Complex numbers: Roots and polynomials
Complex square roots
We already know that the complex square function \(f(z)=z^2\) exists and that the equation \(z^2=-1\) has the complex solutions \(z=\pm\mathrm{i}\). We opted to equate \(\sqrt{-1}\) with the imaginary unit \(\mathrm{i}\). The imaginary unit \(\ii\) corresponds to the point \((0,1)\) in the complex plane and has absolute value \(1\) and principal argument \(\tfrac{1}{2}\pi\). In other words, \(\mathrm{i}\) is the principal value of \(\sqrt{-1}\). But can one also extract the square root of a complex number? For example, can one determine \(\sqrt{2+\mathrm{i}\cdot 2\sqrt{3}}\)? We answer this question as an example of the general case.
Example of extraction of a complex square root Suppose \[\alpha=2+\mathrm{i}\cdot2\sqrt{3}\] We want to solve the equation \[z^2=\alpha\] in the set of complex numbers. For this, we first write \(\alpha\) in polar form: because \(\alpha=4\cdot(\tfrac{1}{2}+\ii\cdot \tfrac{1}{2}\sqrt{3})\) it follows from the special trigonometric function values that \[\alpha = 4\,e^{\tfrac{1}{3}\pi\,\mathrm{i}+k\,2\,\pi\,\mathrm{i}}\] for integers \(k\). Note that we use all possible arguments of \(\alpha\) in the calculation.
We also write \(z\) in polar form, say \[z=r\, e^{\,\mathrm{i}\,\varphi}\] Then De Moivre's formula (o,r if you prefer, the product rule for imaginary power of \(e\)) yields \[z^2=r^2\, e^{\,\mathrm{i}\,2\,\varphi}\] So we must find real numbers \(r>0\) and \(\varphi\) in the interval \((-\pi, \pi]\) such that \[r^2= 4\qquad\text{and}\qquad 2\varphi=\tfrac{1}{3}\pi+2k\pi\] Without the restriction of \(\varphi\) to the interval \((-\pi, \pi]\) we find \[r=2\qquad\text{and}\qquad\varphi=\tfrac{1}{6}\pi+k\,\pi\] integers \(k\). The solutions within the restriction of \(\varphi\) are therefore: \[[r=2,\varphi= \tfrac{1}{6}\pi]\qquad\text{or}\qquad [r=2,\varphi= -\tfrac{5}{6}\!\pi]\] In polar form we have found complex solutions of the equation \(z^2=\alpha\): \[z=2\,e^{\tfrac{1}{6}\!\pi\,\mathrm{i}}\qquad\text{or}\qquad z=2\,e^{-\tfrac{5}{6}\!\pi\,\mathrm{i}}\] As principal value of \(\sqrt{\alpha}\) we select the first solution.
The general case is similar.
The complex square root For any complex number \(z\) we can define the complex square root in polar form: \[\begin{aligned}\sqrt{z}&=\sqrt{|z|}\cdot e^{\,\mathrm{i}\tfrac{1}{2}\mathrm{arg}(z)}\\ \\ \text{principal branch of }\sqrt{z}&=\sqrt{|z|}\cdot e^{\,\mathrm{i}\tfrac{1}{2}\mathrm{Arg}(z)}\end{aligned}\] In terms of the complex exponential function and complex logarithm we can simply define the complex square root function as \[\sqrt{z}=e^{\tfrac{1}{2}\ln(z)}\]
So, if \(z=r\cdot e^{\,\mathrm{i}\,\varphi}\), with \(r>0\) and \(-\pi<\varphi\le\pi\), then the principal value of \(\sqrt{z}\) is equal to \(\sqrt{r}\cdot e^{\,\mathrm{i}\tfrac{1}{2}\varphi}\).
