Principal Square Root of a Negative Number
For any positive real number $a$, the principal square root of $-a$ is defined as:
$\sqrt{-a} = i\sqrt{a}$This follows from $i^2 = -1$. Be careful: the usual product rule $\sqrt{x}\cdot\sqrt{y} = \sqrt{xy}$ does not apply when both $x$ and $y$ are negative.
Quick Reference
| Expression | Value | Reasoning |
|---|---|---|
| $\sqrt{-1}$ | $i$ | Definition: $i^2 = -1$ |
| $\sqrt{-4}$ | $2i$ | $\sqrt{4} \cdot i = 2i$ |
| $\sqrt{-9}$ | $3i$ | $\sqrt{9} \cdot i = 3i$ |
| $\sqrt{-7}$ | $i\sqrt{7}$ | $\sqrt{7} \cdot i$ |
| $\sqrt{-4} \cdot \sqrt{-9}$ | $-6$ | $= 2i \cdot 3i = 6i^2 = -6$ (not $+6$) |
Definition
Since $i^2 = -1$ and $(-i)^2 = (-1)^2 \cdot i^2 = -1$, both $i$ and $-i$ are square roots of $-1$.
Definition. The principal square root of $-1$ is $i$, written $\sqrt{-1} = i$.
For any positive real number $a$, the principal square root of $-a$ is:
$\sqrt{-a} = i\sqrt{a}$For example:
$ \sqrt{-25} = i\sqrt{25} = 5i, \qquad \sqrt{-3} = i\sqrt{3}, \qquad \sqrt{-\tfrac{1}{4}} = \tfrac{1}{2}i $Example 1. Simplify $3\sqrt{-16} - 2\sqrt{-9}$.
Solution.
$ \begin{aligned} 3\sqrt{-16} - 2\sqrt{-9} &= 3(4i) - 2(3i) \\ &= 12i - 6i \\ &= 6i \end{aligned} $
A Critical Warning: Products of Negative Square Roots
For positive numbers $a$ and $b$:
$ \sqrt{-a} \cdot \sqrt{-b} = (i\sqrt{a})(i\sqrt{b}) = i^2 \sqrt{ab} = -\sqrt{ab} $Warning. The product rule $\sqrt{x} \cdot \sqrt{y} = \sqrt{xy}$ is only valid when $x \geq 0$ and $y \geq 0$. It fails when both numbers are negative:
$\sqrt{-a} \cdot \sqrt{-b} \neq \sqrt{(-a)(-b)} = \sqrt{ab}$The correct result is $\sqrt{-a} \cdot \sqrt{-b} = -\sqrt{ab}$.
Example 2. Compute $\sqrt{-4} \cdot \sqrt{-9}$.
Correct approach:
$\sqrt{-4} \cdot \sqrt{-9} = (2i)(3i) = 6i^2 = 6(-1) = -6$Incorrect approach (do not use):
$\sqrt{-4} \cdot \sqrt{-9} \neq \sqrt{(-4)(-9)} = \sqrt{36} = 6$The correct answer is $-6$, not $+6$.
Example 3. Simplify $\sqrt{-2} \cdot \sqrt{-8}$.
Solution.
$\sqrt{-2} \cdot \sqrt{-8} = (i\sqrt{2})(i\sqrt{8}) = i^2 \sqrt{16} = -1 \cdot 4 = -4$Alternatively, $\sqrt{-2} \cdot \sqrt{-8} = -\sqrt{2 \cdot 8} = -\sqrt{16} = -4$.
Simplifying Expressions with $\sqrt{-a}$
When simplifying expressions involving square roots of negative numbers, always convert to the $i\sqrt{a}$ form first, then carry out arithmetic on $i$ as usual.
Example 4. Compute $(3 + \sqrt{-4})(1 - \sqrt{-9})$.
Solution. Convert first: $\sqrt{-4} = 2i$ and $\sqrt{-9} = 3i$.
$ \begin{aligned} (3 + 2i)(1 - 3i) &= 3 - 9i + 2i - 6i^2 \\ &= 3 - 7i - 6(-1) \\ &= 9 - 7i \end{aligned} $Frequently Asked Questions
What is $\sqrt{-1}$ equal to?
$\sqrt{-1} = i$, the imaginary unit. It is defined as the principal square root of $-1$, meaning the one that we designate as positive (by convention). The other square root of $-1$ is $-i$, since $(-i)^2 = i^2 = -1$.
