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Let $f:G\to\mathbb{C}$, $G\subset\mathbb{C}$ be a complex function, differentiable at $z_0\in G$, where $z_0$ is an limit point of $G$. Now, the derivative at this point is defined to be the limit $$\lim_{z\to z_0}\dfrac{f(z)-f(z_0)}{z-z_0}$$ Now, consider any function $\mathbf{g}:\mathbb{R^2}\to\mathbb{R^2}$, and say $g$ is differentiable at some limit point $\mathbf{x_0}=(x_0,y_0)$ of $\mathbb{R^2}$. In this case, we don't define the derivative to be the limit $$\lim_{\mathbf{x}\to\mathbf{x_0}}\dfrac{\mathbf{g(x)}-\mathbf{g(x_0)}}{\mathbf{x}-\mathbf{x_0}}$$ because if we did, then this definition would also apply to any function $h:\mathbb{R^2}\to\mathbb{R}$, which is not the correct definition.

So, my question is, what are the differences between the complex derivative and the multivariable derivative? Why do we define the complex derivative like this?

codetalker
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2 Answers2

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We define the complex derivative that way because we can.

Think of how the ordinary derivative of a real function is defined: $$f'(x_0) = \lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0} $$ Notice, on the right hand side, that we are dividing by $x-x_0$, which we can do because division is defined in the real numbers.

Now turn to complex numbers, with the derivative defined as you stated: $$f'(z_0) = \lim_{z \to z_0} \frac{f(z)-f(z_0)}{z-z_0} $$ Once again, on the right hand side we are dividing by $z-z_0$, which we can do because division is defined in the complex numbers.

But now, consider a function $g : \mathbb{R}^2 \to \mathbb{R}^2$. The reason we do not define the derivative as $$g'(x_0) = \lim_{x \to x_0} \frac{g(x)-g(x_0)}{x-x_0} $$ is not because "it is not the right definition", but because we cannot define it this way. And we cannot define it this way because, on the right hand side, the denominator is a vector in $\mathbb{R}^2$, and division by vectors is not defined. The same reason holds for a function $h : \mathbb{R}^2 \to \mathbb{R}$.

This raises a different question: How shall we generalize the perfectly nice definition of the derivative of a function of a single real or complex variable to get a derivative of a function of multiple real or complex variables? But that's another issue.

Did
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Lee Mosher
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Complex differentiation can be seen as a restricted form of multivariate differentiation in $\mathbb{R}^2$.

If you use your definition of complex differentiation at $z_0$, the output is a complex number (if the limit exists). It may seem strange, but we could think of this complex number, call it $c$, as a linear map from $\mathbb{C} \rightarrow \mathbb{C}$ given by $a \mapsto a c$. This map changes as $z_0$ changes, just like the derivative changes as we supply it different values. It seems silly to think of a number as a linear map at first glance, but a very natural generalization of differentiation in higher dimensions does exactly this, and it's called the Jacobian.

Complex differentiation just forces the Jacobian to become complex multiplication, just like we saw above. To see why, write $f(x, y) = (f_x(x, y), f_y(x, y))$ by breaking $f$ up into its component functions, and then form the standard Jacobian:

\begin{bmatrix} \frac{\partial f_x}{\partial x}(x, y) & \frac{\partial f_x}{\partial y}(x, y)\\ \frac{\partial f_y}{\partial x}(x, y) & \frac{\partial f_y}{\partial y}(x, y) \end{bmatrix}

The Cauchy–Riemann equations restrict our Jacobian so that it looks like

\begin{bmatrix} u(x, y) & -v(x, y)\\ v(x, y) & u(x, y) \end{bmatrix}

where $u(x, y) = \frac{\partial f_x}{\partial x}(x, y)$ and $v(x, y) = \frac{\partial f_y}{\partial y}(x, y)$.

This Jacobian is given by complex multiplication. To see what I mean, let $(a, b)$ in $\mathbb{R}^2$ be $a + b i$ in $\mathbb{C}$. If we multiply our Jacobian by (the column vector) $(1, 0)$, we get $u(x, y) + i v(x, y)$. If we multiply it by $(0, 1)$, we get $-v(x, y) + i u(x, y)$, just as if we multiplied $u(x, y) + i v(x, y)$ by $1$ and $i$ respectively. So, by linearity, our Jacobian from $\mathbb{R}^2$ to $\mathbb{R}^2$ is identical to the linear map \begin{equation} a + b i \mapsto (a + b i) \cdot (u(x, y) + i v(x, y)) \end{equation} given by multiplying any complex number by $u(x, y) + i v(x, y)$.

A natural question then is what happens if we don't restrict the Jacobians with the Cauchy–Riemann equations? Well, a great example of this in action is the complex conjugate. As a function on $\mathbb{C}$ it's not even differentiable, but on $\mathbb{R}^2$ it's analytic!

Otto...
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