# Complex n-th roots

1. Jan 19, 2012

### autre

1. The problem statement, all variables and given/known data

Suppose z is a nonzero complex number $z=re^{i\theta}$ . Show that z has exactly n distinct complex n-th roots given by $r^{(1/n)}e^{i(2\pi k+\theta)/n}$ for $0\leq k\leq n-1$.

3. The attempt at a solution

My attempt: $z^{n}=(r\cos\theta+i\sin\theta)^{n}=r^{m}(\cos \theta+i\sin\theta)^{n}=r^{m}(\cos(n \theta)+i\sin(n \theta))=r^{m}e^{i\theta n}$ ...Not sure where to go from here.

2. Jan 19, 2012

### A. Bahat

Note that you took $z$ to the $n$th power; you don't want to do that. Rather, you want to find $a\in\mathbb{C}$ such that $a^n=z$ (see the difference?). To do this, put $a$ in exponential form also, i.e. write $a=\rho e^{i\varphi}$. Now you have to find $\rho$ and $\varphi$ such that
$$\rho^n(\cos(n\varphi)+i\sin(n\varphi))=r(\cos(θ)+i\sin(\theta)).$$

3. Jan 19, 2012

### autre

So I take $a=\rho e^{i\varphi} \rightarrow a^{n}=(\rho e^{i\varphi})^{n}=\rho^{n}(\cos(n\varphi)+i\sin(n \varphi)$. Let $\rho^{n}=r so r^{1/n}=\rho$. Let $n\varphi=\theta\rightarrow\varphi=\theta/n$ . Then roots a have the form $r^{1/n}(\cos(\frac{\theta}{n})+i\sin(\frac{\theta}{n})) \rightarrow r^{1/n}e^{i/n}$... not sure if I'm doing this the right way?

Last edited: Jan 19, 2012
4. Jan 19, 2012

### A. Bahat

That looks good (don't forget the $i$ in that last trigonometric expression for $a$). Now you have one of the roots. To find the other $n-1$ of them, exploit the periodicity of the trigonometric or exponential functions. That is, either use $e^{i(θ+2\pi)}=e^{iθ}$ or use $\sin(θ+2\pi)=\sin(θ)$ and $\cos(θ+2\pi)=\cos(θ)$ (it really doesn't matter which you use). The key thing here is that, while $\rho$ must equal $r^{1/n}$, there is more than one angle that works; in fact, there are $n$ of them.

5. Jan 19, 2012

### autre

Thanks for the insight -- didn't really understand the intuition behind there being $n$ roots before your post. Just one more thing, when I use periodicity I get:

$r^{1/n}(\cos(\frac{\theta}{n})+i\sin(\frac{\theta}{n}))=r^{1/n}(\cos(\frac{\theta}{n}+2\pi k)+i\sin(\frac{\theta}{n}+2\pi k))=r^{1/n}e^{i(\frac{\theta+2\pi kn}{n})}.$

The exponent in that last term is $i(\frac{\theta+2\pi kn}{n})$ where it should be $i(\frac{\theta+2\pi k}{n})$. Since n is in N and k is in Z, can I define j in Z s.t. j = kn and use that?

6. Jan 19, 2012

### A. Bahat

You're right that the exponent should be $i(θ+\frac{2πk}{n}).$ The reason for this is that the angle in the exponent doesn't have to be a multiple of $2\pi$. Rather, it is the angle times $n$ that should be a multiple of $2\pi$, so that when you raise $a$ to the $n$th power, then you get that period which leaves your answer unchanged.

7. Jan 19, 2012

### A. Bahat

One final remark: to interpret all of this geometrically, notice that the n nth roots of z are the vertices of an n-gon in the complex plane, where the modulus of each point is r1/n. This is the way I usually think of the different possible angles.

8. Jan 19, 2012

### autre

Thanks A. Bahat!

9. Jan 20, 2012

### A. Bahat

You're welcome! Happy to help.