Let $n\in\mathbb{N}$. We consider, in $\mathbb R[X]$, the polynomial $$C_{n}=X(X+1)^{2n}-2^nX.$$ Determine the values of $n$ for which the complex number $i$ is a root of $C_n$.
My thoughts: note that $$(1+i)^2=2i$$
$i$ is root of $C_n$ then \begin{align} C_{n}(i)&=0\\ i(i+1)^{2n}-2^ni&=0\\ i(2i)^{n}-2^ni&=0 \end{align} if $n$ is even then $\exists k \in \mathbb{N}$ such that $n=2k$ then $$i(2i)^{(2k)}-2^{2k}i=0$$
if $n$ is odd then $\exists k \in \mathbb{N}$ such that $n=2k+1$ then $$i(2i)^{(2k+1)}-2^{2k+1}i=0$$
update
can we see that n multiple of 4 through $$\text{arg} ( (1+i)^{2n} )=n \text{arg}(2i)=n \frac{\pi}{4}$$
You're on the right track. To continue, write $$ C(i)=i(2i)^{n}-2^ni=i2^n(i^n-1) $$ It remains to find those $n$ such that $i^n-1=0$, that is, $i^n=1$: they are exacty the multiples of $4$, because the powers of $i$ repeat in a cycle $1,i,-1,-i$.