Suppose $ f(z):\mathbb{C}^{1}\to\mathbb{C}^{1} $, write $ \frac{\partial f}{\partial \overline{z}}=\frac{1}{2}(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}) $. We know that by Cauchy-Riemann equation, the differential property is equivalent to $ \frac{\partial f}{\partial \overline{z}}=0 $. But can we take the derivative directly regarding to $ \overline{z} $ to determine whether the function is differential or not? If we can't do this, then what's the reason to introduce this symbol? And how to determine a function is differentiale without directly expanding the real and imaginary part out and check the Cauchy-Riemann equation?
Understand the notation $ \frac{\partial f}{\partial \overline{z}} $
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Best Answer
Is somewhat of an abuse of notation, but the idea is that if $f:\mathbb C\to\mathbb C$ is real differentiable at $z_0$ -- that is, it is differentiable as a function between 2-dimension real vector spaces -- then we can write $$ f(z) = f(z_0+h) = f(z_0) + F(h) + o(h) $$ where $F(h)$ is a linear transformation of $\mathbb C$, still viewed as a 2-dimensional real vector space. We can expand $F$ in matrix form and do some algebra to see that this can also be written $$ f(z) = f(z_0+a+bi) = f(z_0) + c_1(a+bi) + c_2(a-bi) + o(a+bi) $$ for some (uniquely determined) complex constants $c_1$ and $c_2$.
Since $c_1$ is the coefficient we multiply by $a+bi=z-z_0$ and $c_2$ is the coefficient we multiply by $a-bi = \overline z - \overline{z_0} $, there is a certain justice in calling them $\partial f/\partial z$ and $\partial f/\partial \bar z$.
But don't take this for more than it is; the notation depends on pretending that $z$ and $\bar z$ can vary independently of each other, which is not really the case.