Let $z_1\in \mathbb{R}\setminus\{0\}$ and $z_2\in i\mathbb{R}\setminus\{0\}$. Then $\{z_1,z_2\}$ form a basis for $\mathbb{C}$. This means that any $z\in\mathbb{C}$ can be written as a linear combination $$z=z(z_1,z_2)=\alpha z_1+\beta z_2, \quad \alpha,\beta\in \mathbb{R}.$$ I am interested in making sense of the derivative of $z$ w.r.t. the basis vectors -- i.e., of $\partial_{z_i}z(z_1,z_2)$ for $i=1,2$. For example, let's focus on $\partial_{z_1}z(z_1,z_2)$. Intuitively, as we "vary" the basis element $z_1$, I expect $z$ to vary non-trivially to according to something like $$\partial_{z_1}z(z_1,z_2)=\alpha\neq 0.$$ On the other hand, it is easy to see that $\alpha=\text{Re}(z)/z_1$ and $\beta=i\text{Im}(z)/z_2$. Then, $$\partial_{z_1}z(z_1,z_2)=\alpha(z_1)'z_1+\alpha(z_1)=-\frac{\text{Re}(z)}{z_1^2}z_1+\frac{\text{Re}(z)}{z_1}=0.$$ The two equations are in contradiction. I feel like the problem is trivial, but somehow I am stuck on it and I may be just overthinking it...
Any suggestions on which interpretation is the good one?
You are considering $Re(z)$ as constant function. First, note that $Re(z)= \alpha z_1$ is a function of $z_1$, therefore \begin{align*} \partial_{z_{1}} \alpha(z_1)&= \partial_{z_1} \frac{Re(z)}{z_1} \\ &= \frac{z_1 \partial_{z_1}Re(z)-Re(z)\partial_{z_1}(z_1)}{z_1^2} \\ &= \frac{z_1 \partial_{z_1}(\alpha z_1)-\alpha z_1\cdot 1}{z_1^2} \\ &= \frac{\alpha z_1 -\alpha z_1}{z_1^2} \\ &= 0. \end{align*}