$(z-1)^5 = z^5$ for complex $z$ So the question asks to find $z \in \mathbb{C}$ such that $(z-1)^5 = z^5$.
I argued that:
$(\frac{z}{z-1})^5 = 1 = e^{2k \pi i}$
so $\frac{z}{z-1} = e^{\theta i}$ 
where $\theta = \frac{2k \pi}{5}$
so rearranging gives
$z = \frac{e^{i \theta}}{e^{i \theta} -1} = \frac{e^{i \theta}}{e^{i \phi}(e^{i \phi} - e^{-i \phi})}$ where $\phi = \frac{\theta}{2} = \frac{k \pi}{5}$
$z = \frac{e^{i \phi}}{2i \sin (\phi)} = \frac{ \cos \phi + i \sin \phi}{2i \ sin \phi} = \frac{1}{2} - \frac{i}{2} \cot \phi = \frac{1}{2} - \frac{i}{2} \cot \frac{k \pi}{5}$ for k = 1, ...4
However the answer says that it should be +cot, not -, but I can't see where I have dropped a negative.
Can anyone help me?
Many thanks

Nothing has been dropped and everything is  fine.  Since
\begin{align*}
\cot\left(\frac{\pi}{5}\right)
&=\frac{\cos\left(\frac{\pi}{5}\right)}{\sin\left(\frac{\pi}{5}\right)}
=\frac{-\cos\left(\frac{4\pi}{5}\right)}{\sin\left(\frac{4\pi}{5}\right)}
=-\cot\left(\frac{4\pi}{5}\right)\\
\cot\left(\frac{2\pi}{5}\right)
&=\frac{\cos\left(\frac{2\pi}{5}\right)}{\sin\left(\frac{2\pi}{5}\right)}
=\frac{-\cos\left(\frac{3\pi}{5}\right)}{\sin\left(\frac{3\pi}{5}\right)}
=-\cot\left(\frac{3\pi}{5}\right)
\end{align*}

we have
\begin{align*}
\left\{\frac{1}{2}\color{blue}{-}\frac{i}{2}\cot\left(\frac{k\pi}{5}\right)\Bigg|1\leq k \leq 4\right\}
=\left\{\frac{1}{2}\color{blue}{+}\frac{i}{2}\cot\left(\frac{k\pi}{5}\right)\Bigg|1\leq k \leq 4\right\}
\end{align*}

