. You can read more at Wikipedia.
When we add or subtract two complex numbers, we combine and real parts and imaginary parts, just like combining like terms. Here are some examples
Do addition: \((1-2i)+(3+4i)\text{.}\)
Do subtraction: \((1-2i)-(3-2i)\text{.}\)
When we multiply complex numbers, be careful that \(i^2=-1\text{.}\) There is an interesting pattern on different exponents of \(i\text{.}\) Let's look at the first four:
\begin{align*} i^1\amp=i\\ i^2\amp=-1\\ i^3\amp=i^2\cdot i=(-1)\cdot i=-i\\ i^4\amp=i^2\cdot i^2=(-1)(-1)=1 \end{align*}When we calculate higher exponents, we can use \(i^4=1\) and those four products above to simplify our calculations:
\begin{align*} i^5\amp=i^4\cdot i=(1)\cdot i=i\\ i^6\amp=i^4\cdot i^2=(1)\cdot(-1)=-1\\ i^7\amp=i^4\cdot i^3=(1)\cdot(-i)=-i\\ i^8\amp=i^4\cdot i^4=(1)\cdot(1)=1\\ i^9\amp=i^4\cdot i^4 \cdot i=(1)(1)(i)=i\\ \ldots \end{align*}We will organize the results and find a pattern:
| \(i^1=i\) | \(i^2=-1\) | \(i^3=-i\) | \(i^4=1\) |
| \(i^5=i\) | \(i^6=-1\) | \(i^7=-i\) | \(i^8=1\) |
| \(i^9=i\) | \(i^{10}=-1\) | \(i^{11}=-i\) | \(i^{12}=1\) |
Based on the pattern, we can easily calculate the value of \(i^n\text{,}\) as in the next example.
Calculate \(i^{541}\) and \(i^{542}\text{.}\)
According to the pattern we found above, values of \(i^n\) cycle through \(i, 1, -i, 1\text{.}\) To find which one is the value of \(i^{541}\text{,}\) we do a division:
\begin{equation*} 541\div4=135R1 \end{equation*}Since the remainder is \(1\text{,}\) \(i^{541}\) would line up with \(i^1\text{.}\) This implies \(i^{541}=i\text{.}\)
Similarly, since \(542\div4=135R2\text{,}\) \(i^{542}\) would line up with \(i^2\text{.}\) This implies \(i^{542}=-1\text{.}\)
Let's look at a few examples on multiplying complex numbers.
Do multiplication: \(2i(3-2i)\text{.}\)
Note that we always write a complex number in the format of \(a+bi\text{.}\)
Do multiplication: \((1+2i)(3-4i)\text{.}\)
We will use the distributive property to multiply those two binomials.
\begin{align*} (1+2i)(3-4i)\amp=3(1+2i)-4i(1+2i)\\ \amp=3+6i-4i-8i^2\\ \amp=3+6i-4i-8(-1)\\ \amp=3+2i+8\\ \amp=11+2i \end{align*}Expand \((3-4i)^2\text{.}\)
We will use the FOIL method to expand this perfect square.
Multiply \((3+4i)(3-4i)\text{.}\)
We will use the Generic Rectangle Method to multiply those two binomials.
It is critical that we can efficiently use the Difference of Squares Formula
\begin{equation*} (a+b)(a-b)=a^2-b^2 \end{equation*}In the last example, it would be easier if we use the formula to multiply \((3+4i)(3-4i)\text{:}\)
\begin{align*} (3+4i)(3-4i)\amp=(3)^2-(4i)^2\\ \amp=9-16i^2\\ \amp=9-16(-1)\\ \amp=9+16\\ \amp=25 \end{align*}Adding, Subtracting and Multiplying Complex Numbers
When \(i\) is in the denominator, we must remove it like when we rationalize the denominator. We use the property \(\sqrt{x}\cdot\sqrt{x}=x\) when we rationalize the denominator, while we use the property \(i\cdot i=-1\) when we divide complex numbers. Let's compare these two problems \(\frac{2}{\sqrt{2}}\) and \(\frac{2}{i}\text{:}\)
When the denominator is \(a+bi\text{,}\) like \(\frac{1}{4+3i}\text{,}\) we need to use the Difference of Squares Formula to remove \(i\) from the denominator, just like when we rationalize the denominator for \(\frac{1}{4+3\sqrt{2}}\text{.}\) Let's compare those two problems.
To use the Difference of Squares Formula, we multiply \(4+3\sqrt{2}\) with \(4-3\sqrt{2}\text{,}\) and we have:
\begin{align*} (4+3\sqrt{2})(4-3\sqrt{2})\amp=(4)^2-(3\sqrt{2})^2\\ \amp=16-(9\cdot2)\\ \amp=16-18\\ \amp=-2 \end{align*}The solution is:
\begin{align*} \frac{1}{4+3\sqrt{2}}\amp=\frac{1\multiplyright{(4-3\sqrt{2})}}{(4+3\sqrt{2})\multiplyright{(4-3\sqrt{2})}}\\ \amp=\frac{4-3\sqrt{2}}{-2}\\ \amp=-\frac{4-3\sqrt{2}}{2} \end{align*}To use the Difference of Squares Formula, we multiply \(4+3i\) with \(4-3i\text{,}\) and we have:
\begin{align*} (4+3i)(4-3i)\amp=(4)^2-(3i)^2\\ \amp=16-(9\cdot i^2)\\ \amp=16-9(-1)\\ \amp=16+9\\ \amp=25 \end{align*}The solution is:
\begin{align*} \frac{1}{4+3i}\amp=\frac{1\multiplyright{(4-3i)}}{(4+3i)\multiplyright{(4-3i)}}\\ \amp=\frac{4-3i}{25}\\ \amp=\frac{4}{25}-\frac{3}{25}i \end{align*}Note that we always write a complex number in the format of \(a+bi\text{.}\)
Let's look at a more complicated example.
Do division in \(\frac{1+2i}{2-4i}\text{.}\)
To remove \(i\) in denominator, we multiply the numerator and denominator by \(2+4i\text{:}\)
\begin{align*} \frac{1+2i}{2-4i}\amp=\frac{(1+2i)\multiplyright[]{(2+4i)}}{(2-4i)\multiplyright[]{(2+4i)}}\\ \amp=\frac{1(2+4i)+2i(2+4i)}{(2)^2-(4i)^2}\\ \amp=\frac{2+4i+4i+8i^2}{4-16i^2}\\ \amp=\frac{2+8i+8(-1)}{4-16(-1)}\\ \amp=\frac{-6+8i}{20}\\ \amp=\frac{-6}{20}+\frac{8i}{20}\\ \amp=-\frac{3}{10}+\frac{2}{5}i \end{align*}Dividing Complex Numbers
