Issue finding a unitary matrix which diagonalizes a Hermitian 
Given $$H = \begin{pmatrix}2&3-i\\3+i&-1\end{pmatrix}$$ find a unitary matrix $U$ such that $$U^\dagger H U = D$$ where $D$ is the diagonal matrix containing the eigenvalues of $H$.

I assumed $U$ would therefore be the eigenvector matrix of $H$ as that normally gives a diagonal matrix containing the eigenvalues. I found the eigenvalues to be $\lambda = 4,-3$ and the respective eigenvectors to be 
$$\begin{pmatrix}\frac{3-i}{2}\\1\end{pmatrix}, \quad \begin{pmatrix}\frac{i-3}{5}\\1\end{pmatrix}$$
However the eigenvector matrix isn't unitary. How do I find another matrix that fits the relationship, but is unitary?

Edit:
After normalising the eigenvector matrix the diagonal matrix doesnt give the original eigenvalues, but a scaled down version:

 A: You have correctly calculated two eigenvectors of $H$. 
In order to produce an orthonormal basis of eigenvectors (to get the corresponding unitary matrix), you want your eigenvectors $v^{(1)}, v^{(2)}$ to satisfy:
$$\left \langle v^{(1)}, v^{(2)} \right \rangle := v^{(1)}_1 \overline{v^{(2)}_1} + v^{(1)}_2 \overline{v^{(2)}_2} = 0 \ \ \ \text{ and} $$
$$ \left\|v^{(1)}\right\| = \left\|v^{(2)}\right\| = 1$$
Your current eigenvectors already satisfy the first equation (they are orthogonal). All you need to do is to scale them by the appropriate constants so that their norm becomes $1$. 
That is, can you find $c_1, c_2$ such that $\left\|c_1\begin{pmatrix} \frac{3-i}{2}\\ 1 \end{pmatrix}\right\| = \left\|c_2\begin{pmatrix} \frac{i-3}{5}\\ 1 \end{pmatrix}\right\| = 1$?
A: You have done everything correct except for the small mistake in the last step. While normalising the vector, you have multiplied by the norm instead of dividing it and hence you are getting different values. For the first column, instead of $\frac{\sqrt{14}}{2}$ the term should be $\frac{2}{\sqrt{14}}$. And similarly for the other column. That should fix the error.
