4.02a–4.02c - The language of complex numbers

The imaginary unit

Multiplying by $-1$ turns the number line through half a turn about 0: it swaps $3$ with $-3$, $5$ with $-5$, and every number with its negative.

So here is the question this lesson answers: is there a number that turns the number line through a quarter turn, so that doing it twice is the same as multiplying by $-1$? Such a number would have square $-1$, and no number on the number line does: positive and negative numbers both have positive squares, and $0^2 = 0$. If it exists, it must live somewhere off the line. By the end of this lesson you will be able to say exactly where.

A number whose square is minus one

Start with a case you know. The equation $x^2 = 9$ has two roots, $3$ and $-3$. The equation $x^2 = -9$ has no real roots, because the square of any real number is positive or zero.

Now suppose there is a number $i$ whose square is $-1$. Then

and $(-3i)^2 = -9$ as well, so $x^2 = -9$ has the two roots $3i$ and $-3i$.

Write a surd before the $i$ with a clear gap, as in $2\sqrt{3}\,i$, or write $2i\sqrt{3}$, so that nobody reads it as $2\sqrt{3i}$.

Powers of the imaginary unit

Powers of $i$ repeat in a cycle of four, because $i^4 = (i^2)^2 = (-1)^2 = 1$.

To find a high power, divide the power by 4 and keep the remainder: $23 = 4 \times 5 + 3$, so $i^{23} = i^{20} \times i^3 = 1 \times (-i) = -i$. The quotient 5 does not matter, because $i^{20} = (i^4)^5 = 1$.

With $i$ available, every negative number has square roots, and calculations with $i$ follow the ordinary rules of algebra, with $i^2$ replaced by $-1$ wherever it appears.

Cartesian form and the conjugate

Adding a real number to a real multiple of $i$ gives a complex number, and every complex number is described by two real numbers: its real part and its imaginary part.

Every real number is a complex number whose imaginary part is 0, so $\mathbb{R} \subset \mathbb{C}$. A number such as $5i$, whose real part is 0 and whose imaginary part is not, is called purely imaginary. So $z$ is real exactly when $\mathrm{Im}(z) = 0$, and purely imaginary when $\mathrm{Re}(z) = 0$ and $\mathrm{Im}(z) \ne 0$. Exam questions write complex numbers either as $x + iy$ or as $a + bi$; both mean the same thing.

When a complex number is zero

A complex number is zero only when both of its parts are zero. Try to break this rule and see what goes wrong: suppose $x + iy = 0$, with $x$ and $y$ real and $y \ne 0$. Then $iy = -x$, so $i = -\dfrac{x}{y}$, which is a real number. But no real number squares to $-1$, so this is impossible. Therefore $y = 0$, and then $x = 0$ as well.

The condition that $x$ and $y$ are real matters. For example, $1 + i \times i = 1 - 1 = 0$, yet neither 1 nor the "imaginary part" $i$ is zero, because $i$ is not a real number.

The complex conjugate

Changing the sign of the imaginary part gives a partner number that you will use in almost every complex calculation.

To use any of these words, first get the number into cartesian form: then the real part, the imaginary part and the conjugate can all be read straight off.

Worked example 2 changes one thing: the parts now contain an unknown real number, and you are told that the whole complex number is zero. That matters because one complex condition becomes two real equations, and the answer must satisfy both.

In the next lesson the same idea, applied to two complex numbers that are equal, is how you solve equations in $z$ and $z^*$.

Modulus and argument

A complex number $x + iy$ is fixed by two real numbers, so you can draw it as the point $(x, y)$, and that picture gives every complex number a size and a direction.

The picture has a surprising history. In 1797 the Norwegian-born surveyor Caspar Wessel, who worked on the mapping of Denmark, presented a paper to the Royal Danish Academy of Sciences on describing directions with numbers. He drew complex numbers as arrows from a fixed point and calculated with their lengths and angles. Almost nobody noticed, and the paper was only rediscovered about a hundred years later. By then the same picture carried the name of Jean-Robert Argand, who published it independently in 1806.

Take $z = 3 + 4i$. Its point is 3 units to the right of the origin $O$ and 4 units up. By Pythagoras its distance from $O$ is $\sqrt{3^2 + 4^2} = 5$, and the line from $O$ to $z$ makes an angle $\theta$ with the positive real axis.

[DIAGRAM: asset_name: The complex number 3 + 4i as a point; asset_slug: h245_4_02_complex_number_point; caption: The point z = 3 + 4i is 5 units from O at angle theta to the positive real axis, and its conjugate 3 - 4i is its reflection in the real axis; description: Axes labelled Re and Im with origin O. A blue line runs from O to the point z = 3 + 4i and is labelled r = |z| = 5. Dashed green lines mark x = 3 and y = 4, and a blue arc at O marks the angle theta. A clay line runs from O to the conjugate z* = 3 - 4i, with the angle -theta marked below the real axis. A side panel gives Re(z) = 3, Im(z) = 4 and |z| = root(3 squared + 4 squared) = 5 as the distance from O by Pythagoras, and says arg z = theta is measured anticlockwise from the positive real axis in radians. A note says z* is the reflection of z in the real axis, with |z*| = 5 and arg z* = -theta]

The distance and the angle have names.

For a real number the modulus is the absolute value you already know, and only $z = 0$ has modulus 0.

For a number on one of the axes you can read the argument straight from the picture.

For other numbers the quadrant tells you roughly where the argument is. A number with positive real and imaginary parts, such as $3 + 4i$, has an argument between 0 and $\frac{\pi}{2}$. A number with negative real part and positive imaginary part has an argument between $\frac{\pi}{2}$ and $\pi$. Working out the exact argument of a number such as $-5 + 12i$ is part of converting between forms, which has its own lesson, as does drawing Argand diagrams.

One point has many arguments. A further full turn of $2\pi$ brings you back to the same direction, so $i$ also has arguments $\frac{5\pi}{2}$ and $-\frac{3\pi}{2}$. The last part of this lesson shows how to choose one.

The conjugate $z^* = x - iy$ is the point $(x, -y)$, the reflection of $z$ in the real axis. Reflection keeps the distance from $O$ and reverses the angle, so $|z^*| = |z|$, and if $z$ has argument $\theta$ then $z^*$ has argument $-\theta$. In the picture, $z^* = 3 - 4i$ also has modulus 5.

Modulus-argument form

Because the modulus and the argument fix a point, you can write any complex number using them instead of its real and imaginary parts.

Look again at the point $z = x + iy$ at distance $r$ from $O$ with argument $\theta$. The right-angled triangle in the picture has hypotenuse $r$, so its horizontal side is $x = r\cos\theta$ and its vertical side is $y = r\sin\theta$. These are the definitions of cosine and sine on a circle of radius $r$, so they hold in every quadrant, not just the first. Substituting,

A number is in modulus-argument form only when it matches this pattern exactly: a number that is not negative in front, then $\cos\theta + i\sin\theta$ with the cosine first, a plus sign, and the same angle in both places. Then you can read off the modulus and an argument with no working.

Each pair of identities repairs one of the faulty forms without changing the value of the number.

Worked example 4 changes one thing: the sine and cosine are the wrong way round. That matters because examiners report candidates treating $\sin\theta + i\cos\theta$ as if it were $\cos\theta + i\sin\theta$, which gives the wrong argument.

So every faulty form has a repair: move a minus sign into the angle, swap sine and cosine with the complementary angle, or absorb a negative number in front with a half turn.

The principal argument

Every non-zero complex number has infinitely many arguments, so to give each number one standard argument you agree on an interval one full turn wide.

Look at the number $z$ with modulus 2 in the picture. Turning clockwise from the positive real axis, you reach it after $\frac{2\pi}{3}$, so $-\frac{2\pi}{3}$ is an argument. Turning anticlockwise, you reach it after $\frac{4\pi}{3}$. The two angles differ by $2\pi$, and adding or subtracting further full turns gives more arguments still, such as $\frac{10\pi}{3}$ and $-\frac{8\pi}{3}$.

[DIAGRAM: asset_name: One point with two principal arguments; asset_slug: h245_4_02_principal_argument; caption: The point z with modulus 2 is reached by turning 2pi/3 clockwise or 4pi/3 anticlockwise, so its principal argument is -2pi/3 or 4pi/3 depending on the interval used; description: Axes labelled Re and Im with origin O. A blue line of length 2 runs from O to the point z in the third quadrant. A green arc with an arrow turns clockwise from the positive real axis to z and is labelled -2pi/3. A purple arc with an arrow turns anticlockwise from the positive real axis to z and is labelled 4pi/3. Side text says that -2pi/3 is the argument in the interval greater than -pi and up to pi, that 4pi/3 is the argument in the interval from 0 up to but not including 2pi, and that the two values differ by 2pi, one full turn, as do all the other arguments of z]

From now on $\arg z$ means the principal argument. To find it, add or subtract $2\pi$ as many times as you need to land in the interval; each $2\pi$ is a full turn, so the point does not move. Here is all the notation of this lesson for one number, $z = -2i$:

Worked example 6 changes one thing: the number in front is negative, so the number is not yet in modulus-argument form. That matters because the half turn adds $\pi$ to the argument, which often pushes it out of the interval the question fixes.

The conjugate is the reflection in the real axis, so in $(-\pi, \pi]$ its principal argument is $\arg z^* = -\arg z$. In $[0, 2\pi)$ the angle $-\theta$ is outside the interval, so add a full turn: $\arg z^* = 2\pi - \arg z$ when $z$ is not real.

The interval matters in the exam as much as the angle does.

Where the quarter turn lives

You can now answer the question from the start of the lesson. The number $i$ is not on the real line: it is the point one unit above $O$, with modulus 1 and argument $\frac{\pi}{2}$, so

The powers of $i$ from the first part sit where the circle of radius 1 crosses the axes, a quarter turn apart.

[DIAGRAM: asset_name: Powers of i as quarter turns; asset_slug: h245_4_02_powers_of_i; caption: The numbers 1, i, -1 and -i lie on the circle of radius 1 a quarter turn apart, and each multiplication by i moves one to the next; description: Axes labelled Re and Im with origin O and a dashed green circle of radius 1. Blue points mark 1 with argument 0, i with argument pi/2, i squared = -1 with argument pi and i cubed = -i with argument -pi/2, all with modulus 1. Four clay arrows run anticlockwise round the circle from each point to the next, each labelled times i. A side panel says that each multiplication by i keeps the modulus 1 and increases the argument by pi/2, a quarter turn, that two quarter turns give i squared = -1, a half turn, and that four quarter turns give i to the power 4 = 1, back to the start]

Each multiplication by $i$ keeps the modulus 1 and adds $\frac{\pi}{2}$ to the argument, a quarter turn about $O$. Two quarter turns give $i^2 = -1 = [1, \pi]$, the half turn you started with, and four give $i^4 = 1$, which is why the powers of $i$ repeat every four. In the lesson on working in modulus-argument form you will prove that multiplying any complex number by $i$ turns it through a quarter turn.

Quick Recap

• $i^2 = -1$, so $\sqrt{-k} = i\sqrt{k}$ for $k > 0$, and the powers of $i$ repeat every four.
• In cartesian form $z = x + iy$, $\mathrm{Re}(z) = x$ and $\mathrm{Im}(z) = y$ are both real, $z^* = x - iy$, and $z = 0$ only when $x = 0$ and $y = 0$.
• $|z| = \sqrt{x^2 + y^2}$ is the distance of $z$ from $O$; $\arg z$ is the angle from the positive real axis, in radians; 0 has no argument.
• Modulus-argument form is $z = r(\cos\theta + i\sin\theta) = [r, \theta]$ with $r = |z| \ge 0$; repair a minus sign, a swapped sine and cosine, or a negative number in front before you read off $r$ and $\theta$.
• The principal argument lies in $(-\pi, \pi]$ or $[0, 2\pi)$; add or subtract $2\pi$ to reach the interval the question asks for.

Where the quarter turn lives: $i = \left[1, \frac{\pi}{2}\right]$ is one unit above $O$, and each multiplication by $i$ turns through $\frac{\pi}{2}$, so two quarter turns give $i^2 = [1, \pi] = -1$.