Existence of square root matrix $B \in \mathbb{C}^{2\times 2}$ for any $A \in \mathbb{C}^{2\times 2}$, where $A^2\neq 0$ I am trying to prove that  for any $A \in \mathbb{C}^{2\times 2}$ with $A^2\neq 0$, there exists $B \in \mathbb{C}^{2\times 2}$ with $BB=A$.
I have tried the approach of a general matrix A andB with variable entries
$$B = \begin{bmatrix}
       a & b  \\          
       c & d           
     \end{bmatrix}$$
   $$A = \begin{bmatrix}
       \alpha & \beta  \\          
       \gamma & \delta           
     \end{bmatrix}$$
and assuming $BB=A$, I get the equations
$$a^2+bc=\alpha$$
$$b(a+d)=\beta$$
$$c(a+d)=\gamma$$
$$d²+cb=\delta$$
However, here I am stuck since I do not know whether any of those variables is $0$, so I cannot operate with those equations. 
I have seen a solution on Wikipedia, however to me it seems to fall from the sky, especially the restrictions it makes.
I have also found a thread, which shows that without the restriction $A^2\neq0$ this statement is false, however I fail to see how this is the critical restriction.
Explanations, clarifications or hints on any of the things I mentioned are most welcome.
 A: Either $A$ is similar to a diagonal matrix, of which we can easily find a square root. Or it is similar to an upper triangular matrix $\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$. Note that $x^2=\lambda$ implies $\begin{pmatrix}x&y\\0&x\end{pmatrix}^2=\begin{pmatrix}\lambda&2xy\\0&\lambda\end{pmatrix}$, so $y=\frac1{2x}$ gives us a solution (as $x\ne0$).
A: We have the equations 
\begin{eqnarray*}
a^2+bc= A \\
b(a+d)=B \\
c(a+d)=C \\
bc+d^2=D
\end{eqnarray*}
Multiply the first equation by $(a+d)^2$ and use the second & third we have
\begin{eqnarray*}
a^2(a+d)^2 +BC=A(a+d)^2 \\
d= -a +\sqrt{\frac{BC}{A-a^2}}.
\end{eqnarray*}
Now subtract the first & the fourth 
\begin{eqnarray*}
a^2-d^2=A-D \\
\end{eqnarray*}
Substitute for $d$ and we have
\begin{eqnarray*}
a^2-A+D= \left( -a +\sqrt{\frac{BC}{A-a^2}} \right)^2 \\
D-A-\frac{BC}{A-a^2}=-2a \sqrt{\frac{BC}{A-a^2}}
\end{eqnarray*}
Square this & we have a quadratic in $a^2$
\begin{eqnarray*}
a^4((D-A)^2+4BC)+a^2(-2A(D-A)^2-2BC(A-D)-4ABC)+A^2(D-A)^2-2ABC(D-A)+B^2C^2=0
\end{eqnarray*}
Note that this has discriminant $ \Delta=4B^2C^2(AD-BC)$. This gives
\begin{eqnarray*}
a^2= \frac{A(A-D)^2+BC(3A-D) \mp 2BC \sqrt{AD-BC}}{(a-d)^2+4BC} \\
=A-\frac{BC}{A+D \pm 2 \sqrt{AD-BC}}
\end{eqnarray*}
Once the dust has settled ... 
\begin{eqnarray*}
\sqrt{\left[
\begin{array}{cc}
A & B  \\
C & D\\
\end{array}
\right]}=\left[
\begin{array}{cc}
\sqrt{A-\frac{BC}{A+D \pm 2 \sqrt{AD-BC}} } & \frac{B}{\sqrt{A+D \pm 2 \sqrt{AD-BC}}}   \\
\frac{C}{\sqrt{A+D \pm 2 \sqrt{AD-BC}}} & \sqrt{D-\frac{BC}{A+D \pm 2 \sqrt{AD-BC}} } \\
\end{array}
\right]
\end{eqnarray*}
