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}iExample 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) = -6Incorrect approach (do not use):
\sqrt{-4} \cdot \sqrt{-9} \neq \sqrt{(-4)(-9)} = \sqrt{36} = 6The 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 = -4Alternatively, \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.
