Introduction to Complex Numbers

The equation $x^2 + 1 = 0$ has no real solution, because no real number squares to give $-1$. Complex numbers fix this by adding one new number, $i$, with $i^2 = -1$, and then allowing everything that ordinary arithmetic can build from it. This note covers the vocabulary (real part, imaginary part, conjugate), the arithmetic of adding, subtracting and multiplying, the rule that two complex numbers are equal only when both parts match, and solving quadratics whose discriminant is negative. Every other complex-numbers question on P3 rests on these skills, so they need to be fast and error-free.

The number $i$

In P1 a quadratic with $b^2 - 4ac < 0$ had "no real roots", and that was the end of it. The trouble is the square root of a negative number. If we simply define a number whose square is $-1$, the problem disappears.

Nothing else changes: $i$ obeys all the usual rules of algebra (brackets expand, terms collect, the order of multiplication does not matter), with the single extra rule that $i^2$ is replaced by $-1$ whenever it appears.

Powers of $i$

Because $i^2 = -1$, the powers of $i$ repeat in a cycle of four:

$$i^1 = i, \qquad i^2 = -1, \qquad i^3 = i^2 \cdot i = -i, \qquad i^4 = \left(i^2\right)^2 = 1, \qquad i^5 = i, \ \ldots$$

To simplify $i^n$, divide $n$ by $4$ and keep the remainder. For example $23 = 4 \times 5 + 3$, so $i^{23} = \left(i^4\right)^5 i^3 = -i$.

What a complex number is

So $\operatorname{Re}(4 - 7i) = 4$ and $\operatorname{Im}(4 - 7i) = -7$. Every real number is also a complex number (with imaginary part $0$), so the complex numbers contain the real numbers, just as the real numbers contain the integers. The set of complex numbers is written $\mathbb{C}$.

Both $x + iy$ and $x + yi$ are fine; Cambridge papers use both. With surds it is clearer to put the $i$ first: $2 + i\sqrt{3}$ cannot be misread as $2 + \sqrt{3i}$.

Equality: one equation becomes two

Two complex numbers are the same only if they agree in both "directions" at once.

The reason is that a real number can never equal a non-zero imaginary one. If $a + ib = c + id$ then $a - c = i(d - b)$. The left side is real; the right side is imaginary unless $d - b = 0$. So $d = b$, and then $a = c$.

This rule is used constantly: in finding square roots, in solving equations involving $z^*$, and in finding unknown coefficients of a polynomial with a given complex root.

Addition, subtraction and multiplication

Addition and subtraction work part by part, like collecting like terms. Multiplication is expanding brackets: there are four terms, and the $i^2bd$ term becomes $-bd$ and joins the real part. Do not memorise the multiplication formula. Expand, replace $i^2$, collect.

Multiplying by a real number scales both parts: $3(2 - 5i) = 6 - 15i$.

The conjugate

Some books write $\bar{z}$; Cambridge uses $z^*$. The conjugate has three properties you will use again and again:

The product is the important one:

$$zz^* = (x + iy)(x - iy) = x^2 - ixy + ixy - i^2y^2 = x^2 + y^2.$$

It is the "difference of two squares" with a sign flip, and it is the key to division: multiplying by the conjugate turns a complex number into a real one. In the Argand diagram you will see that $x^2 + y^2 = |z|^2$.

Quadratic equations with real coefficients

With $i$ available, every quadratic equation has two roots (possibly equal). When $b^2 - 4ac < 0$, write $\sqrt{b^2 - 4ac} = i\sqrt{4ac - b^2}$ and carry on with the formula.

Notice what happens to the roots: the formula gives $-\dfrac{b}{2a} \pm \dfrac{\sqrt{4ac - b^2}}{2a}\,i$. The two roots are conjugates of each other. This is the simplest case of a general fact: non-real roots of a polynomial with real coefficients come in conjugate pairs (see complex roots of polynomials).

Since the roots are $\alpha$ and $\alpha^*$, the quadratic is $(z - \alpha)(z - \alpha^*) = z^2 - (\alpha + \alpha^*)z + \alpha\alpha^*$. For $\alpha = p + iq$:

Worked examples

Building a cube as "square, then multiply once more" keeps each step to a single bracket expansion and is much safer than the binomial expansion with powers of $i$.

Common mistakes

Exam technique

Summary

Practice

Answers

1. (a) $6 - 2i + 15i - 5i^2 = 6 + 13i + 5 = 11 + 13i$. (b) $(1 - i)^2 = 1 - 2i + i^2 = -2i$, so $(1 - i)^3 = -2i(1 - i) = -2i + 2i^2 = -2 - 2i$.

2. (a) $23 = 4 \times 5 + 3$, so $i^{23} = i^3 = -i$. (b) $50 = 4 \times 12 + 2$, so $i^{50} = i^2 = -1$. (c) $(1 + i)^2 = 2i$, so $(1 + i)^6 = (2i)^3 = 8i^3 = -8i$.

3. $(p + qi)(1 + 2i) = p + 2pi + qi + 2qi^2 = (p - 2q) + (2p + q)i$. Real: $p - 2q = 0$. Imaginary: $2p + q = 5$. So $p = 2q$, $5q = 5$: $q = 1$, $p = 2$.

4. With $z = x + iy$: $2x + 2iy - x + iy = x + 3iy$. Real: $x = 3$. Imaginary: $3y = 6$, $y = 2$. So $z = 3 + 2i$.

5. (a) Discriminant $4 - 20 = -16$, $\sqrt{-16} = 4i$, $z = \dfrac{2 \pm 4i}{2} = 1 \pm 2i$. (b) $(z + 2)^2 + 3 = 0$, so $z + 2 = \pm i\sqrt{3}$ and $z = -2 \pm i\sqrt{3}$.

6. $zz^* = 4 + 9 = 13$ and $z + z^* = 4$. Then $z^2 = 4 - 12i + 9i^2 = -5 - 12i$, so $z^2 - 4z + 13 = (-5 - 12i) - (8 - 12i) + 13 = 0$. (Equivalently, $z$ and $z^*$ have sum $4$ and product $13$, so they are the roots of $z^2 - 4z + 13 = 0$.)

7. $(k + 2i)(3 - i) = 3k - ki + 6i - 2i^2 = (3k + 2) + (6 - k)i$. (a) Real part zero: $k = -\tfrac{2}{3}$ (the imaginary part is then $\tfrac{20}{3} \ne 0$). (b) Imaginary part zero: $k = 6$.

8. With $z = x + iy$: $zz^* = x^2 + y^2$, so the equation is $(x^2 + y^2 + 2x) + 2yi = 19 + 8i$. Imaginary: $2y = 8$, $y = 4$. Real: $x^2 + 16 + 2x = 19$, so $x^2 + 2x - 3 = 0$, $(x + 3)(x - 1) = 0$. Hence $z = 1 + 4i$ or $z = -3 + 4i$.

9. (a) $(1 + i)^3 = (1 + i)^2(1 + i) = 2i(1 + i) = 2i + 2i^2 = -2 + 2i$. Then $(1 + i)^{10} = \left((1 + i)^2\right)^5 = (2i)^5 = 32i^5 = 32i$. (b) $(1 + i)^1 = 1 + i$, $(1 + i)^2 = 2i$ and $(1 + i)^3 = -2 + 2i$ are not real, but $(1 + i)^4 = (2i)^2 = -4$ is. So $n = 4$ and the value is $-4$.