I have solve an exercise that ask:
Find the minimum degree polynomial $P(x) \in \Bbb R[x] : P(1+2i) = 0 \land P(i) = i$.
The solution is quite obviously: $P(x) = \frac{1}{10}(x^2 -2x +5)(2i-1)$.
However, the exercise states that the correct solution is $P(x) = \frac{1}{10}(x^2 -2x +5)(2x-1)$.
Of course it works but it's of higher degree and this violate the request of being minimum but moreover I cannot understand how they figured it out (maybe it's a typo).
Since $1+2i$ is it zero so is $1-2i$ and thus $x^2 -2x +5$ must divide $p(x)$ so there is another real polynomial $q(x)$ such that $$p(x) = q(x)(x^2 -2x +5)$$
Now, since $p(i)=i$ we have $$i = q(i)(4-2i)\implies q(i) = {-1+2i\over 10}$$
and thus we see that if $$q(x) = {-1+2x\over 10}$$ we win, so $$ p(x) = {-1+2x\over 10}(x^2-2x+5)$$ would do the job.
Clearly $q(x)$ can not be constatnt polynomial since then we would have nonreal coefficents, thus $q(x)$ must be at least of degree $1$ which we see works.