ARITHMETIC WITH COMPLEX NUMBERS

Whenever you get a new mathematical object (like complex numbers), you need to develop tools to work with the new object.

In this section, you'll learn how to do basic arithmetic and common tasks with complex numbers:

Remember the key fact:

[beautiful math coming... please be patient] $\,i:=\sqrt{-1}\,$,   so that   [beautiful math coming... please be patient] $\,i^2=-1\,$

SQUARE ROOTS OF NEGATIVE NUMBERS

Complex numbers allow us to compute the square root of negative numbers, like [beautiful math coming... please be patient] $\,\sqrt{-4}\,$.

Notice that [beautiful math coming... please be patient] $\,(2i)(2i) = 4i^2 = 4(-1) = -4\,$.
Thus, $\,2i\,$ is a (complex) number which, when squared, gives $\,-4\,$.
We write:   $\,\sqrt{-4}=2i\,$.

SQUARE ROOTS OF NEGATIVE NUMBERS
Let $\,p\,$ be a positive real number, so that $\,-p\,$ is a negative real number.
Then: [beautiful math coming... please be patient] $$ \sqrt{\strut -p} = i\,\sqrt{\strut p} $$

Some of my students like to think of it this way:
You can slide a minus sign out of a square root, and in the process, it turns into the imaginary number $\,i\,$!

Here are some examples:

TWO DIFFERENT QUESTIONS; TWO DIFFERENT ANSWERS

Recall these two different questions, with two different answers:

Similarly, there are two different questions involving complex numbers, with two different answers:

POWERS OF $\,i\,$

Let's look at powers of $\,i\,$:

[beautiful math coming... please be patient] $i^1 = \LARGE \color{green} i$ $i^2 = \LARGE \color{red} {-1}$ $i^3 = \LARGE \color{blue} {-i}$ $i^4 = \LARGE \color{purple} {1}$
$i^5 = \LARGE \color{green} i$ $i^6 = \LARGE \color{red} {-1}$ $i^7 = \LARGE \color{blue} {-i}$ $i^8 = \LARGE \color{purple} {1}$

Notice that the same four values occur over and over again:   $\,i\,$,   $\,-1\,$,   $\,-i\,$,   and   $\,1\,$.
Multiplying by $\,i\,$ causes a $\,90^{\circ}\,$ counter-clockwise rotation (about the origin) in the complex plane!

This makes it easy to compute the value of $\,i\,$ raised to any (whole number) power:
just write the power as a multiple of four plus remainder,
and notice that the remainder determines the final value, as illustrated below:

[beautiful math coming... please be patient] $i^{2959}$ original expression
$\ \ = i^{4(739)+3}$ How many piles of size $\,4\,$ can we get from $\,2959\,$?
First compute $\,2959/4 = 739.75\,$; you get $\,739\,$ piles, with some left over.
Take away these $\,739\,$ piles of $\,4\,$ to see what remains:
$\,2959 - 4(739) = 3\,$, so the remainder is $\,3\,$.
Thus, $\,2959 = 4(739) + 3\,$. This is the name for $\,2959\,$ that we need!
$\ \ = i^{4(739)}\cdot i^3$ Use an exponent law:   $\,x^{m+n} = x^nx^m\,$
$\ \ = (i^4)^{739}\cdot i^3$ Use an exponent law:   $\,x^{mn} = (x^m)^n\,$
$\ \ = 1^{739}\cdot i^3$ since $\,i^4 = 1\,$
$\ \ = i^3$ Use the special properties of $\,1\,$;
note that the only number remaining in the exponent is the remainder term!
$\ \ = -i$ It's a lot easier to compute $\,i\,$ to the third power, rather than the $\,2959^{\text{th}}\,$ power!
Ah, the power of renaming!

(A careful reader may be concerned about using the laws of exponents with non-real numbers;
rest assured that their use here is fine.)

ADDING AND SUBTRACTING COMPLEX NUMBERS

Adding and subtracting complex numbers is easy:
just work with the real and imaginary parts separately.
Here are some examples, where real terms are colored red and imaginary terms are colored blue:

[beautiful math coming... please be patient] $ \displaystyle \begin{gather} (\color{red}{3}\color{blue}{+ 4i}) + (\color{red}{-7}\color{blue}{+ 5i})\ \ =\ \ \color{red}{3-7} \color{blue}{+ 4i + 5i} \ \ =\ \ \color{red}{-4} \color{blue}{+ 9i} \cr\cr \color{blue}{i} \color{red}{- 3} \color{blue}{+ 2i - 5i} \color{red}{+ 6} \ \ =\ \ \color{red}{-3 + 6} \color{blue}{+ i + 2i - 5i} \ \ =\ \ \color{red}{3} \color{blue}{- 2i} \end{gather} $
When doing problems like this, it might be easiest to make two passes through the expression:
on the first pass, combine all the real parts;
on the second pass, combine all the imaginary parts.

Notice how the distributive law is used in the next example:

[beautiful math coming... please be patient] $ \displaystyle (2+3i) - (1-4i) \ \ =\ \ 2+3i - 1 + 4i \ \ =\ \ 1 + 7i $

Precisely, we have:

ADDING AND SUBTRACTING COMPLEX NUMBERS
Let $\,a\,$, $\,b\,$, $\,c\,$, and $\,d\,$ be real numbers, and let $\,i:=\sqrt{-1}\,$.

Then, [beautiful math coming... please be patient] $$ \begin{gather} (a+bi)\, + \, (c+di)\ =\ (a+c) + (b+d)i\cr\cr (a+bi)\, - \, (c+di)\ =\ (a-c) + (b-d)i \end{gather} $$
MULTIPLYING COMPLEX NUMBERS

Multiplying complex numbers is also easy: just use FOIL!
For example,

[beautiful math coming... please be patient] $(3-4i)(2+5i)$ original expression
$\ \ = 6 + 15i - 8i -20i^2$ FOIL
$\ \ = 6 + 7i - 20(-1)$ combine like terms; $\,i^2 = -1\,$
$\ \ = 26 + 7i$ simplify

Since you'll just think of using FOIL, you certainly won't memorize the following result.
It is included here only for completeness:

MULTIPLYING COMPLEX NUMBERS
Let $\,a\,$, $\,b\,$, $\,c\,$, and $\,d\,$ be real numbers, and let $\,i:=\sqrt{-1}\,$.

Then: [beautiful math coming... please be patient] $$ (a+bi)(c+di) \ =\ (ac-bd) + (ad+bc)i $$
FINDING ALL COMPLEX NUMBER SOLUTIONS FOR SIMPLE QUADRATIC EQUATIONS
Question:
Find all complex number solutions of the equation $\,x^2 + 5 = 2\,$.
Be sure to write a nice, clean list of equivalent equations.
Solution:
[beautiful math coming... please be patient] $x^2 + 5 = 2$
$x^2 = -3$
$x = \pm\sqrt{-3}$
$x = \pm i\sqrt{3}$
Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Equations of Circles


On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
(MAX is 17; there are 17 different problem types.)