Complex Numbers and the Imaginary Unit
The imaginary unit i is defined by \(i^2 = -1\). A complex number \(a+bi\) pairs a real number with a multiple of i, and this small extension is exactly enough to give every quadratic equation a solution.
By the end you'll be able to add and multiply complex numbers, simplify powers of i, and explain what multiplying by i does geometrically.
Predict: starting from \(3 + 0i\), where does the point land after you click × i twice? Click it twice below to check.
Every complex number \(a+bi\) is a point on the complex plane: a along the horizontal (real) axis, b along the vertical (imaginary) axis. Drag the sliders to move the point, and watch its modulus — the length of the dashed segment back to the origin. Click × i and the point rotates 90° counterclockwise around the origin.
Point: 3.0 + 0.0i · modulus = 3.000
No real number squares to a negative, so \(x^2=-1\) has no solution among the reals — i is invented to fill exactly that gap, and \(a+bi\) numbers extend the real line to a full plane.
Think of \(a+bi\) as a point (or arrow) on a plane instead of a single number on a line: a tells you how far along the real axis, b how far up the imaginary axis. Multiplying by i doesn't move you along a number line at all — it spins your arrow 90° counterclockwise around the origin, exactly as you saw in the interactive above. Do that spin twice (multiply by \(i^2\)) and you've turned 180°, landing on the exact opposite point across the origin — which is just what "multiply by \(-1\))" means. That's why \(i^2=-1\) isn't an arbitrary rule; it's the geometry of two quarter-turns.
Add and subtract componentwise: \((a+bi)\pm(c+di) = (a\pm c)+(b\pm d)i\). Multiply with FOIL and replace \(i^2\) by \(-1\):
\((a+bi)(c+di) = ac + adi + bci + bd\,i^2 = (ac-bd) + (ad+bc)i.\)
The conjugate of \(a+bi\) is \(a-bi\); multiplying a number by its own conjugate clears the imaginary part entirely, \((a+bi)(a-bi)=a^2+b^2\), which is exactly the square of the modulus and always a non-negative real number.
Complex numbers aren't just an algebra trick — electrical engineers track AC current and voltage as complex numbers because multiplying by i (a 90° phase shift) is exactly how a capacitor or inductor behaves in a circuit, and quadratics with complex roots (a negative discriminant) show up whenever a physical system oscillates instead of settling smoothly.
Multiply \((2+3i)(1-2i)\): FOIL gives \(2\cdot1 + 2\cdot(-2i) + 3i\cdot1 + 3i\cdot(-2i) = 2 - 4i + 3i - 6i^2\). Replace \(i^2=-1\): \(-6i^2 = 6\), so this is \(2 - 4i + 3i + 6\). Combine real and imaginary parts separately: \((2+6) + (-4+3)i = \) \(8 - i\).
Multiply \((4-i)(2+3i)\): FOIL gives \(4\cdot2 + 4\cdot3i + (-i)\cdot2 + (-i)\cdot3i = 8 + 12i - 2i - 3i^2\). Replace \(i^2=-1\), so \(-3i^2 = \) ____, and combine terms to get the final answer.
Reveal the answer
\(-3i^2 = -3(-1) = 3\), so the expression is \(8 + 12i - 2i + 3\). Combine real parts \(8+3=11\) and imaginary parts \(12i-2i=10i\): \(11 + 10i\). The demo above can't multiply two arbitrary complex numbers directly, but it can check the moduli: set a = 4, b = -1 on the sliders to read off the modulus of \(4-i\), which is \(\sqrt{17}\); then set a = 2, b = 3 for \(2+3i\)'s modulus, \(\sqrt{13}\). Moduli multiply under complex multiplication, so \(\sqrt{17}\cdot\sqrt{13} = \sqrt{221}\) — exactly the modulus of the answer \(11+10i\), since \(11^2+10^2=221\).
More info — why this fills the gap left by the reals
Solving \(x^2+4x+13=0\) by completing the square gives \((x+2)^2=-9\). Among the reals that's a dead end — nothing squares to \(-9\) — but with i available, \(\sqrt{-9}=\sqrt{9}\cdot\sqrt{-1}=3i\), so \(x=-2\pm3i\). Checking \(x=-2+3i\) back in the original equation confirms it works, and the two roots \(-2\pm3i\) are conjugates of each other, a pattern that always holds when a real-coefficient quadratic has complex roots. See the Paul's Online Notes link in Dive deeper for more worked examples, and the discriminant link in the previous lesson's Dive deeper for when this happens.
Check your understanding
Simplify \((5 - 2i) - (3 + 4i)\).
What is \(i^{17}\)? (Powers of i cycle every 4: \(i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1\).)
Why does multiplying \(a+bi\) by its conjugate \(a-bi\) always give a real, non-negative number?
Start at \(2 + 5i\). Add \(1 - 2i\) to it, then multiply the result by \(i\). Where do you land?
Recap
- i is defined by \(i^2=-1\); a complex number \(a+bi\) has real part a and imaginary part b.
- Add/subtract componentwise: \((a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i\).
- Multiply with FOIL, then replace \(i^2\) with \(-1\): \((a+bi)(c+di) = (ac-bd) + (ad+bc)i\).
- The modulus of \(a+bi\) is \(\sqrt{a^2+b^2}\), the distance from the origin on the complex plane.
- Multiplying by i rotates a point 90° counterclockwise around the origin; two rotations (multiplying by \(i^2\)) send a point to its opposite — the geometric picture behind \(i^2=-1\).
Dive deeper
- Paul's Online Notes — Complex Numbers Define i and operate on a+bi form.
- Math is Fun — Complex Numbers Build intuition for real and imaginary parts.
Sources
- Complex Numbers and the Imaginary Unit