n-th roots of complex numbers

Complex numbers: Roots and polynomials

N-th roots of complex numbers

We already know that complex square roots and cube roots exist and then the step to $n$ throots of compelx numbers with $n\gt 3$ not so big anymore. The general case will go in the same way.

Suppose $n$ is a natural number greater than or equal to 2.
Any complex number $z=r\cdot e^{\,\mathrm{i}\,\varphi}$ with $r=|z|>0$ has exactly $n$ $n$ th roots, namely $\sqrt[n]{z}=e^{(\tfrac{1}{n}\pi+\tfrac{2}{n}k\,\pi)\,\mathrm{i}}\quad\text{for }k=0,1,\ldots,n-1$

For any complex number $z$ we can define the complex $n$ th root in polar form: $\begin{aligned}\sqrt[n]{z}&=\sqrt[n]{|z|}\cdot e^{\,\mathrm{i}\tfrac{1}{n}\mathrm{arg}(z)}\\ \\ \text{principal branch of }\sqrt[n]{z}&=\sqrt[n]{|z|}\cdot e^{\,\mathrm{i}\tfrac{1}{n}\mathrm{Arg}(z)}\end{aligned}$ In terms of the complex exponential function and logarithm, we can simply define the complex root function as $\sqrt[n]{z}=e^{\tfrac{1}{n}\ln(z)}$

So, if $z=r\cdot e^{\,\mathrm{i}\,\varphi}$ , with $r>0$ and $-\pi<\varphi\le\pi$ , then the principal value of $\sqrt[n]{z}$ is equal to $\sqrt[n]{r}\cdot e^{\,\mathrm{i}\tfrac{1}{n}\varphi}$ .

Below are the seven 7th roots $\sqrt[7]{\alpha}$ of $\alpha=1.7+1.5\,\ii$ drawn.
The following holds: $\begin{aligned}|\alpha|&=\sqrt{1.7^2+1.5^2}\approx 2.267\\ \arg(\alpha)&=\arctan\bigl(\frac{1.5}{1.7}\bigr)+2k\pi\approx 0.723+2k\pi\quad\text{for integers }k\tiny.\end{aligned}$ So $\begin{aligned}\left|\sqrt[7]{\alpha}\right|&=\sqrt[7]{\left|\alpha\right|}\approx 1.124\\ \arg(\sqrt[7]{\alpha})&=\tfrac{1}{7}\arg(\alpha)\approx 0.103+\tfrac{2k\pi}{7}\quad\text{for }k=0,1,\dots 6\tiny.\end{aligned}$ In summary: $\sqrt[7]{1.7+1.5\,\ii}\approx 1.124\,e^{(0.103+\frac{2k\pi}{7})\,\ii}\quad\text{for }k=0,1,\dots 6\tiny.$ All drawn roots can claim the format $\sqrt[7]{\alpha}$ , but in the first quadrant of the complex plane is a special root (lying closest to $1$ ) that we call the principal value of the 7th root.

Calculate exactly the principal value of $\sqrt[3]{-2+2\mathrm{i}}$ in polar form.

Note that $-2+2\mathrm{i}=2\sqrt{2}\, e^{\,\tfrac{3}{4}\pi\,\mathrm{i}}$ and thus $\sqrt[3]{-2+2\mathrm{i}}=\sqrt{2}\, e^{(\tfrac{1}{3}\cdot \tfrac{3}{4}\pi\,\mathrm{i})}=\sqrt{2}\, e^{\,\tfrac{1}{4}\pi\,\mathrm{i}}$

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