Division of $ f= X^4+X^3+X^2+X+2$ by $g(X)=X-\cos(\alpha)+i \sin(\alpha)$ We have the polynomial $ f= X^4+X^3+X^2+X+2$ with $f\in \Bbb C[X] $, it asks to determine the quotient of the division of the polynomial $f$ by the polynomial $g$, $g(X)=X-\cos(\alpha)+i \sin(\alpha) \in \Bbb C[x] $, $α \in(0,π/2)$, knowing that $r$(remainder)${}=1+i(1+\sqrt{2})$. Now, what I've tried is doing long division but it seems like that might not be the first step to start with. So I was looking for a solution.

By remainder theorem,
$$1 + i(1 + \sqrt{2}) = f(\cos \alpha - i \sin \alpha) = f(e^{-i\alpha}).$$
Therefore
\begin{align*}
&(e^{-i\alpha})^4 + (e^{-i\alpha})^3 + (e^{-i\alpha})^2 + e^{-i\alpha} + 2 = 1 + i(1 + \sqrt{2}) \\
\iff \, &(e^{-i\alpha})^4 + (e^{-i\alpha})^3 + (e^{-i\alpha})^2 + e^{-i\alpha} + 1 = i(1 + \sqrt{2}) \\
\iff \, &\frac{1 - e^{-5i\alpha}}{1 - e^{-i\alpha}} = i(1 + \sqrt{2}). \tag{$\star$}
\end{align*}
Now, let's switch to geometry. Let $O$ be the origin, $P$ be the point $e^{-5i\alpha}$, $Q$ be the point $e^{-i\alpha}$, and $R$ be the point $1 + 0i$. Note that $P, Q, R$ all lie on the circle of radius $1$ with centre $O$. The triangle $PRQ$ contains a right angle at $R$, and is contained in this circle, which implies, by circle geometry, that $PQ$ is a diameter. Specifically, this tells me that
$$e^{-5i\alpha} = -e^{-i\alpha},$$
or in other words,
$$e^{4i\alpha} = -1 = e^{i\pi}.$$
Solving this in the usual way, we get four possible solutions:
$$\alpha = \pm \pi/4, \pm 3\pi/4.$$
I didn't end up using all the information, so I think some of these are false solutions. If you substitute them into $(\star)$, you'll find that
$$\alpha = -\frac{\pi}{4}$$
is the only solution.
