Prove the orthogonality of the derivatives of complex-valued function Let
$$
f(x,y,z)=\prod_{n=1}^N\left(a_n-q^n\right) +  \prod_{n=1}^N\left(b_n-p^n\right) 
$$
$$
q= e^{2 \pi i (x+zi)}, p= e^{2 \pi i (y+zi)}
$$
where $z>=0$; and $a_n,b_n$ are constants (complex numbers),
How to prove (I assume it is true)
$$
\left( \frac{\partial \textrm{Re}(f)}{\partial x} ,\frac{\partial \textrm{Re}(f)}{\partial y},
\frac{\partial \textrm{Re}(f)}{\partial z} \right)
 \cdot
\left( \frac{\partial \textrm{Im}(f)}{\partial x},
\frac{\partial \textrm{Im}(f)}{\partial y} ,
\frac{\partial \textrm{Im}(f)}{\partial z} \right)
=0
$$ 
i.e. the two vectors (the derivative) are orthogonal?
Similarly, does the orthogonality holds for the function
$$
f(x,y,z) = \sum_{n=1}^N \left(  \frac{a_n q^n}{1- q^n} +\frac{b_n p^n}{1- p^n}\right)
$$
$$
q= e^{2 \pi i (x+zi)}, p= e^{2 \pi i (y+zi)}
$$
Is there is a general method to show the orthogonality in such functions?
 A: Dividing the original function $f(x,y,z)$ into $f(x,z)+f(y,z)$ leads to (analytical complex function is conformal)
$$
\frac{\partial \textrm{Re}f(x,z)}{\partial x}\frac{\partial \textrm{Im}f(x,z)}{\partial x} + \frac{\partial \textrm{Re}f(x,z)}{\partial z}\frac{\partial \textrm{Im}f(x,z)}{\partial z}=0
$$
$$
\frac{\partial \textrm{Re}f(y,z)}{\partial y}\frac{\partial \textrm{Im}f(y,z)}{\partial y} + \frac{\partial \textrm{Re}f(y,z)}{\partial z}\frac{\partial \textrm{Im}f(y,z)}{\partial z}=0
$$
Together with 
$$
\frac{\partial \textrm{Re}(f(x,z)+f(y,z))}{\partial z}=\frac{\partial \textrm{Re}f(x,z)}{\partial z}+\frac{\partial \textrm{Re}f(y,z)}{\partial z}
$$
$$
\frac{\partial \textrm{Im}(f(x,z)+f(y,z))}{\partial z}=\frac{\partial \textrm{Im}f(x,z)}{\partial z}+\frac{\partial \textrm{Im}f(y,z)}{\partial z}
$$
one can find out that
$$
\left( \frac{\partial \textrm{Re}(f)}{\partial x} ,\frac{\partial \textrm{Re}(f)}{\partial y},
\frac{\partial \textrm{Re}(f)}{\partial z} \right)
 \cdot
\left( \frac{\partial \textrm{Im}(f)}{\partial x},
\frac{\partial \textrm{Im}(f)}{\partial y} ,
\frac{\partial \textrm{Im}(f)}{\partial z} \right)=
\frac{\partial \textrm{Re}f(x,z)}{\partial z}\frac{\partial \textrm{Im}f(y,z)}{\partial z}+\frac{\partial \textrm{Re}f(y,z)}{\partial z}\frac{\partial \textrm{Im}f(x,z)}{\partial z} \neq 0
$$
Thus, the assumption on orthogonality is false.
