Where we are on the ladder
In the last guide you saw the secret engine of complex multiplication: in polar form a number z = r e^(i theta) carries a length r and an angle theta, and multiplying two numbers multiplies the lengths and adds the angles. That single fact, plus Euler's formula e^(i theta) = cos theta + i sin theta, is everything we need today. We are going to ask one innocent-looking question — what are the solutions of z^n = 1? — and watch geometry hand us a perfect answer.
Then we close the whole rung with a change of scenery. The flat complex plane has one annoying gap: numbers can run off to infinity in a thousand directions and never come back. We will fix that by gluing on a single extra point and curling the plane up into a ball — the Riemann sphere. By the end, z^n = 1 will feel like clockwork, and infinity will be just another dot you can point at.
Solving z^n = 1 by rotation
Write a candidate solution in polar form, z = r e^(i theta), and raise it to the n-th power. By de Moivre's theorem the length and angle each just get multiplied by n, so z^n = r^n e^(i n theta). We want this to equal 1, which sits at length 1 and angle 0. Matching the lengths forces r^n = 1, and since r is a real length r must be exactly 1 — every solution lives on the unit circle, no exceptions.
The angles are where the magic happens. We need n theta to land on the angle of 1, but angle 0 and angle 2 pi and angle 4 pi all describe the same direction — angles only matter up to a full turn of 2 pi. So n theta = 2 pi k for any whole number k, which gives theta = 2 pi k / n. Stepping k = 0, 1, 2, ..., n - 1 walks the angle around in equal jumps of 2 pi / n, and once k reaches n we are back where we started. There are exactly n distinct answers.
z_k = e^(i * 2 pi k / n) = cos(2 pi k / n) + i sin(2 pi k / n), k = 0, 1, ..., n - 1
The perfect ring: roots of unity
These n numbers are the roots of unity, and as a picture they are breathtaking: n equally spaced points around the unit circle, the corners of a regular n-gon. The first one, k = 0, is always plain old 1. The second, omega = e^(2 pi i / n), is the smallest positive rotation — and every other root is just a power of it, since z_k = omega^k. One number generates the whole ring by repeated rotation.
Take n = 3 to feel it. The cube roots of 1 are 1, then e^(2 pi i / 3) = -1/2 + i (sqrt 3)/2, then e^(4 pi i / 3) = -1/2 - i (sqrt 3)/2 — an equilateral triangle with one corner nailed at 1. Two of them are stubbornly complex, which is exactly why we couldn't see them back in real-number land: on the line there is only one cube root of 1, but in the plane there are three, sitting 120 degrees apart.
Two facts fall out for free and both come from the geometry. First, the roots of unity always sum to zero (for n at least 2): they are symmetric arrows pointing out in balanced directions, so they cancel — the centroid of a regular n-gon centred at the origin is the origin. Second, to take a root of any number, not just 1, you find one n-th root of that complex number and then multiply by each root of unity in turn; the n answers are that same regular n-gon, simply rotated and rescaled to fit.
Reading off all the roots
Here is the whole recipe in one breath, for any z^n = w. It is just the rotation argument run once more, this time landing on w instead of 1.
- Write w in polar form, w = R e^(i phi), reading off its length R and one argument phi.
- Take the ordinary real n-th root of the length to fix every solution's length: each root has modulus R^(1/n).
- Divide the angle by n to get a starting angle phi / n — this is one particular root.
- Add multiples of the spacing 2 pi / n: the full set of angles is phi / n + 2 pi k / n for k = 0, 1, ..., n - 1.
Adding a point at infinity
Now the change of scenery. In the plane, what happens to z as |z| grows without bound? In real calculus you would say it 'goes to plus or minus infinity', two different escapes. But a complex z can race off in any of infinitely many directions, and there is no honest way to tell those escapes apart — they all just leave the picture. The clean fix is to declare a single new symbol, the point at infinity, written infinity, that every escaping path arrives at, regardless of direction.
The plane plus this one extra point is called the extended complex plane. To see that infinity really is just an ordinary point in disguise, picture a globe resting on the plane, touching it at the origin (the south pole). Draw a straight line from the north pole through any plane-point z; it pierces the sphere at exactly one spot. That correspondence is stereographic projection, and it is a perfect one-to-one match between the sphere-minus-north-pole and the whole flat plane.
The only spot on the sphere left without a plane-partner is the north pole itself — and that is precisely the point z would have to climb toward as |z| runs off to infinity. So we simply assign the north pole to be infinity. The plane plus its single point at infinity wraps up, with no gaps and no edges, into the Riemann sphere: a smooth ball where every point, infinity included, is utterly unremarkable.
Why the sphere is worth the trouble
Treating infinity as a real point is not just tidiness — it changes what counts as a single object. Stereographic projection has a gorgeous property: it sends every circle on the sphere to either a circle or a straight line in the plane, and a plane-line is exactly a 'circle through infinity'. So on the sphere, lines and circles stop being two different things; they are one family. This is the natural home for the Mobius transformations you will meet later, which shuffle the sphere around and treat infinity like any other point.
Looking up the ladder, this single move pays off again and again. Asking how a function behaves 'at infinity' becomes asking how it behaves at the north pole — an ordinary local question. The whole rich theory of Riemann surfaces, where multivalued functions like the logarithm get a many-storeyed home of their own, grows out of exactly this instinct: when a space has an awkward gap, do not patch over it — add the missing point and curl the space until everything is smooth.