THE SQUARE ROOT OF A NEGATIVE NUMBER

Square roots of negative numbers were introduced in a prior web exercise:
Arithmetic with Complex Numbers in the Algebra II materials.
For your convenience, that material is repeated (and extended) here.

Firstly, recall some information from beginning algebra:

The Square Root of a Nonnegative Real Number

For a nonnegative real number $\,k\,$, $$\overbrace{\ \ \ \sqrt{k}\ \ \ }^{\text{the square root of $\,k\,$}} := \ \ \text{the unique nonnegative real number which, when squared, equals $\,k\,$} $$

(Note: when the distinction becomes important in higher-level mathematics,
then $\,\sqrt{k}\,$ is called the principal square root of $\,k\,$.)
Thus, the number that gets to be called the square root of $\,k\,$ satisfies two properties: For example, $\,\sqrt{9} = 3\,$ since the number $\,3\,$ satisfies these two properties:

So, what is (say) $\,\sqrt{-4}\ $?
There does not exist a nonnegative real number which, when squared, equals $\,-4\ $.
Why not?
Every real number, when squared, is nonnegative:   for all real numbers $\,x\,$,   $\,x^2 \ge 0\,$.
Complex numbers to the rescue!

The Square Root of a Negative Number

Complex numbers allow us to compute the square root of negative numbers, like [beautiful math coming... please be patient] $\,\sqrt{-4}\ $.
Remember the key fact:   [beautiful math coming... please be patient] $\,i:=\sqrt{-1}\,$,   so that   [beautiful math coming... please be patient] $\,i^2=-1\,$

THE SQUARE ROOT OF A NEGATIVE NUMBER
Let $\,p\,$ be a positive real number, so that $\,-p\,$ is a negative real number. Then: [beautiful math coming... please be patient] $$ \overbrace{\ \ \ \sqrt{\strut -p}\ \ \ }^{\text{the square root of negative $\,p\,$}} \ :=\ i\,\sqrt{\strut p} $$
(Note: when the distinction becomes important in higher-level mathematics,
then $\,\sqrt{\strut -p}\,$ is called the principal square root of $\,-p\,$.)

Observe that $\,i\sqrt{\strut p}\,$, when squared, does indeed give $\,-p\ $: $$ (\sqrt{\strut -p})^2 \ =\ (i\sqrt{\strut p})^2 \ =\ i^2(\sqrt{\strut p})^2 \ =\ (-1)(p) \ =\ -p $$

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:

Square Roots of Products

The following statement is true:   for all nonnegative real numbers $\,a\,$ and $\,b\,$, $\,\sqrt{ab} = \sqrt{a}\sqrt{b}\ $.
For nonnegative numbers, the square root of a product is the product of the square roots.

Does this property work for negative numbers, too?
The answer is NO, as shown next.

Certainly, anything called ‘the square root of $\,-4\,$’ must have the property that, when squared, it equals $\,-4\ $.
Unfortunately, the following incorrect reasoning gives the square as $\,4\,$, not $\,-4\,$: $$ \text{? ? ? ? }\ \ \ \ (\sqrt{-4})^2 \ \ =\ \ \sqrt{-4}\sqrt{-4} \overbrace{\ \ =\ \ }^{\text{this is the mistake}}\ \ \sqrt{(-4)(-4)}\ \ =\ \ \sqrt{16}\ \ =\ \ 4\ \ \ \ \text{? ? ? ? } $$ Here is the correct approach: $$ (\sqrt{-4})^2 \ \ =\ \ \sqrt{-4}\sqrt{-4}\ \ =\ \ i\sqrt{4}\ i\sqrt{4} \ \ =\ \ i^2(\sqrt{4})^2 \ \ =\ \ (-1)(4) \ \ =\ \ -4 $$

Similarly, $\,\sqrt{-5}\sqrt{-3}\,$ is NOT equal to $\,\sqrt{(-5)(-3)}\,$.
Instead, here is the correct simplication: $$ \sqrt{-5}\sqrt{-3} \ \ =\ \ i\sqrt{5}\ i\sqrt{3} \ \ =\ \ i^2\sqrt{5}\sqrt{3} \ \ =\ \ (-1)\sqrt{(5)(3)} \ \ =\ \ -\sqrt{15} $$ Be careful about this!

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
the complex conjugate
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
1 2 3 4 5 6 7 8 9 10 11 12 13
AVAILABLE MASTERED IN PROGRESS

(MAX is 13; there are 13 different problem types.)