One line is not enough
A real number is a single position on a number line: 3 is three steps right of 0, and −2 is two steps left.
A complex number such as 3 + 2i has two parts, a real part 3 and an imaginary part 2. A position on a line can record only one number, so no point of the line can stand for 3 + 2i.
The real numbers on a line. Each is one position: 3 is three steps right of 0 and −2 is two steps left.
Two directions: across and up
Give the two parts a direction each. Across measures the real part and up measures the imaginary part, so 3 + 2i is the point 3 across and 2 up, the point (3, 2).
The horizontal axis is the real axis and the vertical axis is the imaginary axis. The plane they make is the Argand diagram. The number line is still there: it is the real axis.
Six numbers on the Argand diagram. 3 + 2i is 3 across and 2 up. −2 + i is 2 to the left and 1 up, −1 − 3i is 1 to the left and 3 down, and 2 − 2i is 2 across and 2 down. The real number 4 sits on the real axis and 3i on the imaginary axis.
Reading a point
The signs of the parts say which quarter of the plane a point is in. −2 + i has a negative real part and a positive imaginary part, so it is up and to the left. −1 − 3i, with both parts negative, is down and to the left. 2 − 2i is down and to the right.
A real number has imaginary part 0, so 4 = 4 + 0i sits on the real axis. A number such as 3i = 0 + 3i has real part 0 and sits on the imaginary axis, 3 up from the origin. The number 0 is the origin.
A number as an arrow
A complex number can also be drawn as an arrow from the origin to its point. The arrow for 3 + 2i goes 3 across and 2 up. Drawn this way, the arithmetic of the earlier lessons becomes a set of moves on the plane.
Adding is arrows end to end. Draw 3 + 2i, then start the arrow for 1 + 4i where the first one ends. Across 3 then 1 makes 4, and up 2 then 4 makes 6, so the second arrow ends at 4 + 6i, the sum. The across steps and the up steps add separately, as the real and imaginary parts do.
The arrow for 1 + 4i starts where the arrow for 3 + 2i ends. Together they reach 4 + 6i, and the gold arrow from the origin is the sum.
The conjugate and the negative
The conjugate of 2 + 3i is 2 − 3i: the same distance across, and as far below the real axis as 2 + 3i is above it. It is the reflection of 2 + 3i in the real axis. That is why the roots of sit one above the real axis and one below it, as mirror images.
The negative, −(2 + 3i) = −2 − 3i, changes both signs. It is on the opposite side of the origin, a half turn about the origin from 2 + 3i.
The gold point, the conjugate 2 − 3i, is the reflection of 2 + 3i in the real axis. −2 − 3i, the negative, is a half turn about the origin away from 2 + 3i.
Multiplying by i is a quarter turn
. On the plane, the point (3, 2) has moved to (−2, 3): a quarter turn counterclockwise about the origin.
Multiply by i again: i(−2 + 3i) = −3 − 2i, another quarter turn. Then i(−3 − 2i) = 2 − 3i, and i(2 − 3i) = 3 + 2i, back at the start after four quarter turns. Two quarter turns make a half turn, which is multiplying by , and four make a full turn, which is multiplying by .
Each multiplication by i turns the point a quarter turn counterclockwise about the origin: 3 + 2i, then −2 + 3i, −3 − 2i and 2 − 3i, and back to 3 + 2i.
The usual mistakes
Swapping the axes. 2 + 3i is 2 across and 3 up, not 3 across and 2 up: across measures the real part.
Putting both parts on one line. 2 + 3i is not the point 5 on the real axis: the 3i needs its own direction.
Putting an imaginary number on the real axis. 3i is 3 up the imaginary axis, at (0, 3), not at (3, 0).
A floor robot
In the first application below, a robot on a floor turns a quarter turn left before each 1 m drive. With east as 1 and north as i, a quarter turn left is multiplying by i, so after n commands it faces i to the power n, and its position is the sum of its drives, arrows end to end.
Worked example: A Floor Robot That Turns a Quarter-Turn Left 2026 Times: Its Heading and Where It Ends Up
Question A floor robot starts at the origin facing east. Each command makes it turn 90° to the left and then drive 1 m forward. On an Argand diagram east is 1 and north is i, and a quarter-turn to the left is multiplication by i, so after n commands the robot faces the direction in. (a) Which way does it face after 2026 commands? (b) Where does it end up after those 2026 commands?
1.The powers of i repeat every four: i1 = i, i2 = −1, i3 = −i and i4 = 1, and then i5 = i again.
Each left turn multiplies the heading by i: east, north, west, south, and east again. 2.Divide 2026 by 4: 2026 = 4 × 506 + 2, so i2026 = (i4)506 × i2 = 1 × (−1) = −1.
Only the remainder on dividing by 4 matters: 2026 = 4 × 506 + 2. 3.(a) The robot faces the direction −1, which is west.
(a) i2026 = i2 = −1: the robot faces west. 4.The nth drive moves the robot 1 m in the direction in, so its position is i + i2 + i3 + ⋯ + i2026. Each block of four is i − 1 − i + 1 = 0: four commands bring it back to the start.
Four commands drive round a square and back to the start. 5.(b) The first 2024 = 4 × 506 drives cancel, which leaves i2025 + i2026 = i + (−1) = −1 + i. The robot ends 1 m west and 1 m north of the origin. Check: after the first two commands it is at i − 1 as well.
(b) The last two drives, i and −1, leave the robot at −1 + i.
Answer: (a) west, because i2026 = −1; (b) at −1 + i, which is 1 m west and 1 m north of the start
Common mistakes
- Working out 2026 ÷ 4 = 506.5 and trying to use the .5. What decides the power is the remainder, 2, so i2026 = i2 = −1.
- Giving i2026 = −1 as the position as well as the heading. The heading is the last power alone, but the position is the sum of every move the robot has made.
Impedances in series
In the second, each part of an AC circuit has an impedance R + iX ohms, an arrow on the Argand diagram, and impedances in series add. The total is the three arrows end to end.
Worked example: A Pump Motor on a Long Cable: The Impedance of Three Parts in Series, and the Capacitor That Cancels Its Reactance
Question In an AC circuit the impedance of each part is written Z = R + iX ohms, where R is its resistance and X is its reactance: positive for a coil and negative for a capacitor. A pump motor has impedance 5 + 9i ohms. It is fed through a cable of impedance 1 + i ohms and an ideal capacitor, which has no resistance, of impedance −2i ohms, all in series, and impedances in series add. (a) Find the total impedance of the circuit. (b) The electrician replaces the capacitor with another ideal capacitor that makes the total impedance a real number. What must the new capacitor's impedance be, and what is the total impedance then?
1.Engineers write j for the imaginary unit, because i already stands for current; here it is i throughout, with i2 = −1. The three impedances are 5 + 9i, 1 + i and 0 − 2i ohms.
Each impedance is an arrow on the Argand diagram, and a series circuit adds them head to tail. 2.Add the real parts, which are the resistances: 5 + 1 + 0 = 6 ohms.
The real parts, the resistances, add to 5 + 1 + 0 = 6 ohms. 3.Add the imaginary parts, which are the reactances: 9 + 1 − 2 = 8 ohms.
The imaginary parts, the reactances, add to 9 + 1 − 2 = 8 ohms. 4.(a) The total impedance is Z = 6 + 8i ohms.
(a) The total impedance is Z = 6 + 8i ohms. 5.(b) Without a capacitor, the motor and the cable give (5 + 1) + (9 + 1)i = 6 + 10i ohms. The new capacitor must add −10i to bring the imaginary part to 0, so its impedance is −10i ohms, and the total is then 6 + 10i − 10i = 6 ohms. Check: 5 + 1 + 0 = 6 and 9 + 1 − 10 = 0.
(b) A capacitor of −10i ohms brings the total down onto the real axis: Z = 6 ohms.
Answer: (a) 6 + 8i ohms; (b) −10i ohms, which makes the total 6 ohms
Common mistakes
- Adding every number as if it were one plain quantity, 5 + 9 + 1 + 1 − 2 = 14 ohms. Resistance and reactance are the real and imaginary parts of one complex number, and they are added separately.
- Choosing −8i for the new capacitor because the total in (a) has imaginary part 8. That total already includes the old capacitor's −2i, which is taken out, so the new capacitor must cancel 9 + 1 = 10.