THE SQUARE ROOT OF A NEGATIVE NUMBER

 
Square roots of negative numbers were introduced in a prior lesson:
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] $$ \overbrace{\ \ \ \sqrt{-p}\ \ \ }^{\text{the square root of negative $\,p\,$}} \ :=\ i\,\sqrt{\vphantom{h}p} $$
(Note: when the distinction becomes important in higher-level mathematics, then $\,\sqrt{-p}\,$ is called the principal square root of $\,-p\,$.)

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

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:

It is conventional to write $\,2i\,$, not $\,i2\,$.
When there's no possible misinterpretation (see below), write a real number multiplier before the $\,i\,$.

  • [beautiful math coming... please be patient] $\sqrt{-5} = i\sqrt{5}$

    In this situation, it's actually better to leave the $\,\sqrt{5}\,$ after the $\,i\,$.
    Why?   If you pull the $\,\sqrt{5}\,$ to the front, it can lead to a misinterpretation, as follows:
    The numbers $\,\sqrt{5}i\,$ and $\,\sqrt{5i}\,$ are different numbers—look carefully!
    In $\ \sqrt{5}i\ $, the $\,i\,$ is outside the square root.
    In $\ \sqrt{5i}\ $, the $\,i\,$ is inside the square root.
    Unless you look carefully, however, you might mistake one of these for the other.
    By writing $\,i\sqrt{5}\,$, you eliminate any possible confusion.

  • 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.)