4.02e - Arithmetic in cartesian form

Adding, subtracting and multiplying

You add, subtract and multiply complex numbers just as you handle expressions in $x$, and that simple idea settles a puzzle from the sixteenth century.

In 1572 the Italian engineer Rafael Bombelli published a calculation that seemed to need the square root of a negative number, even though the answer it was meant to give was the ordinary number 4.

He was solving $x^3 = 15x + 4$. You can check that $x = 4$ works: $4^3 = 64$ and $15 \times 4 + 4 = 64$. The graph of $y = x^3 - 15x - 4$ crosses the $x$-axis three times, so all three roots are real. Yet the formula for cubic equations that Gerolamo Cardano had published in 1545 gives

[DIAGRAM: asset_name: Bombelli's cubic has three real roots; asset_slug: h245_4_02_bombelli_cubic; caption: The cubic from Bombelli's puzzle crosses the x-axis three times, once at x = 4, yet Cardano's formula for it contains the square root of -121; description: Graph of y = x cubed minus 15x minus 4 for x from about -4.6 to 4.9, crossing the x-axis at x = -2 - root 3 (about -3.73), x = -2 + root 3 (about -0.27) and x = 4, with the three roots marked as dots and a box showing Cardano's formula with the cube roots of 2 + root(-121) and 2 - root(-121)]

So here is the question this lesson answers: how can a calculation that passes through $\sqrt{-121}$ end at the honest answer 4? Bombelli's idea was to stop calling $\sqrt{-1}$ impossible and simply calculate with it. By the end of this lesson you will be able to do his calculation yourself.

Calculating in cartesian form

Collect the real parts and the imaginary parts separately, and replace $i^2$ by $-1$ whenever it appears. To add or subtract, combine the real parts and combine the imaginary parts:

To multiply, expand every pair of terms. The product of the two imaginary terms contains $i^2$, which is the real number $-1$, so that product joins the real part. This single step is the whole difference between complex algebra and the algebra you know.

Worked example 2 changes one thing: instead of multiplying two different numbers, you square one number and then use it inside a longer expression. That matters because the middle term of the square and the minus sign in front of $4z$ are where most marks are lost.

You have just shown that $2 - 3i$ is a root of the quadratic equation $z^2 - 4z + 13 = 0$. Finding roots like this directly, and seeing why $2 + 3i$ must be a root as well, comes in the lesson on solving equations with complex roots.

Multiplication also gives you a way to check any complex answer: multiply back or substitute, exactly as you would check a real solution. Do it every time; it takes seconds.

Complex conjugates

Every complex number has a partner called its conjugate, and multiplying the two together always gives a real number, which is exactly what you need in order to divide.

A real number is its own conjugate, for example $6^* = 6$, and taking the conjugate twice gives back the number you started with: $(z^*)^* = z$.

Try a concrete case first. Multiply $4 + 7i$ by its conjugate:

The answer is real, and it equals $4^2 + 7^2$. Squaring does not do this: $(4 + 7i)^2 = 16 + 56i - 49 = -33 + 56i$, which still has an imaginary part. In general $z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy$, which is real only when $x = 0$ or $y = 0$.

The cancellation with the conjugate happens for every complex number. With $z = x + iy$,

So $zz^*$ is real and never negative, and it is 0 only when $z = 0$. It is also the square of the modulus: for $4 + 7i$, $zz^* = 65$ and $|z| = \sqrt{65}$. The sum $z + z^*$ is real and the difference $z - z^*$ is purely imaginary (or 0). A one-mark question asking you to show that $z + z^* = 2\,\mathrm{Re}(z)$ is a genuine exam type, and the full answer is one line: let $z = a + bi$ with $a$ and $b$ real, so $z^* = a - bi$ and $z + z^* = 2a = 2\,\mathrm{Re}(z)$.

These facts come back at the end of this lesson, in Bombelli's puzzle, and again when you solve quadratic equations with complex roots: a number and its conjugate always have a real sum and a real product.

Dividing complex numbers

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

A fraction such as $\dfrac{7 + 4i}{2 - i}$ is not yet in the form $a + bi$, because its denominator is complex. Multiplying the top and bottom of a fraction by the same non-zero number does not change its value, and choosing $(2 - i)^* = 2 + i$ makes the new denominator $(2 - i)(2 + i) = 2^2 + 1^2 = 5$, which is real. This is the result $zz^* = x^2 + y^2$ from the previous part doing its job.

Worked example 4 changes one thing: the division is hidden inside an equation, so you have to get the $z$ term on its own before you divide. That matters because the order is the same as for a real linear equation: undo the addition first, then divide by the coefficient.

Division is the last of the four operations, so you can now simplify any expression built from complex numbers into the single form $a + bi$.

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 equation between complex numbers gives you two equations between real numbers.

Why it is true

You already know the special case where one side is 0: if $x$ and $y$ are real, $x + iy = 0$ only when $x = 0$ and $y = 0$, because otherwise $i$ would be a real number. Now suppose $a + bi = c + di$. Subtracting $c + di$ from both sides gives $(a - c) + (b - d)i = 0$, where $a - c$ and $b - d$ are real, so both are zero. Therefore $a = c$ and $b = d$.

The proof only works because $a$, $b$, $c$ and $d$ are real. That is the condition to remember: before you equate parts, every unknown must be a real number, so replace a complex unknown $z$ by $x + iy$ first.

Worked example 6 changes one thing: $z^*$ is multiplied by $i$. That matters because multiplying by $i$ swaps the roles of the parts, so $x$ and $y$ both appear in each real equation and you must solve them simultaneously.

This method works for any linear equation in $z$ and $z^*$. Later in the course the same comparison of real and imaginary parts is how you find the square roots of a complex number.

Bombelli's puzzle solved

You can now answer the question from the start of the lesson, using only multiplication and conjugates. First simplify the square root: $\sqrt{-121} = 11i$, so Cardano's formula reads $x = \sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i}$. Bombelli showed that $2 + i$ is a cube root of $2 + 11i$ by multiplying it out:

The second number under a cube root, $2 - 11i$, is the conjugate of the first. Repeating the same multiplication with $-i$ in place of $i$, which also squares to $-1$, changes the sign of every imaginary part:

So a cube root of $2 + 11i$ is $2 + i$, the matching cube root of its conjugate $2 - 11i$ is $2 - i$, and

The imaginary parts cancel because the two cube roots are conjugates: $z + z^* = 2\,\mathrm{Re}(z)$, and here $2 \times 2 = 4$. The calculation travelled through complex numbers but arrived at a real answer, and results like this began the slow acceptance of complex numbers by mathematicians.

Quick Recap

• Add, subtract and multiply as in ordinary algebra, collecting real and imaginary parts and replacing $i^2$ by $-1$.
• $zz^* = x^2 + y^2 = |z|^2$ is real and never negative; $z + z^* = 2\,\mathrm{Re}(z)$ and $z - z^* = 2i\,\mathrm{Im}(z)$.
• Divide by multiplying the top and bottom by the conjugate of the denominator, so the denominator becomes the real number $c^2 + d^2$.
• Two complex numbers are equal only when their real parts are equal and their imaginary parts are equal; write $z = x + iy$ with $x$ and $y$ real to turn one complex equation into two real ones.

Bombelli's answer: $(2 + i)^3 = 2 + 11i$ and $(2 - i)^3 = 2 - 11i$, so Cardano's formula gives $(2 + i) + (2 - i) = 4$.