Roots of Polynomials and Roots of Unity
A polynomial of positive degree n with complex coefficients has n complex roots counting multiplicity. For real coefficients, non-real roots occur in conjugate pairs. The nth roots of unity are evenly spaced on the unit circle.
What a learner can do afterwards
- Find all roots of a real cubic using a known real root, polynomial division and the remaining quadratic
- Find all nth roots of unity for small n and place them on the Argand plane
- State what the fundamental theorem of algebra promises and what it does not
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The fundamental theorem of algebra applies to polynomials of degree n ≥ 1 with complex coefficients. There are exactly n complex roots counting multiplicity. Real numbers are complex numbers too. Multiplicity counts repetitions: a factor (x - 8)³ counts the root 8 three times. The theorem gives a count, not the root values. It does not require the roots to be real or different.
You have used a root r to find the factor x - r. Divide that factor out of a cubic, then solve the remaining quadratic. Set each factor equal to zero. For c > 0, x² + c² = 0 gives ci and -ci, since i² = -1. If all polynomial coefficients are real, every non-real root a + bi has the conjugate a - bi as a root with the same multiplicity.
For f(x) = x³ + 6x² + 25x + 150, f(-6) = 0. Divide by x + 6: subtract x²(x + 6), leaving 25x + 150, then subtract 25(x + 6), leaving 0. Thus f(x) = (x + 6)(x² + 25). Solve the factors to get -6, 5i and -5i. Over the complex numbers, f(x) = (x + 6)(x - 5i)(x + 5i).
A cubic with real coefficients is continuous and has outputs of opposite signs far to the left and right. It must take the value zero between them, so it has a real root. For x³ = A with real A, take the real cube root: x³ = 27 gives x = 3. Since x³ increases, this real solution is unique. To count real roots with multiplicity, subtract the non-real root count from the degree.
Let n be a positive integer. The nth roots of unity solve zⁿ = 1. Write z = r e^(iθ). Then zⁿ = rⁿ e^(inθ), so r = 1 and nθ = 2πk for an integer k. Use θ = 2πk/n for k = 0, ..., n - 1 to list each root once. Neighboring roots are 360/n degrees apart. Euler gives coordinates (cos θ, sin θ), with θ in radians: real part horizontal, imaginary part vertical. Unit-circle points (a, b) satisfy a² + b² = 1.
For z⁸ = 1, use θ = kπ/4 for k = 0 through 7. The eight arguments are listed below. They are 45 degrees apart, forming a regular octagon on the unit circle. At θ = π/4, cos θ = sin θ = √2/2, so the root is √2/2 + (√2/2)i and its coordinates are (√2/2, √2/2). Repeat the cosine and sine calculation for each argument.
Add multiplicities to recover the degree. A factor (x + 2)² gives one distinct root, -2, counted twice. For a degree 10 polynomial, if two of its three distinct roots have multiplicities 6 and 1, the third has multiplicity 10 - 6 - 1 = 3. For a polynomial with real coefficients, a real root of even multiplicity makes the graph touch and turn at the x-axis; odd multiplicity makes it cross.
Count roots with multiplicity, solve factors to find their values, and use equally spaced arguments to locate roots of unity on the unit circle.
The lesson, practice questions, worked solutions and flashcards were written for this dataset.
The readings, extra questions and flashcards were written for this dataset. Video references: Khan Academy.
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Root values and root counts
The degree and the number of different root values need not match. Consider (x - 5)³(x + 1). Its degree is 4. Its two distinct roots are 5 and -1, but 5 has multiplicity 3. Counting each repeated factor gives four roots in total.
The fundamental theorem of algebra gives this total for any polynomial of positive degree with complex coefficients. It does not find the values for you. It also does not promise that the roots are different, or that any are non-real. Real numbers belong to the complex numbers.
Constant polynomials need separate treatment. A nonzero constant never equals zero, so it has no roots. The zero polynomial equals zero at every input and has no degree to insert into the theorem's count.
For real coefficients, non-real roots and their conjugates have matching multiplicities. Real roots are the ones that can appear as x-intercepts of the real graph. At a real root, even multiplicity gives a touch and turn; odd multiplicity gives a crossing.
From an argument to a point
A complex number a + bi is plotted at (a, b). The real part sets the horizontal coordinate; the imaginary part sets the vertical coordinate. For example, -3 + 5i is at (-3, 5), and its conjugate -3 - 5i is at (-3, -5).
Euler's formula writes a unit-circle point as e^(iθ) = cos θ + i sin θ, where θ is in radians. Raising this number to the nth power multiplies its argument by n. The result is 1 when nθ is an integer multiple of 2π.
For a positive integer n, list the arguments θ = 2πk/n with k = 0 through n - 1. The corresponding complex numbers z = e^(iθ) are all the nth roots of unity, with no repeats. Using k = n returns to the same root as k = 0. For tenth roots, the arguments increase by π/5, or 36 degrees, each time.
Calculate cosine and sine to locate each root at (cos θ, sin θ). Every resulting coordinate pair (a, b) has a² + b² = 1. The root order changes the number and spacing of the points, not the circle's radius.
Also distinguish an equation from principal-root notation. The equation z² = -1 has solutions i and -i. The principal square root √(-1) names i.
Finish solving the quadratic
Take f(x) = x³ - 7x² + 16x - 112. At x = 7, its value is 343 - 343 + 112 - 112 = 0. The factor theorem therefore gives the factor x - 7.
Divide by x - 7. The first quotient term is x². Subtract x²(x - 7), leaving 16x - 112. The next quotient term is 16. Subtract 16(x - 7), leaving zero. Thus f(x) = (x - 7)(x² + 16).
Now solve each factor. The linear factor gives 7. The quadratic gives x² = -16, so its roots are 4i and -4i. Over the complex numbers the full factorisation is (x - 7)(x - 4i)(x + 4i). Multiplying the last two factors gives x² + 16 because i² = -1.
The non-real roots form a conjugate pair because all coefficients of f are real. This outcome is not automatic after finding one real root of a cubic. Another quadratic quotient could give two real roots or a repeated real root. Solve that quotient before deciding what the remaining roots are.
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