Why every polynomial has a root
Tonight's idea
Take \(p(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_0\) , complex coefficients, \(n \ge 1\). The fundamental theorem of algebra says \(p(z) = 0\) for some \(z\).
Let \(z\) walk once round the circle \(|z| = R\). The image \(p(z)\) is a closed curve. If it misses the origin, count its turns round the origin: its winding number.
Huge circle. \(z^n\) has size \(R^n\) and winds \(n\) times, its angle being \(n\) times that of \(z\) (De Moivre). The rest is at most \(|a_{n-1}|R^{n-1} + \cdots + |a_0|\), smaller than \(R^n\) for large \(R\). So \(p(z)\) is a dog on a leash shorter than the distance from \(z^n\) to the lamppost 0, and goes round \(n\) times too.
Tiny circle. As \(R \to 0\), \(p(z) \to a_0\). For \(a_0 \ne 0\) the image is a tiny loop near \(a_0\) missing 0: winding number zero. (If \(a_0 = 0\), then 0 is a root.)
Now shrink \(R\). The winding number is an integer and changes only when the curve passes through 0. It goes from \(n\) to 0, so some \(z\) has \(p(z) = 0\).
That is an argument, not a proof. A proof must define the winding number (a continuously chosen angle), prove it is an integer, and prove it survives continuous deformation avoiding 0: all real analysis.
Watch one: \(p(z) = z^3 - 1\)
On \(|z| = R\) write \(z = Re^{i\theta}\). Then \(z^3 - 1 = R^3e^{3i\theta} - 1\): a circle of radius \(R^3\) centred at \(-1\), traversed three times. It encloses the origin exactly when \(R^3 > 1\): winding number 3 for \(R > 1\), 0 for \(R \lt 1\). The switch is at \(R = 1\), where the curve passes through 0 and the three roots, the cube roots of unity, sit. A general polynomial's image wiggles, but the count is the same.
The three roots, and which start Newton's method sends where: mathdetours.com/demos/newton.html (basins view).
Your turn
- For \(p(z) = z^2 + 1\), describe the image of \(|z| = 2\) and its winding number about 0, then the same for \(|z| = \tfrac12\). For which \(R\) does the image pass through 0, and where?
- Let \(p(z) = z^3 + 2z + 5\). Why is \(|p(z) - z^3| \le 2R + 5\) on \(|z| = R\)? Find the smallest whole number \(R\) with \(2R + 5 \lt R^3\), and say what the argument then tells you.
- \(q(z) = |z|^2 + 1\) is never zero. Which step of the argument fails for it? And why does it say nothing about \(x^2 + 1\) when \(x\) must be real?
- (a) Show that \(z^5 + z + 1\) has no root with \(|z| \ge 1.2\), by running question 2's inequality the other way. (b) The argument gives one root \(r\). Use the factor theorem to explain why a degree-\(n\) polynomial has \(n\) roots, counted with multiplicity.
Stretch: the winding number counts the roots
Angles add when you multiply, so winding numbers add too. Write \(p(z) = (z - r_1)\cdots(z - r_n)\). On \(|z| = R\), what is the winding number of one factor \(z - r_k\) when \(|r_k| \lt R\)? When \(|r_k| > R\)? Deduce what the winding number of \(p\) counts. How many roots of \((z^2 + 1)(z - 3)\) lie inside \(|z| = 2\)?
Answers
- With \(z = 2e^{i\theta}\), \(z^2 + 1 = 4e^{2i\theta} + 1\): a circle of radius 4 centred at 1, traversed twice. It encloses 0, so the winding number is 2. For \(R = \tfrac12\) the image is a circle of radius \(\tfrac14\) centred at 1, which misses 0: winding number 0. The image passes through 0 when \(R^2 = 1\), so \(R = 1\), at \(z = \pm i\), the two roots.
- Triangle inequality: \(|2z + 5| \le 2|z| + 5 = 2R + 5\). \(R = 2\) fails (\(9 > 8\)); \(R = 3\) works (\(11 \lt 27\)). So on \(|z| = 3\) the image winds three times, and \(p\) has a root inside \(|z| \lt 3\). (In fact every root is inside, by the reasoning of question 4(a).)
- \(q\) has no leading term that takes over. On \(|z| = R\) it is the constant \(R^2 + 1\): the image is a single point, winding number 0 for every \(R\), so the huge-circle step gives nothing. (\(q\) is not a polynomial in \(z\); it involves \(\bar z\).) For real \(x\), the "circle" \(|x| = R\) is two points, not a loop, and there is nothing to wind. The argument needs the plane: the roots of \(x^2 + 1\) are \(\pm i\), and you only find them by leaving the line.
- (a) If \(|z| = R \ge 1.2\) then \(|z + 1| \le R + 1 \lt R^5 = |z^5|\): true at \(R = 1.2\) (\(2.2 \lt 2.488\)) and \(R^5 - R - 1\) only increases from there. So \(z^5 \ne -(z + 1)\), and \(p(z) \ne 0\). Meanwhile the winding argument puts at least one root inside \(|z| \lt 1.2\). (b) Since \(p(r) = 0\), the factor theorem gives \(p(z) = (z - r)\,q(z)\) with \(q\) of degree \(n - 1\). If \(n - 1 \ge 1\), \(q\) has a root by the same theorem; factor it out and repeat. After \(n\) steps \(p(z) = (z - r_1)\cdots(z - r_n)\). A root that appears \(k\) times in the list has multiplicity \(k\).
Stretch. \(z - r_k\) is the arrow from \(r_k\) to \(z\). If \(|r_k| \lt R\), the circle goes round \(r_k\) and the arrow turns through a full \(360°\): winding number 1. If \(|r_k| > R\), the circle does not enclose \(r_k\); the arrow swings one way and back, winding number 0. Adding up, the winding number of \(p\) on \(|z| = R\) is the number of roots inside the circle, with multiplicity. For \((z^2 + 1)(z - 3)\) the roots are \(\pm i\) and 3; two lie inside \(|z| = 2\), so the winding number is 2.