I understand that to determine the vector differential operators based on coordinate system one must create a variable change vector first.
$$\mathbf{\overrightarrow{R}} = x(q_1,q_2,q_3) \hat{i} + y(q_1,q_2,q_3)\hat{j} + z(q_1,q_2,q_3) \hat{k}$$
Then define scale factors ${h_{q_1}},{h_{q_2}},{h_{q_3}}$
$$h_{q1}=\left|\frac{\partial\mathbf{\overrightarrow{R}}}{\partial q_1}\right|,h_{q2}=\left|\frac{\partial\mathbf{\overrightarrow{R}}}{\partial q_2}\right|,h_{q3}=\left|\frac{\partial\mathbf{\overrightarrow{R}}}{\partial q_3}\right|$$ Now, general vector differential operators for gradient, divergence and curl become... $$\nabla f(q_1,q_2,q_3) = \frac{1}{h_{q_1}}\frac{\partial f}{\partial q_1} \mathbf{\hat{q_1}} + \frac{1}{h_{q_2}}\frac{\partial f}{\partial q_2} \mathbf{\hat{q_2}} + \frac{1}{h_{q_3}}\frac{\partial f}{\partial q_3} \mathbf{\hat{q_3}} $$
$$ \nabla \bullet \mathbf{\overrightarrow{U}} = \frac{1}{h_{q_1}h_{q_2}h_{q_3}} \left[ \frac{\partial}{\partial q_1} \left( h_{q1}h_{q_2}h_{q_3} u(q_1,q_2,q_3) \right) + \frac{\partial}{\partial q_2} \left( h_{q1}h_{q_2}h_{q_3} v(q_1,q_2,q_3) \right) + \frac{\partial}{\partial q_3} \left( h_{q1}h_{q_2}h_{q_3} w(q_1,q_2,q_3) \right) \right] $$
$$ \nabla \times\mathbf{\overrightarrow{U}} = \frac{1}{h_{q_1}h_{q_2}h_{q_3}} \left| \begin{matrix} h_{q1}\mathbf{\hat{q_1}} & h_{q2}\mathbf{\hat{q_2}} & h_{q3}\mathbf{\hat{q_3}} \\ \frac{\partial}{\partial q_1} & \frac{\partial}{\partial q_2} & \frac{\partial}{\partial q_3} \\ h_{q_1} u & h_{q_2} v & h_{q_3} w \end{matrix} \right| $$
Thus converting to complex conjugate coordinates $z$ and $\bar{z}$ the variable change vector would be ...
$$\mathbf{\overrightarrow{R}} = \frac{z + \bar{z}}{2} \hat{i} + \frac{z - \bar{z}}{2i}\hat{j} + k \hat{k}$$
where $ z = a+ib$ and $\bar{z} = a-ib$ and $k$ is an arbitrary real valued third coordinate. where $q_1 = z , q_2=\bar{z} , q_3= k$
Similarly, converting to a polar coordinate system would use a variable change vector ... $$\mathbf{\overrightarrow{R}} = r cos{\theta} \hat{i} + r sin{\theta}\hat{j} + k \hat{k}$$ where $q_1 = r , q_2=\theta , q_3= k$
$\mathbf{So,\ to\ transform\ to\ complex\ polar\ coordinates\ is\ the\ variable\ change\ vector...?}$
$$\mathbf{\overrightarrow{R}} = \frac{ r \left( e^{i\theta} + e^{-i\theta} \right) }{2} \hat{i} + \frac{r \left( e^{i\theta} - e^{-i\theta} \right)}{2i}\hat{j} + k \hat{k}$$
where $q_1 = r , q_2=\theta , q_3= k$
$\mathbf{ Or\ am\ I\ over\ complicating\ things?}$