Let $U_n$ be the subset of $2\times2$ matrices $M_2(\mathbb C)$ such that $T^n=I$ Fixed $n>0$ let $U_n$ be the subset of $2\times2$ matrices $M_2(\mathbb C) \simeq \mathbb C^4 \simeq \mathbb A_{\mathbb C}^4$ such that $T^n=I$.
a) Show that $U_n$ is a Zarisky closed subset of $M_2(\mathbb C)$.
b) Describe the irriducible components of $U_n$ and show that there are $ \binom {n+1} {2}$ of them.
Partial Solution:
a) $U_n$ is given by an ideal $I \subset \mathbb C[a,b,c,d]$ generated by at most 4 polynomials. We are done.
b) ??? I can diagonalize every matrix with eigenvalues $\alpha$, $\beta$ $n$-th roots of $1$. Can't tell more.
 A: For part b, use the characteristic polynomial and examine how it changes on various classes of matrices. Full solution below.

 The assignment of characteristic polynomials is constant on similarity classes of matrices. Using your diagonalization suggestion, every matrix satisfying $T^n=I$ is similar to $A_{p,q}= \begin{pmatrix} \zeta^p & 0 \\ 0 & \zeta^q\end{pmatrix}$ for integers $0\leq p,q \leq n-1$ and $\zeta$ a primitive $n^{th}$ root of unity.

Further,

 The characteristic polynomial of $A_{p,q}$ is $\chi_{p,q}(x)=x^2-(\zeta^p+\zeta^q)+\zeta^{p+q}$. How many possibilities are there for this polynomial based on different choices of $p$ and $q$? If $p\neq q$, then $\chi_{p,q}=\chi_{q,p}$ and no other choices will work, so for $p\neq q$ there are $n(n-1)/2=\frac{n^2-n}{2}$ possible characteristic polynomials. If $p=q$, then there are no choices that are not identical which give the same polynomial, so there are $n$ possible polynomials this way.

And...

 Putting these two cases together, we get $\frac{n^2-n}{2}+n=\frac{n^2-n+2n}{2}=\frac{n^2+n}{2}=\frac{(n+1)n}{2}=\binom{n+1}{2}$.

