Complex square roots

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.

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.

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 $\sqrt{r}\cdot e^{\,\mathrm{i}\tfrac{1}{2}\varphi}$ .