Every quadratic with a negative discriminant left us stranded — "no real roots." Complex numbers are mathematics refusing to be stopped by \( \sqrt{-1} \), and the payoff is a whole new plane to work in.
Inventing \( i \)
The equation \( x^2 = -1 \) has no real answer, because a square is never negative — multiply any real number by itself and you get something positive (or zero). For centuries that was the end of the conversation. The bold move is to stop treating it as a dead end and instead define a brand-new number, called \( i \), whose entire job is to make that equation true:
\[ i^2 = -1. \]Once \( i \) exists, we can build a richer kind of number around it. A complex number is anything of the form
\[ z = a + bi, \]where \( a \) is the real part and \( b \) is the imaginary part (both \( a \) and \( b \) are ordinary real numbers). The familiar real numbers haven't gone anywhere — they're simply the special case \( b = 0 \), sitting quietly inside this larger world.
One small wonder falls out immediately: the powers of \( i \) march around in a four-step cycle and then repeat forever.
\[ i^1 = i, \qquad i^2 = -1, \qquad i^3 = -i, \qquad i^4 = 1. \]After \( i^4 = 1 \) you're back to the start, so \( i^5 = i \), \( i^6 = -1 \), and so on. To simplify any power of \( i \), you only ever need its remainder after dividing by four.
Arithmetic: like terms, with one twist
The comforting news is that complex numbers obey all the usual algebra. To add or subtract, you just combine the real parts with the real parts and the imaginary parts with the imaginary parts — exactly like collecting like terms:
\[ (a + bi) + (c + di) = (a + c) + (b + d)i. \]To multiply, expand the brackets with ordinary FOIL, then use the one new rule — replace every \( i^2 \) with \( -1 \) — and collect what's left:
\[ (a + bi)(c + di) = ac + adi + bci + bd\,i^2 = (ac - bd) + (ad + bc)i. \]That single substitution, \( i^2 = -1 \), is the whole twist. Everything else is the algebra you already know.
A number needs a plane
A real number lives on a line — one axis is enough to pin it down. But a complex number carries two independent pieces of information, its real part and its imaginary part, so a single line can't hold it. It needs a plane: a horizontal real axis and a vertical imaginary axis. This picture is called the Argand plane, and on it we plot \( z = a + bi \) at the point \( (a, b) \).
Once a complex number is a point, two natural measurements appear. Its modulus is its straight-line distance from the origin, which Pythagoras hands us at once:
\[ |z| = \sqrt{a^2 + b^2}. \]Its argument is the angle the line from the origin makes with the positive real axis. Together, modulus and argument describe \( z \) as a distance and a direction — a second language for the same number. Drag the point in the widget below, or step it around, and watch the coordinates, the modulus and the argument update together.
Multiplying by \( i \) is a quarter-turn. Press × i in the widget and watch the point swing a quarter-turn around the origin — multiplying by \( i \) is exactly a 90° rotation. The algebra agrees: \( i \cdot (a + bi) = ai + bi^2 = -b + ai \), which sends the point \( (a, b) \) to \( (-b, a) \), precisely a quarter-turn anticlockwise. Press conjugate to flip the point across the real axis.
The conjugate, and division
For every complex number \( z = a + bi \) there is a mirror twin, its conjugate, written with a bar on top and made by flipping the sign of the imaginary part:
\[ \bar z = a - bi. \]On the Argand plane, \( \bar z \) is simply \( z \) reflected across the real axis. Here is the small piece of magic that makes conjugates so useful — multiply a number by its own conjugate and the imaginary part vanishes completely:
\[ z\bar z = (a + bi)(a - bi) = a^2 - (bi)^2 = a^2 + b^2 = |z|^2. \]The result is a real, non-negative number — in fact exactly the square of the modulus. That single fact is the key to division. We don't really know how to divide by \( a + bi \) directly, but we can always multiply the top and bottom of a fraction by the conjugate of the denominator. That turns the denominator into a plain real number, after which dividing is easy.
- Sum. Combine real with real, imaginary with imaginary: \( (3 + 1) + (2 - 4)i = 4 - 2i \).
- Product. FOIL the brackets: \( (3 + 2i)(1 - 4i) = 3 - 12i + 2i - 8i^2 \).
- Replace \( i^2 \) with \( -1 \). The last term \( -8i^2 \) becomes \( +8 \), so we have \( 3 - 10i + 8 \).
- Collect the real parts: \( 3 + 8 = 11 \), leaving \( 11 - 10i \).
The sum is \( \mathbf{4 - 2i} \) and the product is \( \mathbf{11 - 10i} \).
- The denominator's conjugate is \( 1 + i \). Multiply top and bottom by \( \dfrac{1 + i}{1 + i} \), which is just multiplying by \( 1 \) and changes nothing but the form.
- Numerator. \( (2 + 3i)(1 + i) = 2 + 2i + 3i + 3i^2 = 2 + 5i - 3 = -1 + 5i \).
- Denominator. \( (1 - i)(1 + i) = 1 - i^2 = 1 + 1 = 2 \) — real, exactly as promised.
- Now the division is easy: split the real number across both parts, \( \dfrac{-1 + 5i}{2} = -\dfrac{1}{2} + \dfrac{5}{2}i \).
The quotient is \( \mathbf{-\tfrac{1}{2} + \tfrac{5}{2}i} \).
Practice
Try each one yourself, then reveal the full solution.
1. Simplify \( (5 - i) - (2 + 3i) \).
Subtract real parts and imaginary parts separately, being careful with the signs: the real part is \( 5 - 2 = 3 \), and the imaginary part is \( -1 - 3 = -4 \).
The answer is \( \mathbf{3 - 4i} \).
2. Compute \( (4 + i)(4 - i) \) and \( |4 + i| \).
The two factors are conjugates, so the imaginary part will cancel: \( (4 + i)(4 - i) = 16 - i^2 = 16 + 1 = \mathbf{17} \) — a real number, just as \( z\bar z = a^2 + b^2 \) predicts.
The modulus is the square root of that same quantity: \( |4 + i| = \sqrt{4^2 + 1^2} = \mathbf{\sqrt{17}} \).
3. Solve \( x^2 + 9 = 0 \).
Move the \( 9 \) across: \( x^2 = -9 \). A real square can't be negative, but now we have \( i \).
Take the square root of both sides: \( x = \pm\sqrt{-9} = \pm\sqrt{9}\,\sqrt{-1} = \mathbf{\pm 3i} \). These are exactly the "missing" roots a negative discriminant warned us about — they were waiting off the real line all along.