Prove that if $T^3$ = $T^4$, then $T^2 = T$ and there exists a polynomial $f(x) \in P(C)$ such that $T^* = f(T)$ Let V be a finite-dimensional complex inner product space and $T \in L(V)$ a normal operator. Prove that 
(i) If $T^3$ = $T^4$, then $T^2 = T$
(ii)There exists a polynomial $f(x) \in P(C)$ such that $T^* = f(T)$
$P(C)$ is the polynomials over complex field$
For (i), I am thinking that since V is complex inner product space, eigenvalues exist so let v be an eigenvector of T and so $T^3(v) = T^4(v)$ and hence $\lambda^3(v) = \lambda^4(v)$ and so by doing cancellation, I will get $\lambda^2(v) = \lambda(v)$ and hence $T^2 = T$. I have a feeling that it is not quite right, but I don't really know how to do it.
The second part, I have no idea at all.
 A: Both parts follow from the spectral theorem for normal operators, but since I did not see this theorem until well after I had taken linear algebra, I would like to present a proof that doesn't explicitly cite the spectral theorem (essentially by proving it in the finite dimensional case).  We begin with my favorite observation in linear algebra.
Lemma: If $AB=BA$, then $B$ preserves the eigenspaces of $A$.
Proof: If $Av=\lambda v$, then $A(Bv)=B(Av)=B(\lambda v)=\lambda (Bv)$.
In what follows, $V$ will be a complex finite dimensional vector space with a hermetian inner product, and $T:V\to V$ will be an operator.
Lemma: If $T$ is normal, then there is a nonzero vector $v\in V$ such that $v$ is an eigenvector of both $T$ and $T^*$.  
Proof: Since $\mathbb C$ is algebraically closed, the characteristic polynomial of $T$ has a root, $\lambda$, and the corresponding eigenspace $V_{\lambda}=\{v\in V \mid T(v)=\lambda v\}$ is nontrivial.  By the previous lemma, $V_{\lambda}$ is invariant under $T^*$ because $T$ is normal, and when we restrict $T^*$ to $V_{\lambda}$, we can find an eigenvector for $T^*$.  This is an eigenvector for both $T$ and $T^*$.
Lemma: If $v$ is an eigenvector of both $T$ and $T^*$, then $v^{\perp}=\{w\in V \mid \langle v,w \rangle =0\}$ is preserved by both $T$ and $T^*$.  
Proof:  Let $w\in v^{\perp}$.  Then $\langle v,Tw \rangle = \langle T^*v,w \rangle = \langle \lambda v,w \rangle= 0$ (where $\lambda$ corresponding eigenvalue).  Similarly with $T$ replaced by $T^*$.  
Theorem: If $T$ is normal, then $V$ has an orthogonal basis of vectors which are each simultaneously eigenvectors of both $T$ and $T^*$.  
Proof: We induct on the dimension of $V$.  The statement is trivial if $V$ is $1$-dimensional.  Assume it holds true in $n$ dimensions.  Let $V$ be $n+1$ dimensional.  By our second lemma, we can find a $v$ such that $v$ is an eigenvector of both $T$ and $T^*$.  By our third lemma, $T$ and $T^*$ restrict to $v^{\perp}$, which by our induction hypothesis has a basis of orthogonal common eigenvectors.  Adding $v$ completes this to an orthogonal basis for all of $V$.

Now, let's answer your questions by making use of our eigenbasis.
(i) If $v$ is an eigenvector of $T$ with eigenvalue $\lambda$, then $(\lambda^4-\lambda^3)v=0$, and so $(\lambda^4-\lambda^3)=0$.  However, since we can factor $x^4-x^3=x^3(x-1)$, and since $\mathbb C$ is an integral domain (in fact, a field!), we have
$$ \lambda^3(\lambda-1)=0 \Leftrightarrow \lambda =0\text{ or }\lambda = 1 \Leftrightarrow \lambda(\lambda-1)=0. $$
Therefore, on each of our eigenvectors, $T^2v=Tv$.  If two linear operators agree on a basis, they agree on the entire vector space.
(ii) Here, we use not just that we have an eigenbasis for $T$, but that this basis is also an eigenbasis for $T^*$.  Let $(v_i)$ be the basis, with $Tv_i=\lambda v_i, T^* v_i=\mu_i v_i$.
Lemma If all the $\lambda_i$ are unique, then there exists a polynomial $P$ such that $P(\lambda_i)=\mu_i$ for all $i$.
Proof: Let $f_i(x)=\prod_{j\neq i} (x-\lambda_j)$, and let $g_i(x)=f_i(x)/f_i(\lambda_i)$.  Then $g_i(\lambda_j)=1$ if $i=j$ and $0$ otherwise.  Then $P=\sum \mu_i g_i$ is the desired polynomial.
Using the above interpolation lemma, we can make $P(T)v=T^*v$ on a basis, and therefore on all of $V$, assuming that $T$ has no repeated eigenvalues.  In fact, we can do slightly better: by dropping repeated eigenvalues, we only need that whenever $\lambda_i=\lambda_j$, we also have that $\mu_i=\mu_j$.  However, this follows from calculating 
$$\langle \lambda_i v_i, v_i \rangle =\langle Tv_i,v_i \rangle = \langle v_i, T^* v_i \rangle = \langle v_i, \mu_i v_i \rangle$$
from which it follows that $\mu_i = \overline{\lambda_i}$.
A: For the second part, use that the spectrum of $T$ is a finite set.
On this finite set, you can solve an arbitrary interpolation problem, i.e. for any given values $z_i$ for $\sigma(T)=\{w_1,\dots,w_m\}$, you can find a polynomial $p$ such that $p(w_i)=z_i$. Why does that help you? (Read on for hints to the first part, these are also helpful for the second one).
For the first part, everything you write is fine. More precisely, since $T$ is normal, it can be diagonalized (spectral theorem), i.e. $T =UDU^{-1}$ for some invertible (even unitary $U$) and some diagonal matrix $D$. Now calculate $T^n$ using this expression for $T$.
A: The relation $T^3=T^4$ implies that all the eigenvalues of $T$ belong to the set $\{0,1\}$.
To see this, assume that $Tu=\lambda u$, then $$0=(T^3-T^4)u=(\lambda^3-\lambda^4)u=0.$$
Hence $\lambda^3-\lambda^4=0.$
Next, every normal operator is orthogonally diagonalizable, and thus $T=U^{-1}DU$, where
$U$ is orthogonal, i.e., $U^{-1}=U^*$,
and $D=\mathrm{diag}(d_1,\ldots,d_n)$, with $d_i\in\{0,1\}$.
Thus, 
$$
T^2-T=U^{-1}(D^2-D)U=U^{-1}\mathrm{diag}(d^2_1-d_1,\ldots,d_n^2-d_n)U=0.
$$
But $T^*=(U^*DU)^*=U^*D^*(U^*)^*=U^*DU=T$. Hence, no need for a polynomial.
Note. In general however, if $T$ is normal, then  $p(T)=U^*p(D)U$, for every polynomial.
So if $D=\mathrm{diag}(d_1,\ldots,d_n)$, then  $D^*=\mathrm{diag}(\overline{d}_1,\ldots,
\overline{d}_n)$, and it suffices to find a polynomial $p$ satisfying
$$
p(d_j)=\overline{d}_j, \quad j=1,\ldots,n.
$$
Such does exists, doing a standard interpolation.
