Complex roots and roots of unity

From a root list to factors

Use the ordered-root method for polar cubics and higher powers. Once the complete distinct roots z_j of z^n−w are known for w≠0, the monic factorisation is ∏(z−z_j). Each root is simple. Preserve a nonzero leading scalar if the original polynomial has one

List fourth roots exactly

For z⁴=−16, the target has modulus 16 and argument π. Thus r=2 and θ=π/4+jπ/2 for j=0,1,2,3. In [0,2π), the roots are 2 cis(π/4), 2 cis(3π/4), 2 cis(5π/4), 2 cis(7π/4). Cartesian conversion gives all four sign combinations ±√2±√2 i

Equation

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

Argument convention

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

Give the coefficient of π 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. In the ordered-entry tasks, use increasing coefficients in [0,2), so each field has one specified answer

Which complete factorisation matches the cube roots of 8 shown here?

A polar cubic can have nonreal coefficients

Use the complete polar root list to obtain every linear factor. The coefficients are not all real, so conjugation need not preserve the root set

1. 1Target and roots Take the positive cube-root radius and retain the three directions \(\displaystyle \begin{gathered}z^3=8i=8\operatorname{cis}\frac\pi2\\r=2,\quad\theta_j=\frac\pi6+\frac{2j\pi}{3}\\j=0,1,2\end{gathered}\)
2. 2Cartesian values Use the exact sine and cosine at each angle \(\displaystyle \begin{gathered}2\operatorname{cis}\frac\pi6=\sqrt3+i\\2\operatorname{cis}\frac{5\pi}6=-\sqrt3+i\\2\operatorname{cis}\frac{3\pi}2=-2i\end{gathered}\)
3. 3Complete factorisation Each simple root supplies its monic linear factor \(\displaystyle \begin{aligned}z^3-8i&=(z-\sqrt3-i)\\&\quad\times(z+\sqrt3-i)(z+2i)\end{aligned}\)