Solving (z^N-a^N) = 0 for a^N<0

[WolfOfTheSteps]

Homework Statement

Find the roots for

$$(z^N-a^N) = 0$$

when $a^N<0$ is allowed.

Homework Equations

No idea.

The Attempt at a Solution

I know that the roots are

$$z_k=ae^{i(2\pi k/N)}, k = 0, 1, ... , N-1.$$

How do I find that solution? I'm reading a signals book, and it assumed that this is obvious. It should also be obvious that this is not a homework problem, so if possible please just tell me what I'm missing without hints. I'm studying and in a rush. Thanks!

 

[Dick]

exp(i*2*pi*k/N)^N=exp(2i*pi*k)=1 for k an integer. Take z_k to the Nth power. exp(2i*pi*k)=cos(2*pi*k)+i*sin(2*pi*k).

 

[unique_pavadrin]

couldn't you use the relationship
$$z - \frac{1}{z} = 2{\bf{i}}\sin \theta$$ and De Movire's Theorem?

 

[HallsofIvy]

Homework Statement

Find the roots for

$$(z^N-a^N) = 0$$

when $a^N<0$ is allowed.

Homework Equations

No idea.

The Attempt at a Solution

I know that the roots are

$$z_k=ae^{i(2\pi k/N)}, k = 0, 1, ... , N-1.$$

How do I find that solution? I'm reading a signals book, and it assumed that this is obvious. It should also be obvious that this is not a homework problem, so if possible please just tell me what I'm missing without hints. I'm studying and in a rush. Thanks!

There are two things you are "missing". The first is that you're formula is not correct- the N "N^th roots of a" are given by $z_k= (a^{1/N}e^{i(2\pi k)/N}$ )($a^{1/N}$ here is the "principal" real N^th root of the positive real number a, $\sqrt[N]{a}$) and the fact that you don't seem to have tried to use that formula!
$z_1= \sqrt[N]{a}$ with k= 0.
$z_2= \sqrt[N]{a}e^{2\pi i/N}$ with k= 1.
$z_3= \sqrt[N]{a}e^{4\pi i/N}$ with k= 2.
etc.

 

[WolfOfTheSteps]

There are two things you are "missing". The first is that you're formula is not correct- the N "N^th roots of a" are given by $z_k= (a^{1/N}e^{i(2\pi k)/N}$ )($a^{1/N}$ here is the "principal" real N^th root of the positive real number a, $\sqrt[N]{a}$) and the fact that you don't seem to have tried to use that formula!
$z_1= \sqrt[N]{a}$ with k= 0.
$z_2= \sqrt[N]{a}e^{2\pi i/N}$ with k= 1.
$z_3= \sqrt[N]{a}e^{4\pi i/N}$ with k= 2.
etc.

Thanks for your help, Halls. I posted this the night before a test in a cram.

But now I'm coming back to it and trying to understand, and I still don't get it! I tried using the formula, but the z_k's you have up there make no sense! I just don't see what the complex exponential is doing! And when I plug in sample values for N, that exponential evaluates to crazy things.

I think I'm having a mental block and missing something obvious. Also, was my formula really wrong? Here is exactly what the book says:

http://img444.imageshack.us/img444/432/exampleqe5.jpg

 

[Kummer]

$$z^n = a^n$$ let $$\zeta$$ be the $$n$$-th root of unity and $$w = \sqrt[n]{a} = \exp (\log (a)/n)$$ then the roots are $$w,w\zeta,...,w\zeta^{n-1}$$. (The square root here and log are the complex-valued functions).