6.3 Roots of Complex Numbers

I expect from (6.16) that every point $(\cos\theta,\sin\theta)$ in the unit circle has $n$th roots for all $n\in\mbox{{\bf Z}}_{\geq 1}$, and that in fact

$$\left(\cos\left({\theta\over n}\right)+i\sin\left({\theta\over n}\right)\right)^n=\cos\theta+i\sin\theta.$$

In particular, each vertex of the regular $n$-gon inscribed in the unit circle and having a vertex at $1$ will be an $n$th root of $1$.

6.17   Exercise. The figure below shows the seventeen points $${ \left\{\left( \cos {{2\pi j}\over {17}}+i\sin {{2\pi j}\over {17}}\right)\colon 0\leq j<17\right\}}$$.

Let $${w=\left(\cos {{4\pi}\over {17}}+i\sin {{4\pi}\over {17}}\right)}$$ and $${u=\left(\cos{{10\pi}\over {17}}+i\sin {{10\pi}\over {17}}\right)}$$. Draw the polygons $1$-$w$-$w^2$-$\cdots$-$w^{17}$ and $1$-$u$-$u^2$-$\cdots$-$u^{17}$ on different sets of axes, (i.e. draw segments connecting $1$ to $w$, $w$ to $w^2$, $\cdots$, $w^{16}$ to $w^{17}$, and segnents joining $1$ to $u$, $\cdots$, $u^{16}$ to $u^{17}$.)

6.18   Exercise. The sixth roots of $1$ are the vertices of a regular hexagon having one vertex at $1$. Find these numbers (by geometry or trigonometry) in terms of rational numbers or square roots of rational numbers, and verify by direct calculation that all of them do, in fact, have sixth power equal to $1$.

6.19   Theorem (Polar decomposition.)Let $z\in\mbox{{\bf C}}\backslash\{0\}$. Then we can write $z=ru$ where $r\in\mbox{${\mbox{{\bf R}}}^{+}$}$ and $u\in C(0,1)$. In fact this representation is unique, and

$$r=\vert z\vert,\qquad u={z\over {\vert z\vert}}.$$

I will call the representation

$$z=ru \mbox{ where } r\in\mbox{${\mbox{{\bf R}}}^{+}$},\; u\in C(0,1)$$

the polar decomposition of $z$, and I'll call $r$ the length of $z$, and I'll call $u$ the direction of $z$.

Proof: If $z=ru$ where $r\in\mbox{${\mbox{{\bf R}}}^{+}$}$ and $\vert u\vert=1$, then we have

$$\vert z\vert=\vert ru\vert=\vert r\vert\;\vert u\vert=r\cdot 1=r.$$

This shows that $r=\vert z\vert$, and it then follows that $${u={z\over r}={z\over {\vert z\vert}}}$$. Since $${\left\vert {z\over {\vert z\vert}}\right\vert}$$ $${={\vert z\vert\over {\vert\vert z\vert\vert}}}$$ $=1$, we see $${ {z\over {\vert z\vert}}\in C(0,1)}$$ and $${z=\vert z\vert\left( {z\over {\vert z\vert}}\right)}$$ gives the desired decomposition. $\mid\!\mid\!\mid$

6.20   Notation (Direction.) I will refer to any number in $C(0,1)$ as a direction.

6.21   Example. The polar decomposition for $-1+i$ is

$$\begin{eqnarray*} (-1+i)&=&\vert-1+i\vert\left( {{-1+i}\over {\vert-1+i\vert}}\r... ... &=&\sqrt 2\left(-{1\over {\sqrt 2}}+{i\over {\sqrt 2}}\right). \end{eqnarray*}$$

I recognize from trigonometry that $${\left(-{1\over {\sqrt 2}}+{i\over {\sqrt 2}}\right)=\left(\cos {{3\pi}\over 4}+i\sin{{3\pi}\over 4}\right)}$$.

6.22   Remark. Let $z,w\in\mbox{{\bf C}}\backslash\{0\}$. Let $z=ru$ and $w=sv$ be the polar decompositions of $z,w$, respectively, so $r,s\in\mbox{${\mbox{{\bf R}}}^{+}$}; u,v\in C(0,1)$. Then $zw=rusv=(rs)(uv)$ where $rs\in\mbox{${\mbox{{\bf R}}}^{+}$}$ and $uv\in C(0,1)$. Hence we have

length of product $=$ product of lengths

and

direction of product $=$ product of directions.