Numbers & Functions · 8 Numbers That Shouldn't Exist

8.9 The harvest: every polynomial gets all its roots

One more debt. In Chapter 4 we noticed a degree-n polynomial has at most n roots, and I promised complex numbers would sharpen “at most” into “exactly.” Here it is, one of the great theorems of mathematics, the Fundamental Theorem of Algebra: Every polynomial of degree n has exactly n roots, once you allow complex numbers (counting repeats).

Every quadratic has exactly two roots; every cubic exactly three; every degree100 polynomial exactly one hundred. No polynomial is ever “missing” roots — the missing ones were simply complex all along, waiting off the real line for us to build the plane they live in. The real numbers were a stage too small for algebra; the complex plane is exactly the right size, and nothing bigger is ever needed.

8.10 Where we are now — and where you go next

Look at what just happened in a single chapter. We invented one number, i, with one rule, i2 = −1 — and out of that one bold move, everything we had planted came due at once: • the negative discriminant resolved, and every quadratic gained its roots; • complex numbers turned out to be points in a plane, and multiplication turned out to be rotation; Numbers That Shouldn’t Exist • the growth machine and the circling machine were revealed as the same machine, bridged by i, in Euler’s formula; • the trig addition formulas collapsed out of it in three lines; • and every polynomial got its full set of roots.

That is what this whole book was built to do: not to teach you isolated tricks, but to plant ideas early and let them grow into each other, until a single new number could tie the entire structure together. That habit — expecting the pieces to connect, looking for the deep structure under the surface — is the real thing you’ve been learning.

You started this book able to count. You end it holding the complete number system, the five great families of functions, the operations that reshape and combine them, and the most beautiful equation in mathematics — and, more importantly, knowing why each one had to exist. There’s one machine we kept gesturing at and never built: the one that measures how fast a function is changing, and the one that adds up its accumulated effect. That machine is called calculus, and it’s the next book. Let’s keep going.

Things worth thinking about 1. We said multiplying by i rotates a point 90◦. We checked it on 3. Try it on 1 + i: compute (1 + i) × i and plot both. Did it turn a quarter circle? (Remember i2 = −1.)

2. Euler’s identity says eiπ = −1. Using the walker picture, explain why in one sentence: where on the unit circle does the walker stand after going angle π, and what’s its complex address there?

3. The Fundamental Theorem says every degree-n polynomial has exactly n roots over the complex numbers. x2 + 1 has degree 2 — so it must have 2 roots. Find them. (What squares to −1? There are two answers.) Where do they sit?

4. (Looking back over the whole book.) We extended our numbers five times: naturals →integers →rationals →reals →complex. List which operation forced each extension. Then the deep question: is there any operation left that breaks over the complex numbers? (The answer, for ordinary algebra, is no — the complex numbers are “complete” in a way the reals never were. That completeness is why the journey ends here.)

Problems to work Problem 1 — Arithmetic with i.

Compute each, writing your answer as a + bi. The only new rule is i2 = −1. (a) (2 + 3i) + (4 −i); (b) (2 + 3i) −(4 −i); (c) i3 (hint: i2 · i); (d) i4 (do powers of i cycle?); (e) (1 + i)(1 −i) (watch the i2 — the answer is purely real). Problem 2 — Solve a quadratic with no real roots.

Use the quadratic formula to find both complex roots of x2 + 4x + 13 = 0. Write them as a+bi and notice they’re a matched pair — same real part, opposite imaginary parts (“complex conjugates”).

Problem 3 — Multiplication is rotation.

Start with 2 (the point 2 across). Multiply by i, then by i again, then again, then again — four times — writing and plotting each result. Confirm you’ve walked all the way around a circle and returned to 2.

Problem 4 — Use Euler’s formula.

Euler’s formula says eiθ = cos θ + i sin θ. Use it to find eiπ/2 (a quarter turn). What single number do you get, and where does it sit? (Check against the rotation picture — it should be straight up.)

Problem 5 — De Moivre at work.

Use (cos θ + i sin θ)n = cos(nθ) + i sin(nθ) to compute (cos 30◦+ i sin 30◦)3 without multiplying the long way. (Just multiply the angle by 3. What familiar point do you land on?)

⋆Teach it — explain it to someone Problem 6 — Teach it (the whole journey). This is the capstone teaching task for the entire book. Explain to someone — ideally someone who started it skeptical that “imaginary” numbers could be real — why i is no more imaginary than −2 is. Walk them through the pattern: how each kind of number (negatives, fractions, irrationals, and now i) was invented to fix an operation that had broken, and how i in particular turned out to secretly be rotation. If you can make someone see that i earned its place exactly the way negative numbers did — by making the mathematics work — then you’ve understood not just complex numbers, but the single idea this whole book was built around.

Numbers That Shouldn’t Exist Going further Problem 7 — Find the roots of unity yourself.

Following the chapter’s method for the cube roots, find the four fourth roots of unity (x4 = 1). Write x = cos θ + i sin θ, use De Moivre to get x4 = cos(4θ) + i sin(4θ), set it equal to 1 (angle 0, 2π, 4π, 6π), and solve for the four angles. Plot them. What shape? (Check against ordinary algebra: x4−1 = (x2−1)(x2+1) = 0 gives 1, −1, i, −i. Do your angles match?)

Problem 8 — [challenge] The conjugate trick.

The conjugate of a + bi is a −bi (flip the sign of the imaginary part). (a) Show that multiplying any complex number by its conjugate, (a + bi)(a −bi), always gives a real number — and find it (a difference of squares in disguise!). (b) Use this to divide: 1 2+i. Multiply top and bottom by the conjugate 2 −i, and write the answer as a + bi. (This is exactly how you divide complex numbers — by making the bottom real.)

Problem 9 — [challenge] A square root of i.

Everyone knows i is a square root of −1. But does i itself have a square root — a complex number z with z2 = i? Use polar thinking: i sits at distance 1, angle 90◦. A square root should sit at distance 1 and half the angle (so that squaring — which doubles the angle — gets back to 90◦). (a) What angle is that? (b) Write z = cos θ + i sin θ at that angle, using your special-angle values. (c) Square it by hand to confirm you get i. (You’ve just taken a square root of a square root of −1 — and it was easy, because in polar form roots are just dividing angles.) Problem 10 — [challenge] Why exactly n roots, seen on the circle. The equation x6 = 1 has, by the Fundamental Theorem, exactly six roots. (a) Using De Moivre, find all six angles (set 6θ equal to 0, 2π, 4π, . . . , 10π). (b) Plot them — what regular shape do they make? (c) Explain in one sentence why there are exactly six and not more: what happens to the angle if you try 12π? (You’ll find it lands back on a root you already have — the circle only has room for n before it repeats. That’s the geometric reason “degree n” means “exactly n roots.”)

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