Rearranging the Dirac Equation, we find that $$\sum_\mu\gamma^\mu \partial_\mu \psi = -imc\frac{2\pi}{h} \psi.$$ It appears that the wavefunction times the constant is equal to the matrix result of the summation on the left hand side of the equation.
Am I doing something wrong? Keep in mind that I'm going off of little more than my knowledge of differential equations and matrices, as I do not have access to formal education in quantum mechanics.
Your rearranged Dirac equation is $$\sum_\mu\gamma^\mu \partial_\mu \psi = -imc\frac{2\pi}{h} \psi.$$
For this to make sense $\psi$ needs to be $4\times 1$ matrix, i.e. a column vector. Then you have a column vector on both sides of the equation.
To understand this, remember the definition of matrix multiplication. Multiplying a $4\times 4$ matrix with a $4\times 1$ matrix will give a $4\times 1$ matrix. For example: $$\gamma^0\partial_0 \psi =\begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&-1&0\\ 0&0&0&-1\end{pmatrix} \partial_0 \begin{pmatrix}\psi_1\\ \psi_2\\ \psi_3\\ \psi_4\end{pmatrix} =\begin{pmatrix} \partial_0\psi_1\\ \partial_0\psi_2\\ -\partial_0\psi_3\\ -\partial_0\psi_4 \end{pmatrix}$$
The equation you describe is $$\gamma^\mu \partial_\mu \psi = -\frac{imc}{\hbar} \psi$$
where $\hbar \equiv h/2\pi$. The gamma matrices are $4\times 4$ matrices, and the solution $\psi$ is a $4\times 1$ matrix, i.e. a column vector.
In case you are confused by the summation notation, the expanded equation is $$\gamma^0 \partial_0\psi + \gamma^1\partial_1\psi + \gamma^2\partial_2\psi + \gamma^3\partial_3\psi = -\frac{imc}{\hbar}\psi$$
