# Scalar field Derivative. Real vs Complex

From Churchill - Complex Variables and Application it's shown that in order to get Cauchy-Riemman equations, we just asume a derivable function $$w(z)$$ and go for the limit definition.

$$w'(z_0) = \lim_{z \to z_0}\frac{w(z) - w(z_0)}{z - z_0} = L$$

Because a complex function can be seen as a $$\Re^2 \to \Re^2$$ vector field, (from the z-plane to the w-plane) as $$w(z) = w(x + iy) = u + iv$$, with $$u$$ and $$v$$ scalar fields. Thus, we can extend the limit, into the real and the imaginary part. So, here is my problem. For C-R Equations we say that, as $$L$$ exists, we say that is valid, for any arbitrary choice of trajectory that is continous at $$z_0$$. We choose the real and imaginary axis, simplify the limit and get the C-R. This means that for any complex function $$f(z)$$, we can indeed have a derivative $$f'(z)$$ which is not so intuitive from multivariable calculus. If we have an scalar field, and derive it by limit definition and apply the exact same trick, the limit would depend on the choosed trajectory, givin birth to the partial derivatives of $$f$$.

So Why do we assume, in the complex case, that this does not hold. And that if $$f$$ is derivable, at a point $$z = z_0$$, it's "partial derivatives" should be the same.

I know in complex analysis everithing is kind of weird, but I don't see the rationale behind that.

• Is just by definition?
• The gradient in the Real case, would be "analogous" to the complex derivative?
• Is valid due to the division between complex numbers?
• Does this answer your question? Complex differentiability and differentiability in R2 Apr 8 '21 at 0:21
• @MarkS. Not at all. It defines de derivatives, but its does not explains the rationale behind the method. I want to know, why we cannot create an analogous to the complex derivative in multivariable real calc. Apr 8 '21 at 3:12

There seem to be a few different questions, and I'm not certain I've grasped the question fully, but I have a few comments that might help

# Single-Variable Perspective

One thing that may seem odd about the complex derivative of a function $$w:\mathbb C\to \mathbb C$$ is that we are expecting/requiring the derivative to be the same from various directions (e.g. horizontal and vertical directions). This is actually analogous to the case from single-variable real calculus, with a function $$f:\mathbb R\to\mathbb R$$.

In real calculus, we can certainly consider the "right derivative" $$\partial_+f(x_0):={\displaystyle\lim_{x\to x_0^+}}\dfrac{f(x)-f\left(x_0\right)}{x-x_0}$$ and similarly for the "left derivative". For functions like $$f(x)=|x|$$, this is a useful concept since the left and right derivatives exist but are not equal. However, for many functions of interest in calculus, the left and right derivatives are equal, so it's useful to have a single concept of "the derivative" $$f'(x_0)$$.

The situation is similar for a single-variable complex function $$w$$. We could consider, say, the "horizontal derivative" $$\left.\frac{\partial}{\partial x}w(x+iy_0)\right|_{x=x_0}$$ or even the "right horizontal derivative". However, it turns out that, for many functions of interest in calculus of complex functions, the limits along all approaches are equal. Thus, it's useful to have a single concept of "the derivative" $$w'(z_0)$$.

# Differentiability with Multiple Inputs

In multivariable calculus, while discussion of the partial derivatives is common, discussions of "differentiability" still require a limit to be the same from all directions, just as in the complex derivative.

## One output

A function $$f:\mathbb R^2\to\mathbb R$$ is said to be differentiable at $$\mathbf x_0=(x_0,y_0)$$ not when its partial derivatives exist there, but when the following two-dimensional limit holds for a linear "Jacobian" map $$J$$: $$\lim_{\mathbf h\to0}\frac{\left|f(\mathbf x_0+\mathbf h)-f(\mathbf x_0)-J(\mathbf h)\right|}{\Vert\mathbf h\Vert}=0$$ It turns out that such a linear function has a certain form if it exists: $$J\left(h_1,h_2\right)=\left.\begin{bmatrix}\dfrac{\partial f(\mathbf x)}{\partial x}&\dfrac{\partial f(\mathbf x)}{\partial y}\end{bmatrix}\right|_{\mathbf x=\mathbf x_0}\begin{bmatrix}h_1\\ h_2\end{bmatrix}=(\nabla f(\mathbf x_0))^\top \mathbf h=\nabla f(\mathbf x_0)\cdot\mathbf h$$

## Two outputs

To get closer to the complex case where $$w=u+iv$$, we could instead consider $$\mathbf f:\mathbb R^2\to\mathbb R^2$$ where $$\mathbf f=\begin{bmatrix}u&v\end{bmatrix}^\top$$. Now, since $$\mathbf f$$ is a vector which doesn't have an absolute value, we must take the norm $$\Vert\cdot\Vert$$ in the numerator.

Also, the Jacobian is now a linear map $$J:\mathbb R^2\to \mathbb R^2$$, which would be represented by a $$2\times2$$ matrix. The Jacobian matrix (in the standard basis) has all the partials: $$\begin{bmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{bmatrix}$$. This means that $$J(\mathbf h)$$ would be $$\begin{bmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{bmatrix}\begin{bmatrix}h_1\\ h_2\end{bmatrix}=\begin{bmatrix}\nabla u(\mathbf x_0)\cdot\mathbf h\\ \nabla v(\mathbf x_0)\cdot\mathbf h\end{bmatrix}$$

## Complex Differentiability

The limit for complex differentiability can be written similarly to the above. Starting with $${\displaystyle \lim_{z \to z_0}}\dfrac{w(z) - w(z_0)}{z - z_0} = L$$, we can rename $$z-z_0$$ as $$h$$ and subtract $$L$$ from both sides to obtain $${\displaystyle \lim_{h \to 0}}\dfrac{w(z_0+h) - w(z_0) - L*h}{h} = 0$$. We can then take complex absolute values/real norms of both sides, and use continuity to write $${\displaystyle \lim_{h \to 0}}\left|\dfrac{w(z_0+h) - w(z_0) - L*h}{h}\right| = 0$$. By the multiplicative property of complex absolute values (which could be proven via the polar form, say), we can distribute the absolute value bars to the top and bottom, to obtain $${\displaystyle \lim_{h \to 0}}\dfrac{\left|w(z_0+h) - w(z_0) - L*h\right|}{|h|} = 0$$.

In other words, since complex-linear functions $$J:\mathbb C\to\mathbb C$$ have the form $$J(h)=L*h$$ for some complex number $$L$$, $$w$$ is complex-differentiable at $$z_0$$ precisely when there exists a complex-linear function $$J$$ so that we have: $${\displaystyle \lim_{h \to 0}}\dfrac{\left|w(z_0+h) - w(z_0) - J(h)\right|}{|h|} = 0$$.

Written in this way, real differentiability of $$\mathbf{f}$$ and complex differentiability of $$w$$ are nearly identical. The only significant difference is that complex-linearity is a stronger condition on $$J$$ than real-linearity; this is what makes complex analysis work.

# Analogous to Complex Derivative?

For a complex function $$w(x+iy)$$ with complex-valued partial derivatives $$\dfrac{\partial w}{\partial x}$$ and $$\dfrac{\partial w}{\partial y}$$, there are complex-linear combinations of those partials with nice properties. The Wirtinger derivatives are $$\dfrac{\partial w}{\partial z}:=\dfrac12\left(\dfrac{\partial w}{\partial x}-i\dfrac{\partial w}{\partial y}\right)$$ and $$\dfrac{\partial w}{\partial \overline{z}}:=\dfrac12\left(\dfrac{\partial w}{\partial x}+i\dfrac{\partial w}{\partial y}\right)$$. In the special case when $$w'(z)$$ exists, it turns out that $$\dfrac{\partial w}{\partial z}=w'(z)$$ and $$\dfrac{\partial w}{\partial \overline{z}}=0$$. Given $$\mathbf f=\begin{bmatrix}u&v\end{bmatrix}^\top$$ as before, we can calculate analogous quantities to the Wirtinger derivatives for a real vector field.

Corresponding to $$\dfrac{\partial w}{\partial z}$$ would be $$\dfrac12\begin{bmatrix}\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}&\dfrac{\partial v}{\partial x}-\dfrac{\partial u}{\partial y}\end{bmatrix}^\top$$. Some would write the components of this as $$\mathop{\mathrm{div}} \mathbf f$$ and $$\mathop{\mathrm{curl}} \mathbf f$$ (for the divergence and the scalar curl). So in this sense, the complex derivative is like "the divergence and the scalar curl of the corresponding real vector field, packaged together and divided by two".

Corresponding to $$\dfrac{\partial w}{\partial \overline{z}}$$ would be $$\dfrac12\begin{bmatrix}\dfrac{\partial u}{\partial x}-\dfrac{\partial v}{\partial y}&\dfrac{\partial v}{\partial x}+\dfrac{\partial u}{\partial y}\end{bmatrix}^\top$$. I don't know of names for these quantities, but when a real vector field corresponds to a differentiable complex function, these quantities are zero by the Cauchy-Riemann equations.