Consider the function $f:\mathbb{C}\rightarrow\mathbb{C}$ defined by $$\tag{1} f:=e^{iz}-e^{i(\frac{2\pi}{3}-z)} $$
I want to show that $1/f:D'(\pi/3,\pi)\rightarrow \mathbb{C}$ has a pole of order $1$ at $\pi/3$.
My procedure so far:
I prefer moving the pole to zero, so define the funcion $\tilde{f}:D'(0,\pi)\rightarrow\mathbb{C}$ by
$$\tag{2}
\tilde{f}(z):=f(z+\pi/3)=e^{i\left(\frac{\pi}{3}+z\right)}-e^{i\left(\frac{\pi}{3}-z\right)}
$$
I then did the expansions of each term in $\tilde{f}$:
$$
\begin{aligned}
&e^{i\left(z+\frac{\pi}{3}\right)}=e^{i z} e^{i \pi / 3}=(-1)^{1 / 3}+(-1)^{5 / 6} z-\frac{1}{2}(-1)^{\frac{1}{3}} z^{2}-\frac{1}{6}(-1)^{\frac{5}{6}} z^{3}+\mathcal{O}(z)^{4} \\
&e^{i\left(\frac{\pi}{3}-z\right)}=e^{\frac{i \pi}{3}} e^{-i z}=(-1)^{1 / 3}-(-1)^{5 / 6} z-\frac{1}{2}(-1)^{\frac{1}{3}} z^{2}+\frac{1}{6}(-1)^{\frac{5}{6}} z^{3}+\mathcal{O}(z)^{4}
\end{aligned}
$$
So that the expansion of $\tilde{f}$ becomes
$$\tag{4}
\tilde{f}=2(-1)^{5 / 6} z-\frac{1}{3}(-1)^{\frac{5}{6}} z^{3}+\mathcal{O}(z)^{4}
$$
My lecture book states that a function holomorphic in a punctured disk has a pole of order $m$ at $a$ if and only if $$ \lim _{z \rightarrow a}\left\{(z-a)^{m} f(z)\right\}=D $$ where $D$ is a non-zero constant. For $1/\tilde{f}$: $$ \lim _{z \rightarrow 0}\left\{\frac{z}{2(-1)^{5 / 6} z-\frac{1}{3}(-1)^{\frac{5}{6}} z^{3}+\mathcal{O}(z)^{4}}\right\} $$ Since higher-order terms go to zero quicker than $z$ we are 'eventually' left with $$ \lim _{z \rightarrow 0}\left\{\frac{z}{2(-1)^{5 / 6} z}\right\}=\frac{1}{2(-1)^{5 / 6}}. $$ Since this is a non-zero constant (the residue), we can conclude that $1/\tilde{f}$ has a pole of order $m=1$ at $0$. Is there anything wrong with this last argument/proof?