I'm trying the following exercise:
Find a minimum degree polynomial $ P(x) \in \Bbb C[x] : P(1)=2 \land P(i-1)=i \land P(-\pi i)=0 $.
The root lead to $ P(x) = (x + \pi i)Q(x) $
now using one of the two values I can calculate values of $ Q(x) $ i.e.:
$ P(1)=(1+\pi i)Q(1)=2 $
and then:
$Q(1)=\frac{2}{1+\pi i}$
or
$Q(1) = \frac{2(1-\pi i)}{1+\pi^2}$
I can do the same with the second value:
$ P(i-1) = (1-i+\pi i)Q(i-1) = i $
$ Q(i-1) = \frac{i}{(i-1+\pi i)}$
but now I'm stuck here... I cannot understand how to fully define $ P(x) $
Your first step $$ P(x) = (x + \pi i)Q(x) $$ is definitely right. We are seeking the minimal $P(x)$ such that $P(1) = 2$ and $P(i - 1) = i$. As you say, this means that we must have $$ Q(1) = \frac{2}{1 + \pi i} $$ and $$ Q(i - 1) = \frac{i}{1 - i + \pi i} $$ Thus, if we can find a polynomial $Q(x)$ that takes on the prescribed values at $1$ and $i - 1$, we have found a candidate polynomial $P(x)$. It is clear that $Q(x)$ cannot be constant. But, as you can check, $Q(x)$ can be taken to be a linear function (think point slope form). This will yield the minimal polynomial (up to multiplication by a constant) with the properties stated above.