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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.

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

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

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

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

Cheng
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Twenty-six colours
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3 Answers3

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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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The functions you consider are all analytic, therefore your $x$-derivative will coincide with the complex derivative.

Analyticity ensures that the complex derivative coincides with the 'partial' derivatives with respect to the real/imaginary axis. Formally, analyticity means that the complex limit $$\lim_{z\to 0} \frac{f(z_0 + z) - f(z_0)}z$$ is independent of the way that $z\to 0$. In particular you could just take the limit along the real axis (which would just correspond to your $x$-derivative) and obtain the same result. But you could also take the limit along the imaginary axis, and should obtain the same result as before. Analyticity ensures that. The property that those limits are the same is ensured by the Cauchy-Riemann equations, which are equivalent to analyticity (for a continuous function on an open set).

If a function is not analytic, the complex derivative need not be well defined, and the direction in which you take your derivative might influence the outcome. An analogy is the question whether a function $f:\mathbb R^2\to\mathbb R$ is Frechet-differentiable or not. Hence, for a non-analytic function, the complex derivative will not exist, and it plays a role whether you take the $x$-derivative or the $\mathrm{i}y$-derivative.

DominikS
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    Why does being analytic mean that the partial derivative with respect to $x$ (and not $y$) is equivalent to the derivative of the function $z$ at any $(x,y)$ ? – Twenty-six colours Jul 18 '17 at 15:28
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    Referring to edit: I don't think I'm seeing it, as if I were to just take "partial" derivative w.r.t $y$ of the first example, I would have concluded that $f'(z) = \frac{\partial}{\partial y}(x^2 - y^2 + 2ixy) = -2y + 2xi \neq 2z$ – Twenty-six colours Jul 18 '17 at 15:38
  • Analyticity implies that both the derivative wrt. $x$ and $\mathrm{i} y$ are the same as the derivative wrt. general $z$. – DominikS Jul 18 '17 at 15:39
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    Note that you would have to take the derivative along the imaginary axis, hence deriving wrt. $\mathrm{i}y$, not $y$ (if $y$ is the imaginary part), i.e. the limit $\lim_{y\to 0}(f(z+\mathrm{i}y)-f(z))/(iy)$, for real $y$. – DominikS Jul 18 '17 at 15:43
  • "The property that those limits are the same basically comes from the Cauchy-Riemann equations." Is this circular? Doesn't the proof of the C-R equations rely on the "the property that those limits are the same"? –  Jul 18 '17 at 15:48
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    Actually, the CR-equations are equivalent to the analyticity conditions. They hold for a function if and only if it is analytic. – DominikS Jul 18 '17 at 15:51
  • No, they are not: https://en.wikipedia.org/wiki/Looman%E2%80%93Menchoff_theorem. One needs extra assumptions for the function in the statement. –  Jul 18 '17 at 15:53
  • Indeed they are, given the right framework (i.e. within the set of continuous functions, ...). The point above is that complex differentiability on open sets implies the CR-equations, that's a very short calculation. The reverse statement is not really relevant here. – DominikS Jul 18 '17 at 16:05
  • Well, now that you have added "given the right framework" to what you said, which is the point I made ("One needs extra assumptions for the function in the statement."), I agree with you. –  Jul 18 '17 at 19:45
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There is a difference between a general function of two variables f(x,y) and a complex function f(z)=f(x+iy). In the first, the function variations with x and y are completely independent, whereas in a complex function the variable (x+iy) transforms wholly. Accordintly, whatever happens to x, happens to y too (and hence z as well) in terms of the slope, which is the (partial) derivative. Hope this helps.

Masoud

Masoud Vakili
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