Every matrix $A\in M_2(\mathbb{C})$ is similar to one of two forms I'm in trouble at the following exercise:

(Ex. 6, page 143 - Um Curso de Álgebra Linear; Coelho, Flávio Ulhoa)
  Show that if $A\in M_2(\mathbb{C})$, then $A$ is similar (or conjugated) to a matrix of one of the following forms
  $$\left(\begin{array}{cc}a&0\\0&b\end{array}\right) \text{ or } \left(\begin{array}{cc}a&0\\1&a\end{array}\right), \text{ with } a,b\in\mathbb{C}.$$

What I've tried: Be $A\in M_2(\mathbb{C})$. Consider $T:\mathbb{C}^2\to\mathbb{C}^2$ the linear operator whose matrix at the standard basis $\mathscr{C}=\{(1,0),(0,1)\}$ of $\mathbb{C}^2$ is $A$, i.e., $[T]_\mathscr{C}=A$. Since $\mathbb{C}$ is algebraically closed and we know that the characteristic polynomial of $T$ is monic and has degree 2, we have that it is of the form $p_T(x)=(x-a)(x-b)$, for some $a,b\in \mathbb{C}$. Of course $a$ and $b$ are eigenvalues of $T$.  
Case 1: $a\neq b$; Considering the dimensions of the eigenspaces of $T$, we conclude that $T$ is diagonalizable and, therefore, exists a basis $\mathscr{B}$ of $\mathbb{C}^2$ such that
$$[T]_{\mathscr{B}}=\left(\begin{array}{cc}a&0\\0&b\end{array}\right),$$
and, through basis exchange, $A$ is similar to a matrix of the first kind.
Case 2: $a=b$ and $p_T(x)=(x-a)^2$
In this case, if the geometric multiplicity $m_g(a)=2$ then, again, there is a basis $\mathscr{B}$ of $\mathbb{C}^2$ of eigenvectors of $T$ and 
$$[T]_{\mathscr{B}}=\left(\begin{array}{cc}a&0\\0&a\end{array}\right).$$
Now, if $m_g(a)=1$, then the eigenspace $Eig_T(a)$ spanned by the eigenvectors associated to the eigenvalue $a$ has dimension one, say, $Eig_T(a)=[v]$. And then, completing a basis $\mathscr{B}=\{u,v\}$ of $\mathbb{C}^2$, I only know that
$$[T]_{\mathscr{B}}=\left(\begin{array}{cc}a_1&0\\a_2&a\end{array}\right).$$
From here, I've tried to find some suitable $u$ such that $T(u)=au+v$ in a way that $a_1=a$ and $a_2=1$, but solving this system seems impossible (in general) to me! Is this right?
Suggestions? Ideias? Thank you once more, guys...
 A: I think I've got. Just like @DustanLevenstein has said in the comment above, $a_1=a$.
Observe now that $(v)_\mathscr{B}=(0,1)$.
 And, by definition of $[T]_{\mathscr{B}}$, $T(u)=au+a_2v$ and, therefore
$$au-T(u)=-av$$
and writing this on its coordinates relative to $\mathscr{B}$,
$$[aI-T]_\mathscr{B}[u]_\mathscr{B}=-[v]_\mathscr{B}$$
$$\iff \left(\left(\begin{array}{cc}a&0\\0&a\end{array}\right)-\left(\begin{array}{cc}a&0\\a_2&a\end{array}\right)\right)\left(\begin{array}{c}1\\0\end{array}\right)=\left(\begin{array}{c}0\\-1\end{array}\right)$$
$$\iff \left(\begin{array}{cc}0&0\\-a_2&0\end{array}\right)\left(\begin{array}{c}1\\0\end{array}\right)=\left(\begin{array}{c}0\\-1\end{array}\right)$$
$$\iff a_2=1$$
and we are done. ^_^
THIS IS ALL WRONG!  
CORRECTION:
By definition of $[T]_\mathscr{B}$ we have that $T(u)=au+a_2v$, and $T(v)=av$.
See that if $a_2=0$ the matrix is of the first kind. Suppose then $a_2\neq 0$ and consider $\mathscr{B}'=\{u,a_2v\}$. Is not hard to see that $\mathscr{B}'$ is basis of $\mathbb{C}^2$ and, since $T(u)=av+1\cdot(a_2v)$ and $T(a_2v)=a_2T(v)=a_2av=a(a_2v)$, we have
$$[T]_{\mathscr{B}'}=\left(\begin{array}{cc}a&0\\1&a\end{array}\right)$$ and, NOW, we are done. Thank you @DustanLevenstein for correcting me!
