Complex numbers: Roots and polynomials
Complex cube roots
We already know that complex square roots exist, but we can go a step further and introduce the cube roots. But first an example.
Complex cube root of -1 What could \(\sqrt[3]{-1}\) be? We want to solve the equation \[z^3=-1\] in the set of complex numbers. For this we first write \(-1\) in polar form: \[-1 = e^{\pi\,\mathrm{i}+2\,k\,\pi\,\mathrm{i}}\] for integers \(k\). We also write \(z\) in polar form, say \[z=r\, e^{\,\mathrm{i}\,\varphi}\] Then: \[z^3=r^3\, e^{\,\mathrm{i}\,3\varphi}\] So we must find real numbers \(r>0\) and \(\varphi\) in the interval \((-\pi, \pi]\) such that \[r^3=1\qquad\text{and}\qquad 3\varphi=\pi+2k\pi\] Without the restriction of \(\varphi\) to the interval \((-\pi, \pi]\) we find \[r=1\quad\text{and}\quad \varphi=\tfrac{1}{3}\pi+\tfrac{2}{3}k\,\pi\] for integers \(k\). The solutions within the set of restricted values of \(\varphi\) are: \[[r=1,\phi= \tfrac{1}{3}\pi]\quad\text{or}\quad [r=1,\varphi= \pi]\quad\text{or}\quad [r=1,\varphi= -\tfrac{1}{3}\pi]\] We have found complex solutions to the equation in polar form \(z^3=-1\); we can also write them down as \[z=e^{(\tfrac{1}{3}\pi+\!\tfrac{2}{3}k\,\pi)\,\mathrm{i}}\quad\text{for }k=0, 1, 2\] For other integer values of \(k\) we always get one of these solution because of the periodicity of the complex exponential function. In standard form we have the following solutions: \[z=\cos\bigl(\tfrac{1}{3}\pi+\tfrac{2}{3}k\,\pi\bigr)+ \cos\bigl(\tfrac{1}{3}\pi+\tfrac{2}{3}k\,\pi\bigr)\mathrm{i}\quad\text{for }k=0, 1, 2\] For \(k=1\) we have a real solution \(-1\). As principal value of \(\sqrt[3]{-1}\) we choose in this case \(k=0\) and thus the solution \(z=e^{\tfrac{1}{3}\pi\,\mathrm{i}}=\tfrac{1}{2}+\mathrm{i}\,\tfrac{1}{2}\sqrt{3}\) .
The following picture shows where the solutions are on the unit circle:
The points that correspond to the complex cube roots of \(-1\) form the vertices of an equilateral triangle with the origin as center. The radii of the three cube roots of \(-1\) are separated by the angle \(\tfrac{2}{3}\pi\) .
The general case is similar.
The complex cube root For any complex number \(z\) we can define the complex cube root in polar form: \[\begin{aligned}\sqrt[3]{z}&=\sqrt[3]{|z|}\cdot e^{\,\mathrm{i}\tfrac{1}{3}\mathrm{arg}(z)}\\ \\ \text{principal branch of }\sqrt[3]{z}&=\sqrt[3]{|z|}\cdot e^{\,\mathrm{i}\tfrac{1}{3}\mathrm{Arg}(z)}\end{aligned}\] In terms of the complex exponential function and complex logarithm we can simply define the complex cube root function as \[\sqrt[3]{z}=e^{\tfrac{1}{3}\ln(z)}\]
So, if \(z=r\cdot e^{\,\mathrm{i}\,\varphi}\), with \(r>0\) and \(-\pi<\varphi\le\pi\), then it the principal value of \(\sqrt[3]{z}\) is equal to \(\sqrt[3]{r}\cdot e^{\,\mathrm{i}\tfrac{1}{3}\varphi}\) .
If want to solve the equation \(z^3=1\) in \(\mathbb{C}\), you can best write the solution as \(z=\sqrt[3]{1}\), at least when you keep in mind this it actually denotes the following three solutions: \(z=e^{0\,\ii}=1\), \(z=e^{\tfrac{2}{3}\pi\,\ii}=-\tfrac{1}{2}+\tfrac{1}{2}\sqrt{3}\,\ii\) and \(z=e^{\tfrac{4}{3}\pi\,\ii}=-\tfrac{1}{2}-\tfrac{1}{2}\sqrt{3}\,\ii\). After all, \(\arg(1)=2k\pi\ii\) for integers \(k\) .