The Complex Plane and Modulus Advanced. This lesson goes beyond core Algebra II. You can skip it.

Intuition

Take $3 + 2i$ . It is built from two ordinary real numbers, $3$ and $2$ , and a pair of real numbers is exactly what a point of a plane needs: an across-coordinate and an up-coordinate. Plot $3 + 2i$ at the point $(3, 2)$ , and in general plot $a + bi$ at $(a, b)$ . Complex numbers stop being formal expressions and become geometry you can see.

That plane is genuinely needed, not just convenient. The first lesson of this chapter built $i$ because the real number line had no room for it: every real number squares to something at least $0$ , so no point on the line can be a square root of $-1$ . If $i$ has nowhere to live on the line, neither does $3 + 2i$ , so the complex numbers need a second dimension. In that plane, addition turns into sliding and conjugation turns into a mirror reflection. Even the vague question “how big is a complex number” gets an honest answer: the distance from zero, measured by the Pythagorean theorem you have known since Pre-Algebra.

Learning goals

Identify $a + bi$ with the point $(a, b)$
Read addition as a slide completing a parallelogram
See conjugation as a mirror flip across the real axis
Measure $|a + bi|$ as $\sqrt{a^2 + b^2}$ , the distance to zero, and use $|zw| = |z|\,|w|$
Write $|z - c| = r$ for a circle of radius $r$
State the triangle inequality and when equality holds

A number with two coordinates

A complex number is written $z = a + bi$ with $a$ and $b$ real; $a$ is its real part and $b$ its imaginary part. Notice that the imaginary part is the real number $b$ , not the term $bi$ . The plan is to treat the pair $(a, b)$ as coordinates, and the plan works because the pair pins the number down with no ambiguity: two complex numbers are equal exactly when their real parts match and their imaginary parts match. So distinct numbers land on distinct pairs $(a, b)$ , and every pair of real numbers is claimed by exactly one complex number, $a + bi$ . The matching between complex numbers and points of the plane is perfect in both directions.

That perfect matching is the definition of the complex plane: draw the familiar coordinate plane, and let the point $(a, b)$ be the number $a + bi$ . The horizontal axis collects the numbers with imaginary part $0$ , which are exactly the real numbers, so it is called the real axis: your old number line, embedded unchanged inside the bigger picture. The vertical axis collects the numbers with real part $0$ , which are $0$ and the pure imaginary numbers $bi$ , so it is called the imaginary axis. The four quadrants are read the same way as for $(x, y)$ points: $3 + 2i$ sits in the first quadrant, $-2 + 3i$ in the second, and so on.

Six complex numbers plotted as points. The real part is the horizontal coordinate and the imaginary part is the vertical one, so 3 + 2i sits 3 right and 2 up. Real numbers like 2 live on the horizontal axis, and pure imaginary numbers like minus 3i live on the vertical axis.

Worked example 1 From points to numbers and back

Which complex number sits at the point $(-2, 5)$ ? The horizontal coordinate is the real part and the vertical coordinate is the imaginary part, so the number is $-2 + 5i$ .

Where does $4 - i$ sit? Its real part is $4$ and its imaginary part is $-1$ , so start at $0$ , move $4$ right and $1$ down, landing at the point $(4, -1)$ in the fourth quadrant.

Where do $-3$ and $2i$ sit? The number $-3$ is $-3 + 0i$ , so it lies on the real axis, three units left of $0$ , exactly where it has always lived on the number line. The number $2i$ is $0 + 2i$ , so it lies on the imaginary axis, two units above $0$ . The axes hold those two number sets in their places: the real numbers along the real axis, the pure imaginary numbers along the imaginary axis.

Check your understanding

Which point is the complex number $-4 + 3i$ ?

(-4, 3)

(3, -4)

(4, -3)

(-4, -3)

Addition slides points

The previous lesson established the arithmetic rule: to add complex numbers, add the parts separately,

(a + bi) + (c + di) = (a + c) + (b + d)\,i.

Read that rule with coordinate eyes. The point for the sum has horizontal coordinate $a + c$ and vertical coordinate $b + d$ . So adding the fixed number $w = c + di$ sends every point $c$ units across and $d$ units up (negative values meaning the opposite direction). Adding $w$ is a rigid slide of the entire plane, the same shift applied to every point at once; no stretching, no turning.

There is a second picture of the same fact. Draw $z$ not as a dot but as an arrow from $0$ to its point. To form $z + w$ , place a copy of $w$ ‘s arrow so its tail starts at $z$ ‘s tip; the copy’s tip lands on $z + w$ . This is tip-to-tail addition, and it is what physics and later mathematics call adding vectors. Doing it in the other order, $w$ first and then $z$ , traces the other two sides of the same figure, so both routes end at the same corner. The four points $0$ , $z$ , $w$ , and $z + w$ form a parallelogram, and the equation $z + w = w + z$ becomes something you can see.

Adding z = 3 + i and w = 1 + 2i. Solid arrows are z and w from 0; each dashed arrow is a copy of the other number placed tip-to-tail. Both routes end at the same corner z + w = 4 + 3i, so the four points close into a parallelogram.

Subtraction has its own reading. The difference $z - w$ is, by definition of subtraction, the number you add to $w$ to reach $z$ , since $w + (z - w) = z$ . As an arrow, then, $z - w$ is the trip from $w$ to $z$ , picked up and carried back so its tail sits at $0$ . That arrow is what will turn the modulus into a ruler for measuring the gap between any two complex numbers.

Worked example 2 Adding and subtracting, in coordinates and in pictures

Let $z = 2 + 3i$ and $w = 4 - i$ . Adding the parts separately,

z + w = (2 + 4) + (3 - 1)\,i = 6 + 2i,

which is the point you reach from $z$ by sliding $4$ right and $1$ down, exactly the slide $w$ performs on every point of the plane.

For the difference,

z - w = (2 - 4) + (3 + 1)\,i = -2 + 4i.

Check it against the meaning of subtraction: $w + (-2 + 4i) = (4 - 2) + (-1 + 4)i = 2 + 3i = z$ , as required. So the arrow from $w$ to $z$ runs $2$ left and $4$ up. How long is that trip? That is a distance question, and the next two sections build the tool that answers it.

Check your understanding

Let $z = -1 + 2i$ and $w = 3 + i$ . What is $z + w$ ?

2 + 3i

-4 + i

3 + 2i

2 - 3i

Check your understanding

The points $0$ , $z = 2 + 4i$ , and $w = 5 + i$ are three corners of a parallelogram built by tip-to-tail addition. Which point is the fourth corner?

7 + 5i

3 - 3i

10 + 4i

3.5 + 2.5i

Conjugation reflects across the real axis

The previous lesson introduced the conjugate $\overline{z} = a - bi$ of $z = a + bi$ as the sign flip that makes multiplication produce something real. In the plane, conjugation sends the point $(a, b)$ to the point $(a, -b)$ : the horizontal coordinate is untouched and the vertical coordinate changes sign. That is precisely a reflection across the real axis, the mirror image of the point in the horizontal line. A number in the upper half of the plane trades places with its twin in the lower half, directly above or below itself.

Two small facts fall out of the picture immediately, and the algebra confirms both. Reflecting twice puts every point back where it started, and indeed $\overline{\overline{z}} = \overline{a - bi} = a + bi = z$ . And the points that do not move are exactly the points on the mirror: solving $z = \overline{z}$ means $a + bi = a - bi$ , so $2bi = 0$ , so $b = 0$ . The numbers equal to their own conjugate are exactly the real numbers. That gives a clean geometric test for realness: a number $z$ is real precisely when its point sits on the real axis, which is precisely when conjugation fixes it.

Conjugation in the plane. The numbers z = 3 + 2i and its conjugate 3 minus 2i are mirror images across the real axis: same real part, opposite imaginary part, equal distances above and below the mirror.

Check your understanding

The number $z$ lies in Quadrant III of the complex plane. In which quadrant is $\overline{z}$ ?

A Quadrant I B Quadrant II C Quadrant III D Quadrant IV

The modulus is a distance

For a real number $x$ , the first chapter defined the absolute value $|x|$ as the distance from $x$ to $0$ on the number line. Complex numbers now live in a plane, but the same question still makes sense: how far is the point $z$ from $0$ ? That distance is called the modulus of $z$ (also, still, its absolute value), and it is written with the same bars, $|z|$ .

The modulus of z = 3 + 4i is the hypotenuse of a right triangle with legs 3 and 4, so it equals the square root of 9 + 16, which is 5.

The picture shows the whole idea. Drop straight down from $z = 3 + 4i$ to the real axis, and the segment from $0$ to $z$ becomes the hypotenuse of a right triangle with legs $3$ and $4$ . The Pythagorean theorem gives $|z| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$ . The same triangle works for any point off both axes, which is what turns the picture into a formula.

The modulus formula, from the Pythagorean theorem#

Let $z = a + bi$ with $a \ne 0$ and $b \ne 0$ , so the point sits off both axes, exactly like $3 + 4i$ above. Drop a segment from $(a, b)$ straight down (or up) to the real axis; its foot is the point $(a, 0)$ . The three points $0$ , $(a, 0)$ , and $(a, b)$ form a right triangle: the leg along the real axis has length $|a|$ , and the vertical leg has length $|b|$ . The right angle sits at the foot, because a vertical segment meets the horizontal axis squarely. The hypotenuse runs from $0$ to the point, so its length is exactly the distance we want. The Pythagorean theorem gives

|z|^2 = |a|^2 + |b|^2 = a^2 + b^2,

where the bars vanish because squaring erases signs. Distances are never negative, so taking the non-negative square root of both sides yields the formula

|a + bi| = \sqrt{a^2 + b^2}.

If $a = 0$ or $b = 0$ there is no triangle, but the formula still tells the truth: a point on an axis sits $|a|$ or $|b|$ from $0$ , exactly what the formula returns once the other coordinate squares to $0$ . So the formula computes the distance from $0$ for every complex number without exception.

For a real number $z = a$ , the formula gives $|a + 0i| = \sqrt{a^2} = |a|$ : on the real axis, the modulus is the old absolute value. The bars are not being recycled for something new; the same concept, distance from $0$ , has simply grown to cover the whole plane.

Three properties come straight from the geometry, and each is one line of algebra. First, $|z| \ge 0$ always, and $|z| = 0$ exactly when $z = 0$ . A sum of two real squares $a^2 + b^2$ is zero only when $a = b = 0$ , matching the fact that the only point at distance $0$ from the origin is the origin itself. Second, $|\overline{z}| = |z|$ : reflection across the real axis does not change how far a point is from $0$ , and indeed $\sqrt{a^2 + (-b)^2} = \sqrt{a^2 + b^2}$ . Third, $|-z| = |z|$ : the point $(-a, -b)$ sits on the exact opposite side of $0$ at the same distance. One warning while the bars are fresh: the symbols $<$ and $>$ do not compare complex numbers at all, so a statement like $2 + 3i < 3 + i$ is meaningless. What you can compare are the moduli themselves, since $|2 + 3i|$ and $|3 + i|$ are real numbers. Comparing them settles only which point sits farther from $0$ , nothing more: two different complex numbers can share the same modulus, so an equal modulus never means an equal number.

Worked example 3 Three moduli, including the sign traps

Compute $|-5 + 12i|$ . The real part is $-5$ , and the squaring step erases the sign:

|-5 + 12i| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13.

Compute $|3i|$ . Here $a = 0$ and $b = 3$ , so $|3i| = \sqrt{0^2 + 3^2} = 3$ , which matches its position three units up the imaginary axis.

Compute $|-6|$ . As a complex number, $-6 = -6 + 0i$ , so the formula gives $\sqrt{(-6)^2} = 6$ , exactly what the old absolute value said. Notice that no minus sign anywhere can survive the squaring: a modulus is never negative, and $|a - bi| = |a + bi|$ for every $a$ and $b$ .

Check your understanding

What is $|8 - 6i|$ ?

2

14

10

100

The conjugate identity

The modulus and the conjugate look like unrelated gadgets, one a distance and the other a reflection. They are joined by a single identity that powers most complex-number computations from here through the rest of the subject.

The identity $z\,\overline{z} = |z|^2$ #

The previous lesson already proved that multiplying $z = a + bi$ by its conjugate erases $i$ entirely,

z\,\overline{z} = a^2 + b^2,

by expanding the product and watching the cross terms cancel. That real number is exactly the square of the modulus formula above, so for every complex number $z$ ,

z\,\overline{z} = |z|^2.

Pause on what the identity says. The left side is a product of two complex numbers; the right side is a real number that is never negative. Multiplying any number by its conjugate always lands you on the non-negative real axis, and it lands you at a geometrically meaningful spot: the squared distance from $0$ . This is the secret behind the division method of the previous lesson. Multiplying a denominator $w$ by $\overline{w}$ produced the real number $w\overline{w}$ , and now you know that mystery real number by name: it is $|w|^2$ .

The identity also unlocks the most useful multiplication fact about the modulus: for any two complex numbers,

|zw| = |z|\,|w|.

In words, the length of a product is the product of the lengths. Here is why. Conjugating a product multiplies the conjugates, $\overline{zw} = \overline{z}\,\overline{w}$ , so squaring $|zw|$ and applying the conjugate identity three times gives

|zw|^2 = (zw)\,\overline{zw} = (zw)\big(\overline{z}\,\overline{w}\big) = \big(z\,\overline{z}\big)\big(w\,\overline{w}\big) = |z|^2\,|w|^2,

using that complex multiplication can be reordered (commutativity) and regrouped (associativity) however is convenient. Both $|zw|$ and $|z|\,|w|$ are non-negative real numbers, and non-negative numbers with equal squares are equal, so $|zw| = |z|\,|w|$ .

This is a genuine shortcut: it finds the modulus of a complicated product without ever expanding it. It is also a genuine surprise, since nothing about the multiplication rule $(ac - bd) + (ad + bc)i$ makes it obvious at a glance.

Worked example 4 A product's modulus, the fast way and the slow way

Find $|(3 + 4i)(5 - 12i)|$ .

The fast way uses multiplicativity. Each factor’s modulus is a Pythagorean computation:

|3 + 4i| = \sqrt{9 + 16} = 5, \qquad |5 - 12i| = \sqrt{25 + 144} = 13,

so $|(3 + 4i)(5 - 12i)| = 5 \times 13 = 65$ , and no multiplication of complex numbers ever happened.

The slow way checks it. Expanding, $(3 + 4i)(5 - 12i) = 15 - 36i + 20i - 48i^2 = 63 - 16i$ , and

|63 - 16i| = \sqrt{63^2 + 16^2} = \sqrt{3969 + 256} = \sqrt{4225} = 65.

The two answers agree, as the theorem promised. On a product of three or four factors the fast way stays just as short, while expanding the whole thing out grows painful fast.

Check your understanding

Given $|z| = 2$ and $|w| = 6$ , what is $|z^2 w|$ ?

24

12

8

144

Distance between points, and circles

The subtraction picture now pays off. The arrow from $w$ to $z$ is the number $z - w$ , so the distance between two complex numbers is the length of that arrow:

\text{distance from } w \text{ to } z \;=\; |z - w|.

The coordinates confirm that this is the honest geometric distance. With $z = a + bi$ and $w = c + di$ , the difference is $z - w = (a - c) + (b - d)\,i$ , so

|z - w| = \sqrt{(a - c)^2 + (b - d)^2},

which is the Pythagorean theorem applied to the right triangle whose legs are the horizontal gap $|a - c|$ and the vertical gap $|b - d|$ between the two points. Distance from $0$ was the special case $w = 0$ all along.

Worked example 5 How far apart are two complex numbers?

Find the distance between $1 + 5i$ and $4 + i$ .

Subtract to get the arrow between them (either direction works, since $|{-u}| = |u|$ ):

(1 + 5i) - (4 + i) = -3 + 4i, \qquad |{-3 + 4i}| = \sqrt{9 + 16} = 5.

The two points are exactly $5$ apart. One more question with the same numbers: does $4 + i$ lie inside the circle of radius $6$ centered at $1 + 5i$ ? Its distance to the center is $5$ , and $5 < 6$ , so yes, strictly inside.

Once distance is an equation, whole shapes become equations. Fix a center $c$ and a radius $r > 0$ . The points $z$ with

|z - c| = r

are exactly the points at distance $r$ from $c$ : the circle of radius $r$ centered at $c$ . The simplest case is $|z| = r$ , the circle of radius $r$ around $0$ . The inequality $|z - c| < r$ describes the inside of the circle, and $|z - c| > r$ the outside. This one idea turns modulus equations into pictures. To understand the set of solutions of $|z - (1 + i)| = 2$ , you do not solve anything; you read off the center $1 + i$ and the radius $2$ and draw.

The solution set of the equation with modulus of z minus (1 + i) equal to 2 is the circle of radius 2 centered at 1 + i. The marked point 3 + i lies on it because its distance to the center is exactly 2.

Worked example 6 Where a circle meets a horizontal line

Find every complex number with modulus $5$ and imaginary part $3$ .

The conditions say $z = a + 3i$ for some real $a$ , with $|z| = 5$ . Apply the modulus formula and solve:

\sqrt{a^2 + 3^2} = 5 \;\Longrightarrow\; a^2 + 9 = 25 \;\Longrightarrow\; a^2 = 16 \;\Longrightarrow\; a = 4 \ \text{ or } \ a = -4.

So there are exactly two such numbers, $z = 4 + 3i$ and $z = -4 + 3i$ . The picture explains the count: the numbers with modulus $5$ form the circle of radius $5$ around $0$ , and the numbers with imaginary part $3$ form a horizontal line. A horizontal line at height $3$ cuts that radius- $5$ circle in two points, mirror images of each other across the imaginary axis.

The numbers with modulus 5 lie on the circle of radius 5 around 0. The numbers with imaginary part 3 lie on the horizontal line Im(z) = 3. The line cuts the circle at 4 + 3i and minus 4 + 3i, the two answers above.

Check your understanding

What are the center and radius of the circle $|z - (-2 + 3i)| = 4$ ?

A Center

-2 + 3i

, radius

4

B Center

2 - 3i

, radius

4

C Center

-2 + 3i

, radius

16

D Center

4

, radius

-2 + 3i

Check your understanding

Write the equation of the circle with center $3 - i$ and radius $5$ .

|z - (3 - i)| = 5

|z + (3 - i)| = 5

|z - (3 - i)| = 25

|z| - |3 - i| = 5

The triangle inequality

Walk from $0$ to $z$ , then from $z$ onward to $z + w$ . The first leg has length $|z|$ , and the second leg is a translated copy of $w$ ‘s arrow, so it has length $|w|$ . The straight-line distance from $0$ to the final point is $|z + w|$ . A detour through $z$ can never be shorter than the straight route, so the geometry announces the triangle inequality:

|z + w| \;\le\; |z| + |w|.

The name is apt: $0$ , $z$ , and $z + w$ are the corners of a triangle whose side lengths are $|z|$ , $|w|$ , and $|z + w|$ . And the inequality says one side of a triangle never exceeds the other two sides combined. Geometry makes it believable, and the conjugate identity turns it into a full algebraic proof.

Equality happens exactly when the detour is no detour at all: the arrows for $z$ and $w$ point in the same direction (or one of them is $0$ ), so the two-leg route is already a straight line. Any other pair of directions makes the direct route strictly shorter, so $|z + w| < |z| + |w|$ .

The inequality also has a useful reverse reading. Since $z = (z + w) + (-w)$ , applying the triangle inequality to the right side gives $|z| \le |z + w| + |w|$ , so $|z + w| \ge |z| - |w|$ , and swapping the roles of $z$ and $w$ gives $|z + w| \ge |w| - |z|$ as well. A sum can never be shorter than the gap between the two lengths. Together the bounds pin $|z + w|$ into a window. For example, if $|z| = 3$ and $|w| = 4$ , then $|z + w|$ can be as large as $7$ , as small as $1$ , and never anything outside $[1, 7]$ .

Check your understanding

For which pair is $|z + w| = |z| + |w|$ ?

z = 2 + i

w = 4 + 2i

z = 3

w = 3i

z = 1 + i

w = -1 - i

z = 1 + 2i

w = 2 + i

Common mistakes

Swapping the coordinates. The number $2 + 5i$ sits at $(2, 5)$ : real part across, imaginary part up. Plotting it at $(5, 2)$ draws a different number, $5 + 2i$ .
Calling $bi$ the imaginary part. The imaginary part of $a + bi$ is the real number $b$ . The imaginary part of $7 - 3i$ is $-3$ , not $-3i$ ; coordinates are always real numbers.
Shortcutting the modulus. $|3 + 4i|$ is not $3 + 4 = 7$ and not $9 + 16 = 25$ ; it is $\sqrt{9 + 16} = 5$ . Square the parts, add, then take the root, always in that order.
Letting a sign survive the squaring. $|-5 + 12i| = \sqrt{25 + 144} = 13$ , because $(-5)^2 = 25$ . Minus signs never make a modulus smaller, and $|a - bi| = |a + bi|$ always.
Confusing $z^2$ with $z\overline{z} = |z|^2$ . They differ unless $z$ is real: $(3 + 4i)^2 = -7 + 24i$ , a complex number, while $(3 + 4i)(3 - 4i) = 25$ , a non-negative real. Only the conjugate product measures length.
Writing $z < w$ for complex numbers. The symbols $<$ and $>$ do not compare complex numbers; $2 + 3i < 3 + i$ is meaningless. You can compare the real numbers $|z|$ and $|w|$ , but equal moduli never mean equal numbers: $3 + 5i$ and $5 - 3i$ both have modulus $\sqrt{34}$ .
Replacing $|z - w|$ by $|z| - |w|$ . The distance between points is $|z - w|$ . For $z = 3$ and $w = 3i$ the distance is $|3 - 3i| = 3\sqrt{2}$ , while $|z| - |w| = 0$ ; the two are not even close.
Promoting the triangle inequality to an equation. $|z + w| = |z| + |w|$ only when the arrows point the same way, or one of $z$ , $w$ is $0$ . In general only $\le$ holds, with $|z + w|$ possibly as small as the difference of the lengths.

Key takeaways

The complex number $a + bi$ is the point $(a, b)$ : the matching is one-to-one because complex numbers are equal exactly when both parts match. The horizontal real axis is the old number line ( $b = 0$ ); the vertical imaginary axis holds $0$ and the pure imaginaries ( $a = 0$ ).
Addition is a slide: adding $c + di$ moves every point $c$ across and $d$ up, and $z + w$ completes the parallelogram with vertices $0$ , $z$ , $w$ . The difference $z - w$ is the arrow from $w$ to $z$ .
Conjugation is the mirror flip across the real axis: $(a, b) \mapsto (a, -b)$ . Its fixed points are exactly the real numbers, and $\overline{\overline{z}} = z$ .
The modulus $|a + bi| = \sqrt{a^2 + b^2}$ is the distance from $z$ to $0$ , by the Pythagorean theorem. It extends the real absolute value, $|z| \ge 0$ always, $|z| = 0$ only for $z = 0$ , and $|\overline{z}| = |-z| = |z|$ .
The conjugate identity $z\,\overline{z} = |z|^2$ ties the two gadgets together, explains why dividing by $w$ uses $\overline{w}$ , and proves multiplicativity, $|zw| = |z|\,|w|$ .
Distance and circles: the distance between $z$ and $w$ is $|z - w|$ , and the equation $|z - c| = r$ (with $r > 0$ ) describes the circle of radius $r$ centered at $c$ . The inequality $|z - c| < r$ describes that circle’s inside.
Triangle inequality: $|z + w| \le |z| + |w|$ , with equality only when the arrows are aligned, or one of $z$ , $w$ is $0$ , and in reverse $|z + w|$ is at least the difference of the two moduli.

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Level 1 Foundational 12 questions Start → Level 2 Intermediate 12 questions Start → Level 3 Challenge 12 questions Start →

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

Complex numbers (Khan Academy)
The Complex Plane (Math is Fun)
2.4 Complex Numbers (OpenStax)
Complex numbers and the complex plane (Purplemath)

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Why (a, b) determines z uniquely

The Imaginary Unit and Complex Numbers lesson already proved this in full, by contradiction: assuming $a + bi = c + di$ with $b \ne d$ isolates $i$ as a ratio of two real numbers, forcing $i$ itself to be real, which is impossible since no real number squares to a negative one. That contradiction forces $b = d$ , and the equation then collapses to $a = c$ as well.

So the parts of a complex number are uniquely determined: two complex numbers are equal exactly when their real parts match and their imaginary parts match. Distinct numbers therefore land on distinct pairs $(a, b)$ , and conversely every pair $(a, b)$ of real numbers is claimed by exactly one complex number, namely $a + bi$ . The matching between complex numbers and points of the plane is perfect in both directions.

Why conjugation passes through a product

The Arithmetic of Complex Numbers lesson already proved this: writing $z = a + bi$ and $w = c + di$ , both $\overline{zw}$ and $\overline{z}\,\overline{w}$ expand to the same coordinates, $(ac - bd) - (ad + bc)\,i$ , whichever order you conjugate and multiply in. So $\overline{zw} = \overline{z}\,\overline{w}$ : conjugating a product equals multiplying the conjugates.

The full algebraic proof of the triangle inequality

The triangle inequality $|z + w| \le |z| + |w|$ #

Two preliminary facts. First, conjugation respects addition: flipping the sign of the combined imaginary part gives $\overline{z + w} = (a + c) - (b + d)i = (a - bi) + (c - di) = \overline{z} + \overline{w}$ . Second, no complex number’s real part exceeds its modulus: if $u = p + qi$ , then $p \le |p| = \sqrt{p^2} \le \sqrt{p^2 + q^2} = |u|$ , and note also that $u + \overline{u} = 2p$ , twice the real part.

Now expand the squared modulus of the sum with the conjugate identity:

\begin{aligned} |z + w|^2 &= (z + w)\,\overline{(z + w)} = (z + w)\big(\overline{z} + \overline{w}\big) \\ &= z\overline{z} + z\overline{w} + w\overline{z} + w\overline{w}. \end{aligned}

The outer terms are $|z|^2$ and $|w|^2$ . For the middle pair, observe that $w\overline{z}$ is exactly the conjugate of $z\overline{w}$ , since $\overline{z\overline{w}} = \overline{z}\,\overline{\overline{w}} = \overline{z}\,w$ , using the product rule for conjugates and the fact that conjugating twice returns $w$ . So the middle pair is a number plus its own conjugate, which is twice that number’s real part, and by the second preliminary fact,

z\overline{w} + w\overline{z} \;\le\; 2\,|z\overline{w}| = 2\,|z|\,|\overline{w}| = 2\,|z|\,|w|,

where multiplicativity and $|\overline{w}| = |w|$ finish the bound. Putting the pieces together,

|z + w|^2 \;\le\; |z|^2 + 2|z||w| + |w|^2 = \big(|z| + |w|\big)^2,

and since $|z + w|$ and $|z| + |w|$ are both non-negative, comparing squares compares the numbers: $|z + w| \le |z| + |w|$ .

This also pins down equality precisely: it happens exactly when the real part of $z\overline{w}$ equals the full modulus $|z\overline{w}|$ , which happens only when $z\overline{w}$ is a non-negative real number, the algebraic way of saying the arrows for $z$ and $w$ point in the same direction, or one of $z$ , $w$ is $0$ (in which case $z\overline{w} = 0$ , which is itself a non-negative real number).

A bit of history (optional)

For two and a half centuries a fair question had no answer. Everyone agreed on how to calculate with square roots of negatives. Nobody could say where such a number actually sat. An ordinary number has a position on the line, while these had none at all. That is most of the reason they were called impossible.

Wessel was not the first to gesture at an answer. John Wallis had sketched a related geometric picture in 1685, but it did not catch on. The first complete, correct treatment came from outside mathematics entirely. Caspar Wessel was a Norwegian surveyor who spent his life mapping coastline. A surveyor’s daily job is turning a heading and a length into a pair of numbers. So he already saw a quantity as an arrow that points, which was exactly what was needed. In 1797 his paper laying the whole idea out was presented to the Royal Danish Academy on his behalf, since he was not a member and was not there to read it himself. Measure one part along one axis and the other part at right angles, and the number becomes a place you can reach.

His account was complete and correct, and almost nobody read it. Wessel wrote in Danish, in a journal barely read abroad, and his paper was recovered only in the 1890s. So the picture carries the name of Jean-Robert Argand, a bookkeeper who printed the same idea independently in 1806.

The picture is the surveyor’s, however, and you have relied on it all lesson. The number $a + bi$ sits at the point $(a, b)$ , and its distance from zero is its modulus.