calculate $\frac{\partial f }{ \partial \operatorname{Re} (z)} $ from $\frac{\partial f }{ \partial z}$ for meromorphic function $f(z)$ is a meromorphic function, how to derive $\frac{\partial f }{ \partial \operatorname{Re} (z)} $  if the derivative $\frac{\partial f }{ \partial z} $is given?
I guess
$$
\frac{\partial f }{ \partial \operatorname{Re} (z)} =\frac{\partial f }{ \partial z}
$$
$$
\frac{\partial f }{ \partial \operatorname{Im} (z)} =i\frac{\partial f }{ \partial z}
$$
I tested the two equations with a special case $f(z)=(m+ni)z^2$, where $z=a+bi$.
$$
\frac{\partial f }{ \partial \operatorname{Re} (z)} = 2(m+ni)(a+bi)
$$
$$
\frac{\partial f }{ \partial \operatorname{Im} (z)} = 2(-n+mi)(a+bi)
$$
Am I right?
Happy holiday!
 A: Let $z=x+iy$ and $f(x+iy)=u(x,y)+iv(x,y)$. So: $$\frac{\partial f(x+iy)}{\partial x}=\partial_x u(x,y)+i\partial_xv(x,y)=f'(x+iy)$$
Why? Because:
\begin{align}
f'(x+iy) &= \lim_{h\to 0, k=0} \frac{u(x+h,y+k)+iv(x+h,y+k)-u(x,y)-iv(x,y)}{h+ik}\\
&=\lim_{h\to 0} \frac{u(x+h,y)-u(x,y)}{h} +i\frac{v(x+h,y)-v(x,y)}{h}\\
&=\partial_x u(x,y)+i\partial_x v(x,y)
\end{align}
So we may conclude:
\begin{align}
\frac{\partial f(z)}{\partial \operatorname{Re}{(z)}}=f'(z)
\end{align}
So your guess is right! The second expression you gave is also true. Verify it as I done above.
A: Hint: With $z=x+iy$ use
$$\dfrac{\partial}{\partial x}=\dfrac{\partial}{\partial z}+\dfrac{\partial}{\partial \overline{z}}$$
$$\dfrac{\partial}{\partial y}=i\left(\dfrac{\partial}{\partial z}-\dfrac{\partial}{\partial \overline{z}}\right)$$
A: Differentials are much more straightforward. Since $f$ is differentiable,
$$ \mathrm{d} f(z) = f'(z) \mathrm{d} z = f'(z) (\mathrm{d}x + i \, \mathrm{d} y)$$
Assuming $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ are meant to denote the standard tangent vectors in $(x,y)$ coordinates, applying the first sends $\mathrm{d}x \mapsto 1$ and $\mathrm{d}y \mapsto 0$, and the second vice versa, so we get
$$ \frac{\partial}{\partial x} f(z) = f'(z) (1 + i \cdot 0) = f'(z) $$
$$ \frac{\partial}{\partial y} f(z) = f'(z) (0 + i \cdot 1) = i f'(z) $$
Assuming $\frac{\partial}{\partial z}$ means the standard tangent vector in $(z, \bar{z})$-coordinates, we indeed have
$$f'(z) = \frac{\partial}{\partial z} f(z)$$
confirming the conjecture.
