CP 2.2, 2.3 - Complex number arithmetic and the conjugate

The imaginary unit and complex numbers

In 1572 the Italian engineer Rafael Bombelli wrote about a puzzle that ordinary numbers could not explain, and this lesson gives you the tools to solve it.

The equation $x^3 = 15x + 4$ has the solution $x = 4$, because $4^3 = 64$ and $15 \times 4 + 4 = 64$. Yet the formula for cubic equations, published by Gerolamo Cardano in 1545, gives an answer that looks nothing like 4:

How can a formula built from the square root of a negative number produce the ordinary number 4? Bombelli's idea was to treat $\sqrt{-1}$ as a number in its own right and calculate with it. Keep the question in mind: you will answer it at the end of the lesson.

A number whose square is negative

Start with a smaller case. The equation $x^2 = -9$ has no real solution, because the square of every real number is positive or zero. Now suppose there is a number $\mathrm{i}$ with $\mathrm{i}^2 = -1$. Then

so $x^2 = -9$ has two solutions, $x = 3\mathrm{i}$ and $x = -3\mathrm{i}$. In the same way, $\sqrt{-49} = 7\mathrm{i}$ and $\sqrt{-12} = \mathrm{i}\sqrt{12} = 2\mathrm{i}\sqrt{3}$.

This is not the first time numbers have been extended. Whenever an equation had no solution, mathematicians enlarged the number system so that it did. The picture shows the chain you already know, with the complex numbers on the outside.

[DIAGRAM: asset_name: Nested number systems from the natural numbers to the complex numbers; asset_slug: 9fm0-number-systems; caption: Each new number system solves an equation the last one could not; description: Five nested rounded boxes. The innermost is the natural numbers 1, 2, 3 and so on, used for counting. Around it are the integers, which solve x plus 5 equals 2 with x equals minus 3, then the rational numbers, which solve 3x equals 2 with x equals 2 over 3, then the real numbers in blue, which solve x squared equals 2 with x equals plus or minus root 2, and outermost the complex numbers in lavender, which solve x squared equals minus 1 with x equals plus or minus i]

Combining real numbers with real multiples of $\mathrm{i}$ gives every number you need in this course.

Two special cases matter. If $y = 0$, then $z = x$ is real, so every real number is also a complex number. If $x = 0$ and $y \ne 0$, then $z = \mathrm{i}y$ is called purely imaginary, for example $5\mathrm{i}$ or $-\mathrm{i}\sqrt{2}$. Exam questions use this phrase often: “$w$ is purely imaginary” means $\operatorname{Re}(w) = 0$ and $w \ne 0$.

Adding, subtracting and multiplying

Complex numbers follow the algebra you already use: collect like terms, expand brackets, and then replace every $\mathrm{i}^2$ by $-1$.

Now put $\mathrm{i}$ in place of $x$. The expansion is exactly the same; the only new step is $\mathrm{i}^2 = -1$:

Adding and subtracting are simpler still: combine the real parts and combine the imaginary parts, so $(5 + 3\mathrm{i}) + (2 - 7\mathrm{i}) = 7 - 4\mathrm{i}$. A real multiple multiplies both parts: $3(2 - \mathrm{i}) = 6 - 3\mathrm{i}$.

Powers of $\mathrm{i}$

Powers of $\mathrm{i}$ repeat in a cycle of four, because $\mathrm{i}^4 = (\mathrm{i}^2)^2 = (-1)^2 = 1$:

Each multiplication by $\mathrm{i}$ moves one step round the cycle. Later in the course you will see that this is a genuine quarter-turn about 0.

[DIAGRAM: asset_name: The four-step cycle of powers of i; asset_slug: 9fm0-powers-of-i; caption: Multiplying by i moves one step round a cycle of four; description: The numbers 1, i, minus 1 and minus i sit at the right, top, left and bottom of a cross of faint real and imaginary axes, with real numbers in blue and imaginary numbers in lavender. Curved arrows labelled times i go anticlockwise from 1 to i to minus 1 to minus i and back to 1, with labels i to the 1 equals i, i squared equals minus 1, i cubed equals minus i and i to the 4 equals 1. A side note shows i to the 5 equals i up to i to the 8 equals 1, and i to the 27 equals i to the 24 times i cubed equals minus i]

Worked example 2 changes one thing: a power instead of a product. The binomial expansion now does the bookkeeping, and you need both $\mathrm{i}^2$ and $\mathrm{i}^3$, which is where most sign errors happen.

Many complex-number questions carry an instruction about calculators, because some calculators can do complex arithmetic.

The complex conjugate

Changing the sign of the imaginary part of a complex number gives its conjugate, and a number multiplied by its conjugate is always real.

Take $z = 3 + 2\mathrm{i}$. Its conjugate is $z^* = 3 - 2\mathrm{i}$. Combine the two:

The sum and the product are real, and the difference is purely imaginary. In the product the two middle terms cancel, and $-4\mathrm{i}^2 = +4$ adds to the $9$.

The same working with letters shows why the product is always real:

This is the difference of two squares, $(p + q)(p - q) = p^2 - q^2$, with $q = \mathrm{i}y$, so $q^2 = -y^2$. The result $x^2 + y^2$ is a sum of squares of real numbers: it is real, it is never negative, and it is zero only when $z = 0$.

A picture of the conjugate

Plot $z = x + \mathrm{i}y$ as the point $(x, y)$, with real parts across and imaginary parts up; you will work with this picture, the Argand diagram, in CP 2.4. The conjugate is then the mirror image of $z$ in the horizontal axis, and by Pythagoras' theorem the distance from 0 to $z$ is $\sqrt{x^2 + y^2} = \sqrt{zz^*}$.

[DIAGRAM: asset_name: The conjugate as a reflection, with modulus and argument; asset_slug: 9fm0-conjugate-modulus-argument; caption: The conjugate is the reflection of z in the real axis; description: Axes labelled Re across and Im up. The point z equals 3 plus 2i is plotted at (3, 2) and its conjugate z star equals 3 minus 2i at (3, minus 2), joined by a dashed vertical lavender line labelled reflection in the real axis. A clay segment from 0 to z is labelled modulus of z equals root 13, and a clay arc at 0 between the positive real axis and the segment is labelled arg z. A side note says z equals x plus iy is drawn at the point (x, y), the modulus is the distance from 0, the argument is the angle from the positive real axis, and z and z star have the same real part and opposite imaginary parts]

Putting these together gives the results you will use most often.

Conjugating sums and products

Taking the conjugate replaces $\mathrm{i}$ by $-\mathrm{i}$ everywhere. Because $(-\mathrm{i})^2 = -1$ as well, the arithmetic is identical after the swap, so every line of a calculation stays true. So the conjugate of a sum or product is the sum or product of the conjugates: $(z + w)^* = z^* + w^*$ and $(zw)^* = z^*w^*$. For example, with $z = 2 - 3\mathrm{i}$ and $w = 1 + 4\mathrm{i}$ from Worked example 1, $zw = 14 + 5\mathrm{i}$, and

The same holds for powers, $(z^n)^* = (z^*)^n$. In the next lesson you will see that this is why non-real roots of polynomial equations with real coefficients come in conjugate pairs.

Dividing complex numbers

To divide by a complex number, multiply the top and bottom of the fraction by the conjugate of the bottom; the new denominator is real, so the answer splits into a real part and an imaginary part.

Division by $c + d\mathrm{i}$ works the same way, with the conjugate $c - d\mathrm{i}$ playing the part of $3 - \sqrt{2}$. Multiplying the top and bottom by the same non-zero number does not change the value of the fraction, only its appearance, and the bottom becomes $(c + d\mathrm{i})(c - d\mathrm{i}) = c^2 + d^2$.

Worked example 4 changes one feature: the number you want, $z$, is unknown and is multiplied by another complex number. Dividing finds it, and this time the answer has fractions, so the final form matters.

Mark schemes give the first method mark for the conjugate step itself, so make it visible.

Equating real and imaginary parts

Two complex numbers are equal only when their real parts are equal and their imaginary parts are equal, so one complex equation gives you two real equations.

Why is this true? Suppose $x_1 + \mathrm{i}y_1 = x_2 + \mathrm{i}y_2$, where all four numbers are real. Rearranging gives $x_1 - x_2 = \mathrm{i}(y_2 - y_1)$, and squaring both sides gives

The left side is never negative and the right side is never positive, so both are zero: $x_1 = x_2$ and $y_1 = y_2$.

This turns a problem about complex numbers into a pair of simultaneous equations. A neat first use is finding the square roots of a complex number.

Worked example 6 changes one feature: the equation contains both $z$ and $z^*$. The method is the same, but you substitute for both before you equate parts.

Back to Bombelli

You can now answer the question from the start of the lesson. Since $\sqrt{-121} = 11\mathrm{i}$, Cardano's formula says

Look, as Bombelli did, for a cube root of $2 + 11\mathrm{i}$ of the form $a + b\mathrm{i}$ with $a$ and $b$ integers. Expand and equate parts:

In the first equation $a$ must divide 2, and in the second $b$ must divide 11. Try $a = 2$ and $b = 1$: $2(4 - 3) = 2$ and $1(12 - 1) = 11$. So $(2 + \mathrm{i})^3 = 2 + 11\mathrm{i}$, and taking conjugates, $(2 - \mathrm{i})^3 = 2 - 11\mathrm{i}$. The two cube roots are $2 + \mathrm{i}$ and $2 - \mathrm{i}$, and

The imaginary parts cancel and the real solution appears. Complex numbers began like this, as a route to real answers. Every non-zero complex number has three cube roots (you will find them in CP 2.10); Cardano's formula needs a pair whose product is $15 \div 3 = 5$, and $(2 + \mathrm{i})(2 - \mathrm{i}) = 5$.

Examiners also look for the two equations themselves.

Quick Recap

• $\mathrm{i}^2 = -1$, $\sqrt{-k} = \mathrm{i}\sqrt{k}$ for $k > 0$, and the powers of $\mathrm{i}$ repeat every four.
• $z = x + \mathrm{i}y$ has real part $\operatorname{Re}(z) = x$ and imaginary part $\operatorname{Im}(z) = y$, and both are real numbers.
• Add, subtract and multiply as in ordinary algebra, then replace $\mathrm{i}^2$ by $-1$.
• The conjugate $z^* = x - \mathrm{i}y$ gives the real product $zz^* = x^2 + y^2 = |z|^2$; to divide, multiply the top and bottom by the conjugate of the bottom.
• Equal complex numbers have equal real parts and equal imaginary parts, so one complex equation gives two real equations.