Finding a particular non real value I'm trying to give a counter example in an argument I am making, and it boils down to showing that for any nonzero $\alpha\in\mathbb{C}$ and for any $N\in\mathbb{Z}$, there are points $z\in S^1$ so that 
$$\alpha z^N - i\alpha z^{N-1}\notin\mathbb{R}$$
I'm sure there's a slick argument by way of contradiction with some complex analysis, but I was hoping for a solution in terms of $\alpha$ and $N$
 A: Here's a slick argument (because why not?).
Assume for the sake of contradiction that for all $z = \operatorname{cis}(\theta) \in S^1$,
$$\alpha \operatorname{cis}(N\theta) - i\alpha \operatorname{cis}((N-1)\theta) \in \mathbb{R}$$
Then we have a continuous map $\phi : [0, 2\pi] \to \mathbb{R}$ mapping $z \mapsto \alpha z^{N} - i\alpha z^{N-1}$. Moreover, $\phi$ is differentiable, and $\phi(0) = \phi(2\pi)$. By Rolle's Theorem, there must be a $\theta \in [0, 2\pi]$ such that $\phi'(\theta) = 0$. That is,
$$\phi'(\theta) = iN\alpha \operatorname{cis}(N\theta) + (N-1)\alpha\operatorname{cis}((N-1)\theta) = 0$$
This easily implies $|\phi'(\theta)| = 0$.
But using the Reverse Triangle Inequality, $|a+b| \geq |a| - |b|$,
$$
\begin{align}
|\phi'(\theta)| &= \\
|iN\alpha \operatorname{cis}(N\theta) + (N-1)\alpha\operatorname{cis}((N-1)\theta)| &\geq \\
|iN\alpha \operatorname{cis}(N\theta)| - |(N-1)\alpha \operatorname{cis}((N-1)\theta)| &= \\
N|\alpha| - (N-1)|\alpha| &= \\
|\alpha|.
\end{align}
$$
Therefore $|\phi'(\theta)| = 0 = |\alpha|$, implying $\alpha = 0$, contradiction. 
A: Alt.hint:   by homogeneity in $\,\alpha\,$, it is enough to prove the proposition for $\,|\alpha|=1\,$.


*

*If $\,n=1\,$ then (at least) one of $\,z=1\,$ or $\,z=-1\,$ must satisfy $\,\alpha(z-i) \not\in\mathbb{R}\,$.

*Otherwise, if $\,\alpha \in \{\pm1, \pm i\}\,$ then $\,z=1\,$ satisfies $\,\alpha - i\alpha \not\in \mathbb{R}\,$.

*Otherwise, any of the $\,n-1\,$ solutions to $\,z^{n-1} = \overline \alpha\,$ satisfies: 
$$\alpha z^n - i\alpha z^{n-1}=|\alpha|^2 z - i|\alpha|^2 = z-i$$
Since $\,|z|=1\,$ it follows that $\,\operatorname{Im}(z) = i \iff z=i\,$, but that gives $\,\overline \alpha = i^{n-1} \in \{\pm1, \pm i\}\,$, which has been handled separately at the previous step. Therefore $\,\operatorname{Im}(z) \ne i\,$, so $\,z-i \not\in\mathbb{R}\,$.
