Complex roots and roots of unity

Reverse the integral-power theorem

Fix a positive integer and a nonzero target. Seek a root with positive radius. The power equation gives a modulus equation and an angle equation. Dividing an argument without retaining the full-turn term loses roots

State the complete root set

A nonzero target has the following complete set of distinct roots. Changing the argument convention only changes the order or adds whole turns to the listed angles

1. 1Given target The radius is positive and the root order is a positive integer \(\displaystyle \begin{gathered}z^n=w=\rho\operatorname{cis}\phi\ne0\\\rho>0,\quad n\in\mathbb Z_{>0}\end{gathered}\)
2. 2Common modulus Take the unique positive real root \(\displaystyle r=\rho^{1/n}\)
3. 3All roots Retain one complete set of consecutive indices \(\displaystyle \begin{gathered}z_j=\rho^{1/n}\operatorname{cis}\left(\frac{\phi+2j\pi}{n}\right)\\j=0,\ldots,n-1\end{gathered}\)

Separate zero from nonzero targets

A zero target forces the single root at the origin. It has the stated multiplicity and no argument. A nonzero target has distinct simple roots because the derivative cannot vanish at them. Do not confuse the count of points with the total count including multiplicity

How many distinct roots does z⁶=0 have?

Which statement matches the plotted zero-root case?

List fourth roots exactly

The target has modulus sixteen and argument pi. Apply the root-angle rule to four consecutive indices, then keep the arguments in the stated interval. Cartesian conversion gives all four sign combinations

Equation

\(\displaystyle z^4=-16\)

Argument convention

\(\displaystyle 0\le\theta<2\pi\)

Give the coefficient of pi in the largest root argument in the stated interval

Changing the target argument permutes the roots

Writing −16 with argument 3π instead of π does not change its fourth-root set. It starts the angle list at 3π/4; continuing four steps and reducing modulo 2π yields the same four values in a different order

Roots of unity form a reusable pattern

The roots of unity lie on the unit circle. Multiplying by the first angular step moves to the next root and eventually returns to the start. The sum and product statements concern the full set

Find roots on the axes

The real roots of unity are 1 and, when n is even, −1. The purely imaginary candidates are i and −i; they solve z^n=1 exactly when n is divisible by 4. Thus the sixth roots of unity have two real roots and no purely imaginary roots, while the fourth roots have two of each

Find the product of all six sixth roots of unity

Check candidates without losing whole turns

For z⁴=−16, the expressions 2 cis(π/4) and 2 cis(9π/4) represent the same root. The expression 2 cis(3π/4) is a different root. The candidate 2 cis(π/2) is not a root because its fourth power is 16. Check radius and multiplied angle, then remove duplicate values

Equation

\(\displaystyle z^4=-16\)

First candidate

\(\displaystyle 2\operatorname{cis}\frac\pi4\)

Second candidate

\(\displaystyle 2\operatorname{cis}\frac{9\pi}{4}\)

Third candidate

\(\displaystyle 2\operatorname{cis}\frac{3\pi}{4}\)

Fourth candidate

\(\displaystyle 2\operatorname{cis}\frac\pi2\)

How many distinct valid roots occur among these four candidates?

Read the root polygon

For z³=8, the roots are 2, −1+√3 i and −1−√3 i. All have modulus 2 and adjacent angular spacing 2π/3. They form an equilateral triangle centred at the origin. A nonzero target with n≥3 gives a regular n-gon; n=2 gives an antipodal pair, and a zero target gives only the origin

Which parts of this diagram solve z³=8?

Compute chord length and polygon area

For equally spaced roots on a circle, use the cosine rule for an adjacent side. Splitting the polygon into centre triangles gives its area. For the cube roots of eight, the squared side length is twelve and the area is three times the square root of three. Rotating the target preserves these measurements

Give the squared side length for the triangle of cube roots of 8.

Fifth roots need not have simple Cartesian components

For z⁵=−32, the root modulus is 2 and arguments in [0,2π) are π/5,3π/5,π,7π/5,9π/5. Exact polar form is already a complete answer. Decimal Cartesian coordinates are useful for plotting, but they do not replace the exact arguments

What is the adjacent angular spacing in this fifth-root diagram?

A general modulus need not be an integer

For z³=5i, the roots have modulus ∛5 and arguments π/6,5π/6,3π/2. The exact values are ∛5 cis of those three angles. The practice bank uses perfect-power target moduli for rational field entry; the general theorem still uses ρ^(1/n) for every positive real ρ

Use the geometric-series pattern for sums

Apply the geometric-series identity to the full root set. Reciprocal sums require a nonzero target

1. 1Root set Fix the nonzero target and one of its roots \(\displaystyle \begin{gathered}z^n=w\ne0,\quad\alpha^n=w\\z_j=\alpha\omega^j,\quad\omega=\operatorname{cis}\frac{2\pi}{n}\\j=0,\ldots,n-1\end{gathered}\)
2. 2Lower positive powers For these exponents the geometric ratio is not one \(\displaystyle \begin{gathered}1\le m<n,\quad\omega^m\ne1\\\sum_{j=0}^{n-1}z_j^m=\alpha^m\frac{1-\omega^{mn}}{1-\omega^m}=0\end{gathered}\)
3. 3Nth powers Each term equals the target \(\displaystyle \sum_{j=0}^{n-1}z_j^n=nw\)
4. 4Reciprocals For more than one root, inverse unity powers form the same unity-root set \(\displaystyle \sum_{j=0}^{n-1}\frac1{z_j}=0\quad(n>1)\)

The product has a parity sign

Compare constant coefficients in the complete monic factorisation to determine the product sign

1. 1Factorisation Use all roots with their multiplicities \(\displaystyle z^n-w=\prod_{j=0}^{n-1}(z-z_j)\)
2. 2Constant coefficients Each root contributes a minus sign \(\displaystyle \begin{gathered}(-1)^n\prod_{j=0}^{n-1}z_j=-w\\\prod_{j=0}^{n-1}z_j=(-1)^{n+1}w\end{gathered}\)
3. 3Example For all five roots of the displayed equation, apply the sum and product rules \(\displaystyle \begin{gathered}z_j^5=-32\\\sum z_j^2=0,\quad\sum z_j^5=5(-32)=-160\\\prod z_j=-32,\quad\sum z_j^{-1}=0\end{gathered}\)

Give the real value of the fifth-power sum for the five roots of −32.

Prove completeness and distinctness, ungraded

Starting from z^n=w≠0, derive the full polar list and prove that the n listed values are distinct and that every solution is in the list. Then explain the zero-target case separately. Your proof must justify why infinitely many integer j values do not give infinitely many different roots

Model: reduce the integer index modulo n

Taking moduli gives the unique positive radius. Equality of nonzero polar values gives the angle condition, so every solution has the displayed form. Reduce the integer index modulo the root order; the extra whole turns do not change the value. If two indices in that range gave the same root, their angle difference would be a whole turn, forcing the indices to coincide. Thus there are exactly the stated number of distinct nonzero roots. At a zero target, modulus forces the single origin value, with the stated multiplicity and undefined argument

Justify unity sums and products, ungraded

For an arbitrary integer n>1, prove the sum and product formulas for all nth roots of unity. Extend the sum argument to powers m with 1≤m<n. State why a geometric-series division is permitted and why m=n is different

Model: check the denominator before dividing

Check the geometric-series denominator before dividing, and use the constant coefficient for the product

1. 1Unity roots The root order is greater than one \(\displaystyle \begin{gathered}n>1,\quad\omega=\operatorname{cis}\frac{2\pi}{n}\\\omega\ne1,\quad\omega^n=1\end{gathered}\)
2. 2Sum Multiply the finite series and cancel its adjacent terms \(\displaystyle \begin{gathered}(1-\omega)\sum_{j=0}^{n-1}\omega^j=1-\omega^n=0\\\sum_{j=0}^{n-1}\omega^j=0\end{gathered}\)
3. 3Power sums Use the same argument when the new ratio is not one; otherwise add unit terms \(\displaystyle \begin{gathered}\sum_{j=0}^{n-1}\omega^{jm}=0\quad(1\le m<n)\\\sum_{j=0}^{n-1}\omega^{jn}=n\end{gathered}\)
4. 4Product The distinct roots give a monic factorisation, so compare its constant coefficient \(\displaystyle \begin{gathered}z^n-1=\prod_{j=0}^{n-1}(z-\omega^j)\\(-1)^n\prod_{j=0}^{n-1}\omega^j=-1\\\prod_{j=0}^{n-1}\omega^j=(-1)^{n+1}\end{gathered}\)

Construct and explain a root diagram, ungraded

Find all cube roots of 5i in exact polar form and draw them with equal axis scales. Show the common radius, starting direction and angular spacing. Explain the rotation and dilation taking the cube roots of 1 to this set, and state which parts of your construction are not solution points

Model: transform the unity triangle

Multiply each cube root of 1 by ∛5 cis(π/6). This dilates distances by ∛5 and rotates anticlockwise by π/6, producing arguments π/6,5π/6,3π/2. The three vertices are roots. The circle and triangle edges only show the geometry; their other points do not satisfy z³=5i. The complete learner sketch and explanation remain ungraded

Check roots with technology, ungraded

Compute the five roots of −32 using exact polar data, then convert to decimal Cartesian coordinates and raise each result to the fifth power. Compare the residual z⁵+32, the sum of the roots and their product. Explain the effect of rounding and of changing argument conventions. Keep your own calculations or code

Model: distinguish a residual from a proof

Exact values use radius 2 and the five stated rational multiples of π. Raising these exact polar forms to the fifth power gives −32 directly. Rounded Cartesian values may leave small nonzero complex residuals; rounding before exponentiation generally enlarges them. The exact root sum is 0 and product −32. Whole-turn changes in arguments leave the roots unchanged. A finite numerical check neither proves the theorem nor shows that the learner executed the technology task