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:
For a nonnegative real number $\,k\,,$
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(Note: When the distinction becomes important in higher-level mathematics, then $\,\sqrt\,$ 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, $\,\sqrt = 3\,$ since the number $\,3\,$ satisfies these two properties:
So, what is (say) $\,\sqrt\ $? 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!
Complex numbers allow us to compute the square root of negative numbers, like $\,\sqrt\,.$ Remember the key fact: $\,i:=\sqrt\,,$ so that $\,i^2=-1\,.$
THE SQUARE ROOT OF A NEGATIVE NUMBERLet $\,p\,$ be a positive real number, so that $\,-p\,$ is a negative real number. Then:
(Note: When the distinction becomes important in higher-level mathematics, then $\,\sqrt\,$ is called the principal square root of $\,-p\,.$ )
Observe that $\,i\sqrtp>\,,$ when squared, does indeed give $\,-p\,$:
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:
Recall these two different questions, with two different answers:
Similarly, there are two different questions involving complex numbers, with two different answers:
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 $\,-4\,$:
Here is the correct approach:
Similarly, $\,\sqrt\sqrt\,$ is not equal to $\,\sqrt\,.$ Instead, here is the correct simplication: