What Are the n Distinct Complex n-th Roots of a Nonzero Number?

[autre]

Homework Statement

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.

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.

 

[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)).

 

[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?

 

[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.

 

[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?

 

[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.

 

[A. Bahat]

One final remark: to interpret all of this geometrically, notice that the n n^th roots of z are the vertices of an n-gon in the complex plane, where the modulus of each point is r^1/n. This is the way I usually think of the different possible angles.

 

[autre]

Thanks A. Bahat!

 

[A. Bahat]

You're welcome! Happy to help.