[Math] Derivative and partial derivative of complex functions

complex-analysis

I know the formal definition of a derivative of a complex valued function, and how to compute it (same as how I would for real-valued functions), but after doing some problems, I feel as if I could just take the partial derivative w.r.t $x$ of the function to compute the derivative (so it doesn't depend on $y$?) as opposed to taking derivative w.r.t $z$ first then substitute.
That might be a bit obscure, so I'll put in a couple of examples

Examples

All of the examples are analytic (satisfy the Riemann conditions) with $z = x+iy$ and $f(z) = u(x,y)+iv(x,y)$.
1. $f(z) = z = x+iy =u(x,y) + iv(x,y)$.
The derivative is $f'(z) = 1$. Another way would be just to take partials of $f(z)$ w.r.t $x$ to get the result.

  1. $f(z) = z^2 = (x+iy)^2 = x^2 – y^2 + 2ixy$

    $f'(z) = 2z = 2x + 2iy$
    Another way is to just directly take partial derivative of $f$ w.r.t $x$ since $\frac{\partial u}{\partial x} = 2x$ and $\frac{\partial v}{\partial x} = 2y$.

  2. $f(z) = z^3 = (x+iy)^3 = x^3 – 3xy^2 + i(3x^2 y – y^3)$.

$ f'(z) = 3z^2 = 3x^2 -3y^2 + 6ixy$.
This can also be found similarly in other examples since $\frac{\partial u}{\partial x} = 3x^2 – 6xy$ and $\frac{\partial v}{\partial x} = 6xy$.

So it seems that I could just take the partial derivatives with respect to $x$ of the resultant complex number, and ignore the $y$ to find the derivatives. How come this is true?

Best Answer

The relation you observe is exactly how we get to the Cauchy-Riemann equations for the real and imaginary parts of an analytic function.

The complex derivative of a function $f:U\to{\bf C}$ at $z_0\in U$ where $U$ is an open subset of ${\bf C}$ is defined by $$ f'(z_0)=\lim_{z\to z_0:z\in U\backslash\{z_0\}}\frac{f(z)-f(z_0)}{z-z_0}\tag{1} $$ If ${f}$ is complex differentiable at ${z_0}$, then by specialising the limit (1) to variables ${z}$ of the form ${z = z_0 + h}$ for some non-zero real ${h}$ near zero we have $$ \lim_{z\to z_0:z\in U\backslash\{z_0\}}\frac{f(z)-f(z_0)}{z-z_0} =\lim_{h\to 0:h\in{\bf R}\backslash\{0\}}\frac{f((x_0+h)+iy_0)-f(x_0+iy_0)}{h} =u_x(z_0)+v_x(z_0)=:\frac{\partial f}{\partial x}(z_0) $$ where $z_0=x_0+iy_0$ and $f=u+iv$.

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