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I don't understand the explanation for k & n in my text book. What are they trying to say by "... for any n consecutive values of k"? In the 1st formula, it says $k=0,1,2,\ldots,n-1$ then next it says $k=n,n+1,\ldots,2n-1$, that will mean $k=1,2,3,\ldots,2n-1$?

http://i.imgur.com/3HoQd.png

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It means the formula holds when you plug in $k=n$, or plug in $k=n+1$, or $k=n+2$, or $\dots$ $k=2n-1$; it doesn 't matter what $n$ is but it is apparently fixed from the beginning. Note that these are all consecutive integers and there are $n$ of them. –  anon Nov 23 '11 at 0:39
    
So suppose I have $(\cos{\theta}+i\cdot \sin{\theta})^{\frac{1}{6}}$. Then I can set $k=6 \text{ or 7,8,...,11}$. But according to the 1st formula, won't it be best I let $k=0$ then I'd just have $\cos{\frac{\theta}{n}+i\cdot \sin{\frac{\theta}{n}}}$? –  Jiew Meng Nov 23 '11 at 0:47
    
What do you mean by "best"? It's just saying it works for any integer $k$, not saying what's "best." –  anon Nov 23 '11 at 0:52
    
I mean if I let $k=0$, I will cancel out $2k\pi$? –  Jiew Meng Nov 23 '11 at 1:01
    
Well yes, for $k=0$ we have that $2k\pi=0$ and so vanishes out. I don't see the relevance of that to your question though. –  anon Nov 23 '11 at 1:07

2 Answers 2

This answer is far too long (just copied and pasted from something I did before) --- the crucial point is that $n$ consecutive values of $k$ gives $n$ distinct answers. The fundamental theorem of algebra guarantees that there are no more.

More details here: http://irishjip.wordpress.com/2011/11/19/project-maths-a-more-geometric-approach-to-de-moivres-theorem-and-complex-roots/ --- pictures also.

Proposition

Suppose that $z_1$ has magnitude $r$ and argument $\theta$; and $z_2$ has magnitude $ s$ and argument $\alpha$. Then the product of $z_1$ and $z_2$ has magnitude $r\times s$ and argument $\theta+\alpha$.

Proof : We simply compute:

$$z_1z_2=r(\cos\theta+i\sin\theta)\times s(\cos\alpha+i\sin\alpha),$$

$$=rs(\cos\theta+i\sin\theta)(\cos\alpha+i\sin\alpha),$$

$$=rs(\cos\theta\cos\alpha+i\cos\theta\sin\alpha+i\sin\theta\cos\alpha-\sin\theta\sin\alpha),$$

$$=rs(\cos\theta\cos\alpha-\sin\theta\sin\alpha+i(\cos\theta\sin\alpha+\sin\theta\cos\alpha)).$$

Now note the identities:

$$\cos(A+B)=\cos A\cos B-\sin A\sin B,$$

$$\sin(A+B)=\cos A\sin B+\sin A\cos B.$$

Therefore

$$z_1z_2=rs(\cos(\theta+\alpha)+i\sin(\theta+\alpha)).$$

That is $z_1z_2$ has magnitude $rs$ and argument $\theta+\alpha$ $\bullet$

In other words, we can simplify the multiplication of complex numbers to

$$(r,\theta)\times(s,\alpha)=(rs,\theta+\alpha).$$

A lot more elegant in my opinion than the standard approach!

De Moivre's Theorem

This work makes De Moivre's Theorem very straightforward indeed. De Moivre's Theorem for Natural Number Powers

Suppose that $z=(r,\theta)$ and $n\in\mathbb{N}$. Then

$$z^n=(r^n,n\theta).$$

Proof : Well, this isn't really necessary --- it really is obvious with the work done above.

$$z^n=\underbrace{z\times z\times\cdots\times z}_{n\text{ times}},$$

$$\Rightarrow z^n=(r\times r\times \cdots\times r,\theta+\theta+\cdots+\theta)=(r^n,n\theta)\,\,\bullet$$

De Moivre's Theorem for Integer Powers

Suppose that $z=(r,\theta)$ and $n\in\mathbb{Z}$. Then

$$z^n=(r^n,n\theta).$$

Proof : The case of $n\geq 1$ is covered by the last theorem. If $n=0$ we need

$$z^0=(r^0,0\theta).$$

But $z^0=1$, $r^0=1$ and $0\theta=0$ so we just have to show

$$1=(1,0),$$

which is true (draw a diagram).

Now suppose $n<0$, say $n=-m$ for $ m\in\mathbb{N}$.

$$z^{n}=z^{-m}=\frac{1}{z^m}=\frac{(1,0)}{(r^m,m\theta)}.$$

We have seen already seen how to do division:

$$\frac{(1,0)}{(r^m,m\theta)}=\left(\frac{1}{r^m},0-m\theta\right)=(r^{-m},-m\theta)=(r^n,n\theta)$$

as required $ \bullet$

Roots of Unity

Recall the set of complex numbers $\{1,i,-1,-i\}$ --- which can be written in polar form $\{(1,0),(1,\pi/2),(1,\pi),(1,3\pi/2)\}$. What happens when we take any of these to the power of four?

$$(1,0)^4=(1^4,4(0))=1,$$

$$(1,\pi/2)^4=(1,4\pi/2)=(1,2\pi)=(1,0)=1,$$

$$(1,\pi)^4=(1,4\pi)=(1,0)=1,$$

$$(1,3\pi/2)^4=(1,4(3\pi/2))=(1,6\pi)=(1,0)=1.$$

So these are all numbers such that when we raise them to the power of four, we get one --- that is they are fourth roots of $1$. A good question at this point is to ask are there any more complex numbers $z=a+ib=(r,\theta)$ such that $z^4=1$? The natural setting for this question is back in the algebra picture --- it is a deep theorem that every complex polynomial has a root. Now we can use the factor theorem:

Factor Theorem

Suppose that $p:\mathbb{C}\rightarrow\mathbb{C}$ is a polynomial. Then $k\in\mathbb{C}$ is a root of $p$ if and only if $(z-k)$ is a factor of $p$.

Proof : Here http://irishjip.wordpress.com/2010/09/08/an-inductive-proof-of-the-factor-theorem/ $\bullet$

The fact that a polynomial $p$ has a root $k\in\mathbb{C}$ implies that we can then (after polynomial long division) write it in the form

$$p(z)=(z-k)q(z),$$

where $q(z)$ is another polynomial, whose degree is one less than that of $p$.

Inductively, we then have:

Fundamental Theorem of Algebra

Suppose that $p:\mathbb{C}\rightarrow \mathbb{C}$ is a polynomial of degree $n$. Then $p$ can be written in the form

$$p(z)=c(z-\alpha_1)(z-\alpha_2)\cdots(z-\alpha_n).$$

Moreover, the $\alpha_i$ are the roots of $p$.

How is this relevant to the fourth powers of 1? Well finding solutions to $z^4=1$ is equivalent to finding roots of the polynomial $z^4-1$. The Fundamental Theorem of Algebra guarantees that there are only four such roots --- hence if we find four we know that there are no more! We will use this fact in the sequel: roots can be repeated, but if we find $n$ distinct solutions of a degree $n$ polynomial, we know that there are no more solutions.

Now we can get back to roots of unity. We now know that the $q$th roots of unity are solutions of the equation

$$z^q=1.$$

Let $z=(r,\theta)$:

$$(r,\theta)^q=(r^q,q\theta)\overset{!}{=}(1,0).$$

Now $r\geq0$ so the only solution to $r^q=1$ is $r=1$. Now which angles are such that

$$q\theta\equiv 0\text{ mod }2\pi\text{?}$$

Well this is the same as saying

$$q\theta=2k\pi\text{ for }k\in\mathbb{Z},$$

$$\theta=\frac{2k\pi}{q}\text{ for }k\in\mathbb{Z}.$$

So let's us try $k=0,1,2,\dots,q-1$

$$\theta=0,\,\frac{2\pi}{q},\,\frac{4\pi}{q},\,\frac{6\pi}{q},\dots,\frac{2(q-1)\pi}{q}.$$

We know there are no more solutions as the $q$ solutions are the $q$ roots of the degree $q$ polynomial

$$p(z)=z^k-1.$$

In fact all of this can be neatly appreciated in a picture (see link above).

Consider all of these complex numbers brought to the power of eight ($45^\circ=\pi/4$):

$$(1,k\pi/4)^8=(1,8\frac{k\pi}{4})=(1,2k\pi)=(1,0).$$

Hence to find all the $q$-th roots of unity it suffices to draw the unit circle, divide equally into $q$ sectors and these are the $q$-roots (see link above fpr picture)!

Note from our multiplication that the $q$-th roots of unity are the $q$ powers of the what we might call the principal $q$-th root of unity $\xi_q:=\frac{2\pi}{q}$ :

$$(1,2\pi/q)^k=(1,2k\pi/q)\text{ for }k=0,1,2,\dots,q-1.$$

Complex Roots: De Moivre's Theorem for Fractional Powers

It can also be shown that DeMoivre's Theorem holds for fractional powers. This is to solve equations such as

$$z^3=3-3i.$$

In a certain sense, this is the same as solving $z=(3-3i)^{1/3}$; and we can use De Moivre's Theorem to find $z$. Suppose that $z=(r,\theta)$ and we want the $q$-th root of $z$. Using De Moivre's Theorem we get the principal root of $(r\theta)$:

$$(\sqrt[q]{z},\text{Arg}(z)/q).$$

However there are three roots of the polynomial $z^3-(3-3i)$ so there should be three solutions. Where are they? Instead I propose a much slicker way of solving these equations. I write:

$$z^3=(1)(3-3i),$$

$$z=\{\xi^k_3(3-3i)^{1/3}:k=0,1,2\},$$

where $\xi_3$ is the principal cube root of unity and $(3-3i)^{1/3}$ is the principal root of $3-3i$. A quick sketch of $3-3i$ and calculation shows that $(3-3i)=(3\sqrt{2},7\pi/4)$. A root is a solution of the equation:

$$(r,\theta)^3=(\sqrt{18},7\pi/4),$$

$$(r^3,3\theta)=(\sqrt{18},7\pi/4),$$

$$\Rightarrow ((18^{1/2})^{1/3},7\pi/12).$$

Now the trick is that

$$\xi^k(\sqrt[6]{18},7\pi/12)\text{ for }k=0,1,2.$$

are three distinct solutions to the equation --- hence three distinct roots of the degree three polynomial --- and therefore they are all of them.

Hence, extrapolating, the solutions of $z^q=w$ are:

$$\{\xi_q^k(\sqrt[q]{|w|},\text{Arg}(w)/q)\,:\,k=0,1,2,\dots,q-1\}.$$

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Only getting used to it here --- might edit it tomorrow. This answer is far too long (just copied and pasted from something I did before) --- the crucial point is that $n$ consecutive values of $k$ gives $n$ distinct answers. The fundamental theorem of algebra guarantees that there are no more. –  Jp McCarthy Nov 23 '11 at 1:00
    
Wow, thanks for taking the time to write such a long explaination, need to take some time digesting it :) I may not understand fully after that but I think, thanks anyways –  Jiew Meng Nov 23 '11 at 1:03

Nutshell answer: When complex numbers are involved, $x^y$ is multi-valued whenever $y$ is not an integer. In particular, when $y$ is rational and $y=m/n$ in lowest terms, then $x^y$ has $n$ different values. In your case $m=1$, and you get the $n$ different values by plugging in different $k$s.

(If $y$ is not rational, then $x^y$ has infinitely many values).

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