Complex numbers and the imaginary number $i$
As you've worked your way through mathematics since elementary school, you've learned about successively more all-encompassing sets of numbers. First there were the counting numbers (1, 2, 3, ... ), then you added zero and the negative numbers to get the integers. Then you mixed integers to get rational fractions, then real and irrational numbers. Well, all of those fit inside a bigger set, called the complex numbers (below).
Complex numbers are characterized by both a real part (often labeled "Re") and an imaginary part, labeled "Im". The word "imaginary" is a bit unfortunate, because imaginary numbers play a crucial role in all kinds of mathematical modeling of very real things, but we're stuck with it.
The definition of $i$
Imaginary numbers arise when we take square roots of negative numbers - something we haven't been able to do until now. In a way, we just made up some new numbers (a single new number, actually) to help us overcome that shortcoming.
We define the imanginary number i like this →
The imaginary number opens the door to solving negative square roots (see example below), and much more, but that's the root of it all.
Writing complex numbers
We try to write complex numbers in a consistent way. The real part is on the left, and the imaginary part is added to it. Here are some examples:
Complex arithmetic
The key to doing arithmetic with complex numbers is just remembering that $i^2 = -1$. The rest is no different than arithmetic with real numbers.
Adding and subtracting
To add and subtract complex numbers, we just keep the real and imaginary parts separate. Here are two examples:
$$ \begin{align} (2 + 3i) &+ (5 - 4i) \\[4pt] &= (2 + 5) + (3i - 4i) \\[4pt] &= 7 - i \end{align}$$
$$ \begin{align} (-4 - 3i) &- (-1 + i) \\[4pt] &= (-4 - 3i) + (1 - i) \\[4pt] &= (-4 + 1) + (-3i - i) \\[4pt] &= -3 - 4i \end{align}$$
Multiplying complex numbers
To multiply complex numbers, just FOIL like we do any time we multiply two binomials:
$$(a + b)(c + d) = ac + ad + bc + bd$$
Here are three examples:
$$ \begin{align} &(2 + 5i)·(3 + i) \\[4pt] &= 6 + 2i + 15i + 5i^2 \\[4pt] &= 6 + 17i - 5 \\[4pt] &= 1 + 17i \end{align}$$
$$ \begin{align} &(3 - 4i)·(4 + i) \\[4pt] &= 12 + 3i - 16i -4i^2 \\[4pt] &= 12 - 13i + 4 \\[4pt] &= 16 - 13i \end{align}$$
$$ \begin{align} &(5 + 7i)·(5 - 7i) \\[4pt] &= 25 - 35i + 35i + 49 \\[4pt] &= 74 \end{align}$$
Notice that in the last example, when the difference between the complex numbers being multiplied is just the sign of the imaginary part, the result is a real number. The imaginary part vanishes.
Complex conjugates
When we swap the sign of the imaginary part of a complex number, we have made its complex conjugate. The complex conjugate of $(2 + 3i)$ is $(2 - 3i)$. Confirm for yourself that the product is a real number. In many applications involving real phenomena represented by complex numbers (like the electrons in atoms) we'll square those numbers to arrive at real answers.
Division by a complex number: making the denominator real
In expressions like the one below, in which the denominator is a complex number, we might wish to make the denominator real. We now know that we can always get a real number by multipying a complex number by its complex conjugate, so we can take advantage of that. To make the denominator of a complex ratio real, multiply by the denominator divided by itself:
Finally, we ought to consider what complex numbers mean when they appear as roots of polynomial or rational functions. There is no way to plot complex numbers on the Cartesian (x, y) plane, so they aren't actually x-axis intercepts. In fact, a quadratic function with complex roots will not cross the x-axis - ever.
When polynomial functions have complex roots, they are always present in complex-conjugate pairs, and when we find higher roots, like sixth roots of real numbers, there are always some complex roots in complex-conjugate pairs.
In another section, we learn how to construct a complex plane and use it to plot and manipulate complex functions to do some amazing things.
A good read
There is a nice book about the imaginary number i, An Imaginary Tale: The Story of [the Square Root of Minus One] by Paul J. Nahin. Give it a read!
Practice problems
I. Simplify these ratios of complex numbers so that each has a purely-real denominator
| 1. |
$i + (2 + 3i)$ Solution$$ \begin{align} &= 3 + 3i + i \\[4pt] &= 3 + 4i \end{align}$$ |
| 2. |
$5 + 3i - 7 + 4i$ Solution$$ \begin{align} &= 5 - 7 + 3i + 4i \\[4pt] &= -2 + 7i \end{align}$$ |
| 3. |
$-3 + 2i - 3 - 4i$ Solution$$ \begin{align} &= -3 - 3 + 2i - 4i \\[4pt] &= -6 - 2i \end{align}$$ |
| 4. |
$27 + i - (3 - 2i)$ Solution$$ \begin{align} &= 27 + i - 3 + 2i \\[4pt] &= 27 - 3 + i + 2i \\[4pt] &= 24 + 3i \end{align}$$ |
| 5. |
$(2 + 3i)(4 + i)$ Solution$$ \begin{align} &= 8 + 2i + 12i + 3i^2 \\[4pt] &= 8 + 14i - 3 \\[4pt] &= 5 + 14i \end{align}$$ |
| 6. |
$(2x + i)(2x - i)$ Solution$$ \begin{align} &= 4x^2 - 2ix + 2ix -i^2 \\[4pt] &= 4x^2 + 1 \end{align}$$ |
| 7. |
$(8 - 6i)(8 - 4i)$ Solution$$ \begin{align} &= 64 - 32i - 48i - 4i^2 \\[4pt] &= 64 - 80i + 4 \\[4pt] &= 68 - 80i \end{align}$$ |
| 8. |
$(-5 - 4i)(-4 + i)$ Solution$$ \begin{align} &= 20 - 5i + 16i -4i^2 \\[4pt] &= 20 + 11i + 4 \\[4pt] &= 24 + 11i \end{align}$$ |
| 9. |
$(2 + 3i)(4 - i)$ Solution$$ \begin{align} &= 8 - 2i + 12i -3i^2 \\[4pt] &= 8 + 10i + 3 \\[4pt] &= 11 + 10i \end{align}$$ |
| 10. |
$(2x + i)^2$ Solution$$ \begin{align} &= 4x^2 + 2xi + 2xi + i^2 \\[4pt] &= 4x^2 + 4xi - 1 \\[4pt] &= 4x^2 - 1 + 4xi \end{align}$$ |
| 11. |
$(8 - 6i)(3 - 4i)$ Solution$$ \begin{align} &= 24 - 32i - 18i + 24i^2 \\[4pt] &= 24 - 50i - 24 \\[4pt] &= 50i \end{align}$$ |
| 12. |
$(-5 - 3i)(-4 + i)$ Solution$$ \begin{align} &= 20 - 5i + 12i -3i^2 \\[4pt] &= 20 + 7i + 3 \\[4pt] &= 23 + 7i \end{align}$$ |
| 13. |
$(x - 2i)^2$ Solution$$ \begin{align} &= x^2 - 2ix - 2ix + 4i^2 \\[4pt] &= x^2 - 4ix - 4 \\[4pt] &= x^2 - 4 - 4ix \end{align}$$ |
| 14. |
$(x + i)^3$ Solution$$ \begin{align} &= (x^2 + 2xi - 1)(x + i) \\[4pt] &= x^3 + xi + 2x^2i -2x -x -i \\[4pt] &= x^3 - 3x + i(2x^2 + x - 1) \end{align}$$ |
| 15. |
$2 + 3i(4 - 2i)$ Solution$$ \begin{align} &= 2 + 12i - 6i^2 \\[4pt] &= 2 + 12i + 6 \\[4pt] &= 8 + 12i \end{align}$$ |
| 16. |
$i^0, \; i^1, \; i^2, \; i^3, \; i^4, \; i^5$ Solution$$ \begin{align} i^1 &= i \\ i^2 &= i \cdot i = -1 \\[4pt] i^3 &= -1 \cdot i = -i \\[4pt] i^4 &= -i \cdot i = 1 \\[4pt] i^5 &= 1 \cdot i = i \end{align}$$ Sequence repeats ... |
II. Convert each fraction so that it has a real denominator.
| 17. |
Solution$$ \begin{align} &= \left( \frac{2 + i}{3 - 3i} \right) \left( \frac{3 + 3i}{3 + 3i} \right) \\[4pt] &= \frac{6 + 3i + 3i - 3}{9 + 9i - 9i + 9} \\[4pt] &= \frac{3 + 6i}{18} \\[4pt] &= \frac{1 + 2i}{6} \end{align}$$ |
| 18. |
Solution$$ \begin{align} &= \left( \frac{8 + 3i}{-8 - 35i} \right) \left( \frac{-8 + 5i}{-8 + 5i} \right) \\[4pt] &= \frac{-64 + 40i - 24i - 15}{64 - 40i + 40i + 25} \\[4pt] &= \frac{-79 + 16i}{89} \end{align}$$ |
| 19. |
Solution$$ \begin{align} &= \left( \frac{4 + 3i}{4 - 3i} \right) \left( \frac{4 + 3i}{4 + 3i} \right) \\[4pt] &= \frac{16 + 12i + 12i - 9}{16 + 12i - 12i + 9} \\[4pt] &= \frac{7 + 24i}{25} \end{align}$$ |
| 20. |
Solution$$ \begin{align} &= \left( \frac{i - 5}{5 - i} \right) \left( \frac{5 + i}{5 + i} \right) \\[4pt] &= \frac{5i = 1 - 25 - 5i}{25 + 5i - 5i + 1} \\[4pt] &= \frac{-26}{26} = -1 \end{align}$$ |
Cartesian coordinates
Cartesian coordinates are the normal 2-dimensional (2D) or 3-dimensional (3D) coordinate systems we most-frequently use. In two dimensions, we draw x- and y-axes at 90˚ angles to each other, and in 3D we add a third axis, usually the z-axis, perpendicular to the x-y plane.
The location or direction of an point or particle can be described using Cartesian coordinates (x, y) in the 2D plane, or (x, y, z) in 3D.
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