Incorrect use of complex sine causes $\sin\frac {4\pi}{9}$ to evaluate to $0$

Question

I would like to know where I have gone wrong here (apologies for any inconsistent notation)

$$\begin{align} \sin\left(\frac {4\pi}{9}\right) &= \frac{1}{2i}\left( e^{ i\,\frac{4\pi}{9}} - e^{-i\,\frac{4\pi}{9}} \right) \tag1\\[4pt] &= \frac{1}{2i}\left( e^{i\pi\,\frac{4}{9}} - e^{i\pi\,\frac{-4}{9}} \right) \tag2 \\[4pt] &= \frac1{2i}\left((-1)^{4/9} - (-1)^{-4/9} \right) \tag3 \\[4pt] &= \frac1{2i}\left(\sqrt[9]{1} - \sqrt[9]{1} \right) \tag4 \\[4pt] &= \frac{0}{2i} \tag5 \\[4pt] &= 0 \tag6 \end{align}$$

Answer 1

Using the original definition of

$$\sin \theta = \dfrac {e^{i \theta} - e^{-i \theta}}{2i}$$ $\theta$ here is the argument/angle and cannot be split up similar to the real number identity $e^{ab} = e^a \cdot e^b$, since complex numbers can be multivalued.

When you attempted to use $e^{i \pi (4/9)}$, it is not true that this is equal to $-1^{4/9}$, even though $e^{i \pi}$ does equal $-1$. The angle/argument $4\pi/ 9$ cannot be split out; if we tried to use the exponent definition, we'd have $e^{i \pi} e^{4/9}$, which is definitely not true.

Thus, when representing complex numbers in exponential form, the correct representation is either $e^{i (4\pi /9)}$ or $e^{4\pi i/9}$.