The Square Roots of a Negative Real Number

The Square Roots of a Negative Real Number

We already know that $i^2 = -1$. But notice that $(-i)^2 = (-1)^2 i^2 = 1 \cdot (-1) = -1$ as well. So both $i$ and $-i$ are square roots of $-1$. When we write the symbol $\sqrt{-1}$, however, we mean only the principal square root, which is $i$. Both $i$ and $-i$ satisfy the equation $x^2 + 1 = 0$, or equivalently $x^2 = -1$.

Now consider $-3$. We can check:

So the square roots of $-3$ are $\sqrt{3}\,i$ and $-\sqrt{3}\,i$. The symbol $\sqrt{-3}$ is reserved for the principal square root, which is $\sqrt{3}\,i$. That is, $\sqrt{-3} = \sqrt{3}\,i$.

This is the general rule: if $a$ is a positive real number, then $\sqrt{-a} = \sqrt{a}\,i$. The reasoning is simply $\sqrt{-a} = \sqrt{a \cdot (-1)} = \sqrt{a}\,\sqrt{-1} = \sqrt{a}\,i$.

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The Product Rule for Square Roots When One or Both Numbers Are Negative

We know that for positive real numbers $a$ and $b$, the identity $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ holds. This result also holds when exactly one of $a$ or $b$ is negative and the other is positive. But what happens when both $a$ and $b$ are negative?

Let us test the case $a = -1$, $b = -1$. If we blindly apply the product rule, we would get:

But we know that $\sqrt{-1} = i$, so $\sqrt{-1} \times \sqrt{-1} = i \times i = i^2 = -1$. This gives $1 = -1$, a clear contradiction.