In quantum mechanics, we generally deal with unitary matrices. In the IBM cloud computers and Qiskit, the general $2\times 2$ unitary is defined as
$$U(\theta, \phi, \lambda) = \begin{pmatrix} \cos\left(\frac{\theta}{2}\right) & -e^{i\lambda} \sin\left(\frac{\theta}{2}\right) \\ e^{i\phi} \sin\left(\frac{\theta}{2}\right) & e^{i(\lambda + \phi)} \cos\left(\frac{\theta}{2}\right) \end{pmatrix}. $$
Now say I want to find the parameters $\theta, \phi$ and $\lambda$ corresponding to the unitary $$\begin{pmatrix}-\frac{1}{2}+\frac{i}{2}& -\frac{1}{2}-\frac{i}{2}\\ -\frac{1}{2}+\frac{i}{2} & -\frac{1}{2}+\frac{i}{2}\end{pmatrix}.$$ Then I can't apply this code directly as the first component of the unitary does not lie in the range of $\cos$. So I'll first have to divide the whole matrix by $\frac{1+i}{\sqrt{2}}$ first, to get to
$$\begin{pmatrix}\frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}} \\ \frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{pmatrix}.$$
Note that $|\frac{1+i}{\sqrt{2}}|=1$ and so the two unitaries are equivalent in the quantum mechanics sense. Furthermore, now the first component $\frac{1}{\sqrt 2}$ lies within the range of cosine i.e. $[-1,1]$.
Now for any general unitary matrix how do I find the value with which I have to divide it, such that the first component lies within the range of cosine? In this case, the value was $\frac{1+i}{\sqrt 2}$, but how do I find that for a general $2\times 2$ unitary matrix?
Àccording to Wikipedia, the general form of a unitary $Z$ can also be written as $$Z=e^{i\varphi/2}\begin{bmatrix}e^{i\varphi_1}\cos\theta & e^{i\varphi_2}\sin\theta \\ -e^{-i\varphi_2}\cos\theta & e^{-i\varphi_1}\cos\theta \end{bmatrix}$$ where the determinant of the unitary is $\det Z=e^{i\varphi}$. So the required value is the principle value of $\sqrt{\det Z}$ multiplied by the phase factor $e^{i\varphi_1}$ of the first component of $Z$, which can be easily obtained by representing it in polar form.