MISSION · Number sense
Complex Numbers
Inventing a square root for −1, and what it turns out to do
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LEVEL 2 · LEARN
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The idea in one sentence
There is no real number whose square is \(-1\), so mathematics defines one, calls it \(i\), and everything built on top of that definition turns out to work.
\[i^2 = -1\]Not a trick
Inventing a number to fill a gap sounds like cheating, but it’s what happened at every previous stage too. Subtraction demanded negative numbers before anyone had seen a negative sheep. Division demanded fractions. The square root of 2 forced irrationals onto people who didn’t want them.
Each time, the new numbers were resisted, then turned out to be indispensable. “Imaginary” is a leftover insult from that resistance, and it’s a bad name — these numbers describe rotating machinery, alternating current, and quantum states, which are not imaginary at all.
The form
Every complex number is \(a + bi\): a real part \(a\) and an imaginary part \(b\). Real numbers are the ones where \(b = 0\), so they’re a subset, not a separate world.
Arithmetic follows the ordinary rules of algebra, treating \(i\) as a variable, with one extra move: whenever \(i^2\) appears, replace it with \(-1\).
\[(3 + 2i) + (1 - 5i) = 4 - 3i\] \[(3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i^2 = 3 + 14i - 8 = -5 + 14i\]That last step is the whole difference from ordinary expansion.
Powers of i cycle
\[i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1\]And then it repeats, forever, in fours. So \(i^{27}\) is found by dividing 27 by 4 and keeping the remainder, 3, giving \(-i\).
That four-step cycle is a clue about what \(i\) really does, and the picture makes it obvious.
The picture
Plot \(a + bi\) as the point \((a, b)\) — real part across, imaginary part up. That plane is called the Argand diagram, and on it complex numbers stop being formal and start being geometric.
Multiplying by \(i\) rotates a point a quarter turn anticlockwise about the origin. Do it four times and you’re back where you started, which is exactly the cycle above. Multiplying by \(-1\) is a half turn, so \(i\) being “halfway” to \(-1\) is a literal geometric statement.
This is why complex numbers describe anything that rotates or oscillates. The algebra carries a rotation inside it.
Every polynomial finally behaves
Once complex numbers exist, \(x^2 + 1 = 0\) has solutions, \(\pm i\). More than that: every polynomial of degree \(n\) has exactly \(n\) roots, counted properly, with no exceptions ever again. That result is called the Fundamental Theorem of Algebra, and it’s the reason mathematicians stopped treating complex numbers as a curiosity.
A quadratic with a negative discriminant doesn’t lack solutions. Its solutions are complex, and they come as a conjugate pair, \(a + bi\) and \(a - bi\).
Common mix-ups
- Writing \(\sqrt{-9}\) as \(-3\). It’s \(3i\).
- Forgetting \(i^2 = -1\) when expanding, which leaves an \(i^2\) sitting in the answer.
- Adding a real part to an imaginary one. \(3 + 2i\) doesn’t simplify to \(5i\) any more than \(3 + 2x\) simplifies to \(5x\).
- Assuming complex numbers can be ordered. There’s no sensible “bigger than” on the plane.
Check yourself
Simplify \((2 + 3i)(2 - 3i)\), and find \(i^{15}\).
\(4 - 6i + 6i - 9i^2 = 4 + 9 = 13\) — a conjugate pair multiplies to a real number. \(15 \div 4\) leaves remainder 3, so \(i^{15} = -i\).
LEVEL 3 · REFLECT
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