To find the $n^\text{th}$ roots of a complex number, we can first express it in polar form (I'm assuming $r=1$ for brevity; it doesn't matter for my question):

\begin{align} e^{i\theta} &= \cos\theta + i\sin\theta \\ &= \cos\left(\theta + 2\pi k\right) + i\sin\left(\theta + 2\pi k\right), \qquad k\in\mathbb{Z} \end{align}

and then use De Moivre's formula:

\begin{align} \left(e^{i\theta}\right)^\frac{1}{n} &= \cos\left(\frac{\theta + 2\pi k}{n}\right) + i\sin\left(\frac{\theta + 2\pi k}{n}\right), \qquad k\in\mathbb{Z} \end{align}

However, when we express a complex number in polar form we could also shift the $\cos$ and $\sin$ arguments by different multiples of $2\pi$; this is certainly still valid: \begin{align} e^{i\theta} &= \cos\theta + i\sin\theta \\ &= \cos\left(\theta + 2\pi a\right) + i\sin\left(\theta + 2\pi b\right), \qquad a,b\in\mathbb{Z} \end{align}

but it does it not follow that:

\begin{align} \left(e^{i\theta}\right)^\frac{1}{n} &= \cos\left(\frac{\theta + 2\pi a}{n}\right) + i\sin\left(\frac{\theta + 2\pi b}{n}\right), \qquad a,b \in\mathbb{Z} \end{align}

This is obviously false; for example, taking $\theta = 0$, $n=3$, and setting $a=1$, $b=0$, we get $-0.5 + 0i$ as a cube root of unity.

I know that the answer has to be "this is not a correct application of De Moivre's formula", but an explanation of why this is the case would be nice.

  • $\begingroup$ I think because $\cos\frac{\theta+2\pi}3 + i\sin\frac{\theta}3$ is no longer a valid polar form. $\endgroup$ – peterwhy Jul 8 '15 at 21:36
  • $\begingroup$ The DeMoivre formula is for calculating $n$-th powers, not $n$-th roots. $\endgroup$ – André Nicolas Jul 8 '15 at 21:38
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    $\begingroup$ There are $n$ $n$-th roots of unity, and different choices give different roots. Your argument is similar to the ones that "prove" that $-1=1$ by equating two square roots of $1$. $\endgroup$ – Dilip Sarwate Jul 8 '15 at 22:08

Since $z=\cos\frac{\theta+2\pi}3 + i\sin\frac{\theta}3$ is no longer a valid polar form, this $z$ cannot be considered as an answer of the following question:

Find $z$ such that $z^n =\cos\theta + i\sin\theta$.

Because de Moivre's formula only applies to polar forms: $$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$$


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