The The Discriminant told you that $x^2 + 1 = 0$ has no real solution: no real number squares to a negative. Mathematicians made the same move they made long ago with negative numbers. When subtraction got stuck ($3 - 5$), they invented new numbers. Here we invent a number whose square is $-1$.
Idea clave
The imaginary unit $i$ is defined by $i^2 = -1$. For any positive number $a$, $\sqrt{-a} = \sqrt{a}\,i$, so $\sqrt{-9} = 3i$ and $\sqrt{-12} = 2\sqrt{3}\,i$. Powers of $i$ repeat in a cycle of four: $i, -1, -i, 1$.
Por qué funciona
Why is $i^2 = -1$ a useful definition, and not just wishful thinking? Because we keep every ordinary rule of arithmetic (adding, multiplying, distributing) and add only this one fact. Nothing breaks, and something good happens: every quadratic equation now has solutions. There is also a picture. Put numbers on a plane with real numbers across and multiples of $i$ up and down. Multiplying by $-1$ turns an arrow halfway around. Multiplying by $i$ turns it a quarter of the way around. Two quarter turns make a half turn, so $i \cdot i = -1$.
Worked examples
Ejemplo resuelto: Square roots of negatives
Simplify $\sqrt{-49}$ and $\sqrt{-18}$.
- $\sqrt{-49} = \sqrt{49} \cdot \sqrt{-1} = 7i$.
- $\sqrt{-18} = \sqrt{18}\,i = \sqrt{9 \cdot 2}\,i = 3\sqrt{2}\,i$.
Respuesta: $7i$ and $3\sqrt{2}\,i$
Ejemplo resuelto: A large power of $i$
Simplify $i^{23}$.
- Every four factors of $i$ make $i^4 = 1$. Divide: $23 = 4 \cdot 5 + 3$.
- $i^{23} = (i^4)^5 \cdot i^3 = 1 \cdot i^3 = -i$.
Respuesta: $-i$
Ejemplo resuelto: Multiplying two roots of negatives
Simplify $\sqrt{-4} \cdot \sqrt{-9}$.
- Rewrite with $i$ first: $\sqrt{-4} = 2i$ and $\sqrt{-9} = 3i$.
- Multiply: $2i \cdot 3i = 6i^2 = -6$.
Respuesta: $-6$
Error común
Multiplying $\sqrt{-4} \cdot \sqrt{-9}$ as $\sqrt{36} = 6$. The rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ only works when $a$ and $b$ are not negative. Always rewrite each root with $i$ before doing anything else. Check: $2i \cdot 3i$ has an $i^2$ in it, so the answer must be negative.