Find a $\mathbb{C}^{n\times n}$ Schur factorizarion for $A$.
$$
A=\begin{bmatrix}1 & 0 & 2\\0 & 5 & 0\\2 & 0 &1 \end{bmatrix}
$$
I've found the real Schur factorization $QTQ^*$ to be:
$$
Q=\begin{bmatrix}1/\sqrt 2 & 1/\sqrt 2 & 0\\0 & 0 & 1\\-1/\sqrt 2 & 1/\sqrt 2 &0 \end{bmatrix}
$$
and
$$
T=\begin{bmatrix}-1 & 0 & 0\\0 & 3 & 0\\0 & 0 &5 \end{bmatrix}
$$
through my own working, however how do I convert this to $\mathbb{C}^{n\times n}$? I know if $q$ is an eigenvector of $A$ corresponding to some eigenvalue $λ$, so is $(i q)$. However I'm unsure of what to do from here/how to apply my knowledge?
Schur decomposition
Given a matrix $A$ of order $n$, the aim is to determine:
a unitary matrix $U$ of order $n$, i.e. such that $U^{-1} = U^h$, hence $U^h\,U = U\,U^h = I_n$;
an upper-triangular matrix $T$ of order $n$ with the eigenvalues of $A$ on its diagonal;
such that the below identity is verified:
$$ A = U\,T\,U^h\,. $$
This decomposition isn't unique, what matters is that all the conditions listed are respected.
Beyond this aspect, two fundamental cases should be highlighted:
if $A$ has complex components or $A$ has real components and the eigenvalues are all real, then $T$ is effectively upper-triangular and on its principal diagonal there are the $n$ eigenvalues of $A$;
if $A$ has real components and the eigenvalues aren't all real, i.e. there are conjugated complex eigenvalues, then it follows that $T$ is a block upper-triangular matrix, blocks $2 \times 2$ present on the principal diagonal whose eigenvalues are the complex conjugate ones of $A$.
It's precisely in the latter case that the need may arise to obtain an upper-triangular matrix $T$ with also the complex conjugated eigenvalues present in the respective components of the diagonal of $T$.
A classic case could be the square matrix of order $n = 3$:
$$ A = \begin{bmatrix} 0 & 0 & -1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix} $$
where by setting preliminarily:
$$ U = I_n\,, \quad \quad \quad T = A $$
and iterating enough through the instructions (John Francis's shifted QR algorithm):
$$ \begin{aligned} & (Q,\,R) = \text{QRDecomposition}(T - I_n) \\ & U = U\,Q \\ & T = R\,Q + I_n \\ \end{aligned} $$
we get:
$$ U = \begin{bmatrix} 0.57735 & -0.707107 & -0.408248 \\ -0.57735 & 0 & -0.816497 \\ 0.57735 & 0.707107 & -0.408248 \\ \end{bmatrix} \quad \quad T = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0.5 & -0.866025 \\ 0 & 0.866025 & 0.5 \\ \end{bmatrix} $$
where a $2 \times 2$ block is evident on the principal diagonal of $T$.
On the other hand, after having preliminarily set:
$$ U = I_n\,, \quad \quad \quad T = A $$
if we iterate enough these other instructions appropriately perturbed:
$$ \begin{aligned} & (Q,\,R) = \text{QRDecomposition}(T - \exp(\text{i})\,I_n) \\ & U = U\,Q \\ & T = R\,Q + \exp(\text{i})\,I_n \\ \end{aligned} $$
we get:
$$ \begin{aligned} & U = \begin{bmatrix} -0.510291 - 0.270068\,\text{i} & 0.306891 + 0.489031\,\text{i} & 0.0212594 + 0.576959\,\text{i} \\ 0.510291 + 0.270068\,\text{i} & -0.270068 + 0.510291\,\text{i} & 0.510291 + 0.270068\,\text{i} \\ -0.510291 - 0.270068\,\text{i} & -0.576959 + 0.0212594\,\text{i} & 0.489031 - 0.306891\,\text{i} \\ \end{bmatrix} \\ & \\ & T = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0.5 - 0.866025\,\text{i} & 0 \\ 0 & 0 & 0.5 + 0.866025\,\text{i} \\ \end{bmatrix} \end{aligned} $$
where the fact that $T$ is even a diagonal matrix is due to the fact that the chosen matrix $A$ is normal, i.e. such that $A^h\,A = A\,A^h$. Moreover, in the latter case the column vectors of $U$ correspond to the eigenvectors of $A$ and therefore the Schur decomposition coincides with spectral decomposition. On the other hand, if $A$ were positive (semi)definite, then we would fall back into the case of the singular value decomposition.
In conclusion, in the case object of the question I don't see the need to disturb complex numbers, but if it were imposed, it's sufficient to act exactly as just shown. I hope it's clear enough.
ADDENDUM
Everything that has been dealt with so far revolves around a numerical algorithm, which is essentially the standard in the application world, but no one forbids the application of the classic linear algebra procedures.
In particular, considering this other square matrix of order $n=3$:
$$ A = \begin{bmatrix} 0 & 0 & 2 \\ 1 & 0 & -5 \\ 0 & 1 & 4 \\ \end{bmatrix} $$
it follows that:
the eigenvalues of $A$ correspond to the zeros of the characteristic polynomial $p(\lambda) := \det(A-\lambda\,I_n)$, hence $\lambda_1 = 2$, $\lambda_2 = 1$; $\lambda_3 = 1$;
the eigenvectors of $A$ turn out to be a non-trivial solutions of homogeneous linear systems $(A - \lambda_i\,I_n)\,\mathbf{v}_i = \mathbf{0}$, hence $\mathbf{v}_1 = (1,\,-2,\,1)$, $\mathbf{v}_2 = \mathbf{v}_3 = (2,\,-3,\,1)$;
since $A$ isn't diagonalizable, to the two distinct eigenvectors of $A$ we add a third linearly independent vector chosen at will, for example $(0,\,0,\,1)$, and applying the Gram-Schmidt method we determine a set of orthogonal vectors which are placed along the columns of:
$$ U = \begin{bmatrix} \frac{\sqrt{6}}{6} & \frac{\sqrt{2}}{2} & \frac{\sqrt{3}}{3} \\ -\frac{\sqrt{6}}{3} & 0 & \frac{\sqrt{3}}{3} \\ \frac{\sqrt{6}}{6} & -\frac{\sqrt{2}}{2} & \frac{\sqrt{3}}{3} \\ \end{bmatrix}; $$
$$ T = U^h\,A\,U = \begin{bmatrix} 2 & -3\sqrt{3} & \frac{5\sqrt{2}}{2} \\ 0 & 1 & -\frac{\sqrt{6}}{2} \\ 0 & 0 & 1 \\ \end{bmatrix}. $$
Also in this case it isn't strictly necessary to disturb complex numbers, but if it were imposed it's sufficient to consider vectors with complex components at the third step of the process.