Complex numbers

A square root of −1

No real number squares to a negative number: $x^2 \geq 0$ for every real $x$. So $x^2 = -1$ has no real solution, and $\sqrt{-1}$ has no real value (lesson 1.1.1).

We add one new number, $i$, defined by $i^2 = -1$. It is called the imaginary unit. With it, every negative number has square roots: $\sqrt{-16} = 4i$, since $(4i)^2 = 16i^2 = -16$.

A complex number is $z = a + bi$, where $a$ and $b$ are real numbers. $a$ is the real part and $b$ the imaginary part. For $z = 3 - 2i$ the real part is 3 and the imaginary part is $-2$, a real number, not $-2i$.

Two kinds of number sit inside this. With $b = 0$, $a + 0i$ is just the real number $a$, so every real number is complex. With $a = 0$, the number $bi$ is called purely imaginary.

The complex plane

A complex number has two parts, so it needs two coordinates. Draw $a + bi$ as the point $(a, b)$: real part across, imaginary part up.

This is the complex plane. Its horizontal axis holds the real numbers and its vertical axis the purely imaginary ones.

Adding and subtracting work part by part, as for coordinates: $(2 + 3i) + (3 - 2i) = 5 + i$, and $(3 + 4i) - (1 - 6i) = 2 + 10i$. In the plane, adding $3 - 2i$ moves a point 3 across and 2 down.

The conjugate of $z = a + bi$ is $\bar{z} = a - bi$, read "z bar". It has the same real part and the opposite imaginary part. In the plane it is the mirror image of $z$ in the real axis.

The modulus $|z| = \sqrt{a^2 + b^2}$ is the distance from 0 to the point $(a, b)$, by Pythagoras' theorem (lesson 1.7.1). For a real number it is the absolute value of lesson 0.3.3: $|-5| = 5$. It is never negative, and only 0 has modulus 0.

For $z = 4 + 3i$: $\bar{z} = 4 - 3i$ and $|z| = 5$. Adding a number to its conjugate cancels the imaginary parts: $z + \bar{z} = 8$, twice the real part.

Multiplying

Multiply out the brackets as for $(x + 2)(x + 3)$, then replace $i^2$ by $-1$.

Example 1. Multiply $(2 + 3i)(1 - i)$.

The $-3i^2$ became $+3$, which joined the real part. Doing the same with letters gives a rule for every product.

Powers of $i$, and the quarter turn

Multiplying by $i$ again and again gives a table that repeats.

Algebra shows why: $i^3 = i^2 \cdot i = -i$, and $i^4 = (i^2)^2 = (-1)^2 = 1$. After every fourth power you are back at 1, so only the remainder after dividing the power by 4 matters. For example $i^{10} = i^8 \cdot i^2 = -1$.

Multiplying any number by $i$ gives $i(a + bi) = ai + bi^2 = -b + ai$. The point $(a, b)$ moves to $(-b, a)$: a quarter turn anticlockwise about 0. So $i(3 + 2i) = -2 + 3i$, and three more quarter turns lead back to $3 + 2i$.

Dividing

Division rests on one product. Multiplying out, $(a + bi)(a - bi) = a^2 - abi + abi - b^2i^2 = a^2 + b^2$. So a number times its conjugate is real and never negative.

To divide by $c + di$, multiply the top and the bottom by its conjugate $c - di$. The bottom becomes the real number $c^2 + d^2$, and dividing by a real number divides both parts.

Example 2. Divide $3 + 4i$ by $1 + 2i$. The top becomes $(3 + 4i)(1 - 2i) = 3 - 6i + 4i - 8i^2 = 11 - 2i$, and the bottom $1^2 + 2^2 = 5$.

Multiplying back checks it: $(2.2 - 0.4i)(1 + 2i) = 2.2 + 4.4i - 0.4i - 0.8i^2 = 3 + 4i$.

A special case: $\frac{1}{i} = -i$, because $i \cdot (-i) = -i^2 = 1$. Dividing by $i$ is a quarter turn clockwise.

Complex roots of a quadratic

Lesson 1.2.2 stopped at a negative discriminant: no real roots, because the formula needs the square root of a negative number. With $i$ the quadratic formula works anyway.

Example 3. Solve $x^2 - 6x + 13 = 0$. The discriminant is $b^2 - 4ac = 36 - 52 = -16$, and $\sqrt{-16} = 4i$. So

Putting $3 + 2i$ back in checks it. $(3 + 2i)^2 = 9 + 12i + 4i^2 = 5 + 12i$, and $-6(3 + 2i) = -18 - 12i$. Then $5 + 12i - 18 - 12i + 13 = 0$.

For a quadratic with real $a$, $b$ and $c$, complex roots always come as such a conjugate pair. That is what np.roots showed in lesson 1.3.1: $x^3 - 8$ has the real root 2 and the pair $-1 \pm 1.732051i$.

Lesson 1.7.3 describes each complex number by its distance and its angle instead. That makes multiplying, and the quarter turn above, much simpler to see.

Common mistakes

1. The imaginary part is a real number. For $3 - 2i$ it is $-2$, not $-2i$.
2. $i^2$ is $-1$, so $-3i^2$ is $+3$. Replace every $i^2$ after multiplying out: $(2 + 3i)(1 - i)$ is $5 + i$, not $-1 + i$.
3. Python's j goes straight after a number: 1j, not j. On its own, j is a variable name.
4. The modulus squares both parts. $|4 + 3i| = \sqrt{16 + 9} = 5$, not $4 + 3 = 7$. And $|-4 - 3i|$ is also 5: the squares remove the signs.
5. To divide, use the conjugate of the bottom. Multiplying by the conjugate of the top leaves a complex number underneath.
6. $\sqrt{a}\sqrt{b} = \sqrt{ab}$ fails for negative numbers. $\sqrt{-4}\sqrt{-9} = 2i \cdot 3i = -6$, but $\sqrt{36} = 6$. Take the $i$ out first.
7. A negative discriminant does not mean "no roots". It means no real roots, and two complex ones, conjugates of each other.