Back to the on-screen lesson ·
The modulus as a length and as $z\bar z$, its multiplicativity, the triangle inequality that replaces addition, and the circles, discs and annuli a distance condition describes.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to compute a modulus and a squared modulus, use the fact that the modulus multiplies to find the modulus of a product, a power or a quotient without expanding anything, read a condition on a distance as a circle, a disc or an annulus, and state the triangle inequality together with the case in which it becomes an equality.
Complex arithmetic in the form $x + iy$, and the conjugate $\bar z = x - iy$ with the fact that $z\bar z$ is real and never negative. That product is about to be given a name and a picture, and it becomes the measuring instrument of the whole course.
The modulus $|z| = \sqrt{x^{2} + y^{2}}$ is the distance from $z$ to the origin; $|z - w|$ is the distance between $z$ and $w$. A disc is the set of points within a fixed distance of a centre — open when the boundary circle is left out, closed when it is included. A set is bounded when one disc contains all of it. The triangle inequality is the statement $|z + w| \le |z| + |w|$.
Definition and the identity that drives everything. For $z = x + iy$,
$$|z| = \sqrt{x^{2} + y^{2}}, \qquad |z|^{2} = z\bar z.$$
The second form is the useful one. It turns a statement about lengths into a statement about products, and products are what this arithmetic is good at.
The modulus multiplies. $|zw| = |z||w|$, hence $|z^{n}| = |z|^{n}$ and $\left|\dfrac{z}{w}\right| = \dfrac{|z|}{|w|}$. The one-line proof is $|zw|^{2} = zw\overline{zw} = z\bar z\, w\bar w = |z|^{2}|w|^{2}$.
The modulus does not add. In its place stands the triangle inequality
$$|z + w| \le |z| + |w|,$$
with equality exactly when $z$ and $w$ point in the same direction from the origin, and its useful companion $\bigl||z| - |w|\bigr| \le |z - w|$. Every estimate in this course — the bound on a contour integral, the Cauchy estimate, the proof of Liouville's theorem — is the triangle inequality applied carefully.
Distance, and the regions it describes. $|z - w|$ is the distance between two points, so $|z - c| = \rho$ is the circle of radius $\rho$ about $c$, $|z - c| < \rho$ the open disc inside it, and $\rho_1 < |z - c| < \rho_2$ an annulus. Those three regions are where the whole subject happens: a power series converges on a disc and a Laurent series on an annulus.
Another way: picture
Put $z$ and $w$ at the ends of two arrows from the origin and complete the triangle with $z + w$. The inequality is the ordinary fact that one side of a triangle is no longer than the other two together, and equality is the degenerate triangle where the two arrows lie along the same ray. The modulus is the only place in this subject where a complex statement is a real one about lengths, which is exactly why every estimate goes through it.
Another way: steps
Almost every calculation with a modulus is done on $|z|^{2}$ rather than on $|z|$, and it is worth being deliberate about why.
A square root is an obstacle to algebra: it does not expand, it does not distribute, and over the complex numbers it does not even have a single value. The square has none of those problems. $|z|^{2} = z\bar z$ is a product, so it expands, and every identity about moduli — multiplicativity, the parallelogram law, the triangle inequality itself — is proved by squaring both sides and expanding.
And since $|z| \ge 0$ always, squaring loses nothing: $|z| \le |w|$ if and only if $|z|^{2} \le |w|^{2}$. The habit to build is to reach for $z\bar z$ the moment a modulus appears in an equation, and to keep the square root for the last line.
The problems in this lesson draw their numbers from Pythagorean triples, so the square root comes out whole. That is a convenience of the exercises and not of the subject: $|1 + i| = \sqrt{2}$ is entirely typical, and it is one reason a question here asks for the modulus squared, or for a multiplier, rather than for a decimal nobody can finish writing.
The error that survives longest is treating $|z + w|$ as though it were $|z| + |w|$. It is not, and the counterexample is as small as they come: $z = 1$ and $w = -1$ give $|z + w| = 0$ while $|z| + |w| = 2$.
What makes the habit persistent is that multiplicativity is exact — $|zw|$ really is $|z||w|$, with no inequality about it — so the mind generalises from a true rule to a false one. The two operations behave differently because multiplication is a rotation and a stretch, which composes lengths cleanly, while addition moves a point sideways by an amount that depends on the angle between the two arrows.
A second, quieter error is writing $|z| < |w|$ as though it compared the numbers rather than their distances from the origin. It does not: $i$ and $-i$ and $1$ all have modulus $1$, and none of them is larger than another. The modulus is the only thing about a complex number that can be compared at all.
Find $|(3 + 4i)^{5}|$. The modulus of $3 + 4i$ is $\sqrt{9 + 16} = 5$.
One square root, once.
The modulus is multiplicative, so the answer is $5^{5} = 3125$.
The bracket was never expanded.
Which points satisfy $1 < |z - 2i| \le 3$? Read $|z - 2i|$ as the distance from $z$ to the point $2i$.
A modulus is a distance.
So the condition is: further than $1$ from $2i$, and no further than $3$. That is the annulus between the two circles about $2i$, with the inner circle excluded and the outer one included.
Strict and non-strict inequalities decide the boundaries.
The difference is $(5 - 1) + (2 + 1)i = 4 + 3i$.
Distance is the modulus of the difference.
Its modulus is $\sqrt{16 + 9} = 5$.
Let $z = 7 + 24i$. Fill in the table.
| Value | |
|---|---|
| The real part | |
| The imaginary part | |
| The square of the modulus | |
| The modulus | |
| The modulus of the square | |
| The modulus of the cube |
What is $|(24 + 7i)^{4}|$?
Answer:
The open disc $|z + 4| < 1$ meets the real axis in a set of real numbers $x$. Give that set.
This task has no paper form; do it on a device.
Let $z = 3 + 6i$. Plot $z$, then $\bar z$, then $iz$, reading the first coordinate as the real part and the second as the imaginary part.
Plot your answer on the grid:
Select every statement that is true.
This task has no paper form; do it on a device.
For all complex $z$ and $w$, $|z + w|^{2} + |z - w|^{2} = k\left(|z|^{2} + |w|^{2}\right)$ for one fixed number $k$. What is $k$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Let $z = 9 + 12i$. Fill in the table.
| Value | |
|---|---|
| The real part | |
| The imaginary part | |
| The square of the modulus | |
| The modulus | |
| The modulus of the square | |
| The modulus of the cube |
You can turn a modulus into a product with a conjugate, find the modulus of a power without expanding, and read a distance condition as a region of the plane. Say in your own words why the modulus multiplies but does not add. Next: polar form, which gives the argument the same treatment.
9. Your turn: how far apart are $5 + 2i$ and $1 - i$?, step 2