## 47. Roots of complex numbers.

We have not, up to the present, attributed any meaning to symbols such as $$\sqrt[n]{a}$$, $$a^{m/n}$$, when $$a$$ is a complex number, and $$m$$ and $$n$$ integers. It is, however, natural to adopt the definitions which are given in elementary algebra for real values of $$a$$. Thus we define $$\sqrt[n]{a}$$ or $$a^{1/n}$$, where $$n$$ is a positive integer, as a number $$z$$ which satisfies the equation $$z^{n} = a$$; and $$a^{m/n}$$, where $$m$$ is an integer, as $$(a^{1/n})^{m}$$. These definitions do not prejudge the question as to whether there are or are not more than one (or any) roots of the equation.

## 48. Solution of the equation $$z^{n} = a$$.

Let $a = \rho(\cos\phi + i\sin\phi),$ where $$\rho$$ is positive and $$\phi$$ is an angle such that $$-\pi < \phi \leq \pi$$. If we put $$z = r(\cos\theta + i\sin\theta)$$, the equation takes the form $r^{n}(\cos n\theta + i\sin n\theta) = \rho(\cos\phi + i \sin\phi);$ so that $\begin{equation*} r^{n} = \rho,\quad \cos n\theta = \cos\phi,\quad \sin n\theta = \sin\phi. \tag{1} \end{equation*}$

The only possible value of $$r$$ is $$\sqrt[n]{\rho}$$, the ordinary arithmetical $$n$$th root of $$\rho$$; and in order that the last two equations should be satisfied it is necessary and sufficient that $$n\theta = \phi + 2k\pi$$, where $$k$$ is an integer, or $\theta = (\phi + 2k\pi)/n.$ If $$k = pn + q$$, where $$p$$ and $$q$$ are integers, and $$0 \leq q < n$$, the value of $$\theta$$ is $$2p\pi + (\phi + 2q\pi)/n$$, and in this the value of $$p$$ is a matter of indifference. Hence the equation $z^{n} = a = \rho(\cos\phi + i\sin\phi)$ has $$n$$ roots and $$n$$ only, given by $$z = r(\cos\theta + i\sin\theta)$$, where $r = \sqrt[n]{\rho},\quad \theta = (\phi + 2q\pi)/n,\quad (q = 0,\ 1,\ 2,\ \dots, n – 1).$

That these $$n$$ roots are in reality all distinct is easily seen by plotting them on Argand’s diagram. The particular root $\sqrt[n]{\rho}\{\cos(\phi/n) + i\sin(\phi/n)\}$ is called the principal value of $$\sqrt[n]{a}$$.

The case in which $$a = 1$$, $$\rho = 1$$, $$\phi = 0$$ is of particular interest. The $$n$$ roots of the equation $$x^{n} = 1$$ are $\cos(2q\pi/n) + i\sin(2q\pi/n),\quad (q = 0,\ 1,\ \dots, n – 1).$ These numbers are called the $$n$$th roots of unity; the principal value is unity itself. If we write $$\omega_{n}$$ for $$\cos(2\pi/n) + i\sin(2\pi/n)$$, we see that the $$n$$th roots of unity are $1,\quad \omega_{n},\quad \omega_{n}^{2},\ \dots,\quad \omega_{n}^{n-1}.$

Example XXII

1. The two square roots of $$1$$ are $$1$$, $$-1$$; the three cube roots are $$1$$, $$\frac{1}{2}(-1 + i\sqrt{3})$$, $$\frac{1}{2}(-1 – i\sqrt{3})$$; the four fourth roots are $$1$$, $$i$$, $$-1$$, $$-i$$; and the five fifth roots are \begin{aligned} {4} 1,\quad &\tfrac{1}{4} \Bigl[ &&\sqrt{5} – 1 + i\sqrt{10 + 2\sqrt{5}}\Bigr],\quad && \tfrac{1}{4} \Bigl[-&&\sqrt{5} – 1 + i\sqrt{10 – 2\sqrt{5}}\Bigr],\\ &\tfrac{1}{4} \Bigl[-&&\sqrt{5} – 1 – i\sqrt{10 – 2\sqrt{5}}\Bigr],\quad && \tfrac{1}{4} \Bigl[ &&\sqrt{5} – 1 – i\sqrt{10 + 2\sqrt{5}}\Bigr].\end{aligned}

2. Prove that $1 + \omega_{n} + \omega_{n}^{2} + \dots + \omega_{n}^{n-1} = 0.$

3. Prove that $(x + y\omega_{3} + z\omega_{3}^{2}) (x + y\omega_{3}^{2} + z\omega_{3}) = x^{2} + y^{2} + z^{2} – yz – zx – xy.$

4. The $$n$$th roots of $$a$$ are the products of the $$n$$th roots of unity by the principal value of $$\sqrt[n]{a}$$.

5. It follows from Exs. XXI. 14 that the roots of $z^{2} = \alpha + \beta i$ are $\pm \sqrt{\tfrac{1}{2} \{\sqrt{\alpha^{2} + \beta^{2}} + \alpha\}} \pm i\sqrt{\tfrac{1}{2} \{\sqrt{\alpha^{2} + \beta^{2}} – \alpha\}},$ like or unlike signs being chosen according as $$\beta$$ is positive or negative. Show that this result agrees with the result of § 48.

6. Show that $$(x^{2m} – a^{2m})/(x^{2} – a^{2})$$ is equal to $\Bigl(x^{2} – 2ax\cos\frac{\pi}{m} + a^{2}\Bigr) \Bigl(x^{2} – 2ax\cos\frac{2\pi}{m} + a^{2}\Bigr) \dots \Bigl(x^{2} – 2ax\cos\frac{(m – 1)\pi}{m} + a^{2}\Bigr).$

[The factors of $$x^{2m} – a^{2m}$$ are $(x – a),\quad (x – a\omega_{2m}),\quad (x – a\omega_{2m}^{2}),\ \dots\quad (x – a\omega_{2m}^{2m-1}).$ The factor $$x – a\omega_{2m}^{m}$$ is $$x + a$$. The factors $$(x – a\omega_{2m}^{s})$$, $$(x – a\omega_{2m}^{2m-s})$$ taken together give a factor $$x^{2} – 2ax \cos(s\pi/m) + a^{2}$$.]

7. Resolve $$x^{2m+1} – a^{2m+1}$$, $$x^{2m} + a^{2m}$$, and $$x^{2m+1} + a^{2m+1}$$ into factors in a similar way.

8. Show that $$x^{2n} – 2x^{n}a^{n} \cos\theta + a^{2n}$$ is equal to $\begin{gathered} \left(x^{2} – 2xa\cos\frac{\theta}{n} + a^{2}\right) \left(x^{2} – 2xa\cos\frac{\theta + 2\pi}{n} + a^{2}\right) \dots \\ \dots\left(x^{2} – 2xa\cos\frac{\theta + 2(n – 1)\pi}{n} + a^{2}\right).\end{gathered}$

[Use the formula $x^{2n} – 2x^{n}a^{n} \cos\theta + a^{2n} = \{x^{n} – a^{n}(\cos\theta + i\sin\theta)\} \{x^{n} – a^{n}(\cos\theta – i\sin\theta)\},$ and split up each of the last two expressions into $$n$$ factors.]

9. Find all the roots of the equation $$x^{6} – 2x^{3} + 2 = 0$$.

10. The problem of finding the accurate value of $$\omega_{n}$$ in a numerical form involving square roots only, as in the formula $$\omega_{3} = \frac{1}{2}(-1 + i\sqrt{3})$$, is the algebraical equivalent of the geometrical problem of inscribing a regular polygon of $$n$$ sides in a circle of unit radius by Euclidean methods, i.e. by ruler and compasses. For this construction will be possible if and only if we can construct lengths measured by $$\cos(2\pi/n)$$ and $$\sin(2\pi/n)$$; and this is possible (Ch. II, Misc. Exs. II 22) if and only if these numbers are expressible in a form involving square roots only.

Euclid gives constructions for $$n = 3$$, $$4$$, $$5$$, $$6$$, $$8$$, $$10$$, $$12$$, and $$15$$. It is evident that the construction is possible for any value of $$n$$ which can be found from these by multiplication by any power of $$2$$. There are other special values of $$n$$ for which such constructions are possible, the most interesting being $$n = 17$$.

## 49. The general form of De Moivre’s Theorem.

It follows from the results of the last section that if $$q$$ is a positive integer then one of the values of $$(\cos\theta + i\sin\theta)^{1/q}$$ is $\cos(\theta/q) + i\sin(\theta/q).$ Raising each of these expressions to the power $$p$$ (where $$p$$ is any integer positive or negative), we obtain the theorem that one of the values of $$(\cos\theta + i\sin\theta)^{p/q}$$ is $$\cos(p\theta/q) + i\sin(p\theta/q)$$, or that if $$\alpha$$ is any rational number then one of the values of $$(\cos\theta + i\sin\theta)^{\alpha}$$ is $\cos\alpha\theta + i\sin\alpha\theta.$ This is a generalised form of De Moivre’s Theorem (SecNo 45).