<s>[INST] Finding complex solution to $X^2 = A$ 
Let $A=\begin{pmatrix}2&3\\4&-2\end{pmatrix}$.
(i) Find an invertible matrix $P$ such that $P^{-1}AP$ is diagonal.
(ii) Find $A^n$ (for positive integers $n$).
(iii) Find four (complex) solutions to $X^2=A$. Show that there are no other solutions.

I'm having troubles with coming up with a solution to part (iii), an idea was to use:
$$ X^2 = A \iff P^{-1}X^2P = P^{-1}AP = D$$ and note that $P^{-1}X^2P = (P^{-1}XP)^2$, so we can solve $Y^2 = D$ to get 4 solutions $Y_1,Y_2,Y_3,Y_4$ i.e. $$\begin{pmatrix} \pm 2 & 0 \\ 0 & \pm 2i \end{pmatrix}$$ which then leads to four solutions for $X$: $$X = P\begin{pmatrix} \pm 2 & 0 \\ 0 & \pm 2i \end{pmatrix} P^{-1}.$$
I'm having trouble showing that they are no more solutions.
 [/INST] I quite like the approach to finding $X$ which I learned from the paper
Bernard W. Levinger: The Square Root of a $2\times2$ Matrix, Mathematics Magazine, Vol. 53, No. 4 (Sep., 1980), pp. 222-224, DOI: 10.2307/2689616. This paper is among references in the Wikipedia article Square root of a 2 by 2 matrix.
The basic idea is that we are able to find $\newcommand{\Tra}{\operatorname{Tr}}\Tra(X)$ and $\det(X)$. Thus we know the characteristic polynomial and from Cayley-Hamilton theorem we have $X^2-\Tra(X)X+\det(X)I=0$. If we plug $X^2=A$ into the previous equation, we might be able to express $X$ using $A$. (Although the computations will be somewhat different in the case $\lambda_1=\lambda_2$.) At least if you are trying to solve this by hand, not using some software which immediately gives you the Jordan normal form and the matrix $P$.
This works only for $2\times2$ matrices. But I think the calculations with this approach are easier then finding the Jordan form and the corresponding base-change matrix.

Let us denote by $\lambda_1$ and $\lambda_2$ the eigenvalues of the matrix $X$. We know that $A=X^2$ then has the eigenvalues $\lambda_1^2$ and $\lambda_2^2$.
Let us denote $d=\lambda_1+\lambda_2=\Tra(X)$ and $t=\lambda_1\lambda_2=\det(X)$. We know that
$$\det(A)=\lambda_1^2\lambda_2^2 = d^2$$
and 
$$\Tra(A)=\lambda_1^2+\lambda_2^2 = (\lambda_1+\lambda_2)^2-2\lambda_1\lambda_2=t^2-2d.$$
Since the matrix $A$ is given, this is a system of equations with the unknowns $t$ and $d$.
For the given matrix
$$A=\begin{pmatrix}2&3\\4&-2\end{pmatrix}$$
we have $\det(A)=-16$ and $\Tra(A)=0$. Hence
\begin{align*}
d^2&=-16\\
t^2-2d&=0
\end{align*}
i.e.
\begin{align*}
d^2&=-16\\
t^2&=2d
\end{align*}
We have two possible values $d=\pm4i$. For each value of $d$ we can find two possibilities for $t$.
$$
\begin{array}{|c|c|}
\hline
  d & t \\\hline
 4i & 2+2i \\\hline
 4i &-2-2i \\\hline
-4i & 2-2i \\\hline 
-4i &-2+2i \\\hline 
\end{array}
$$
For any of the above numbers we have 
\begin{align*}
X^2-tX+dI&=0\\
A-tX+dI&=0\\
X&=\frac{A+dI}t
\end{align*}
This gives us four possible values for $X$. For example, for $t=2+2i$ and $d=4i$ we get
\begin{align*}
X
&=\frac1{2+2i}\begin{pmatrix}2+4i&3\\4&-2+4i\end{pmatrix}\\
&=\frac{1-i}4\begin{pmatrix}2+4i&3\\4&-2+4i\end{pmatrix}\\
&=\frac14\begin{pmatrix}6+2i&3-3i\\4-4i&2+6i\end{pmatrix}\\
\end{align*}
We can check whether this is indeed a solution of the given equation simply by squaring it. Here is computation in WolframAlpha.
The remaining three solutions can be found in a similar way, simply by plugging the remaining values of $d$ and $t$. It is also clear that if $X$ is a solution, then so is $-X$. So we can simply compute one matrix for $d=4i$ and one matrix for $d=-4i$ and then add the opposite matrices.
We have also seen that the matrices of this form are the only possible candidates for $X$. So we know that there are no other solutions.
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