# Differences between the complex derivative and the multivariable derivative.

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?

• Notice that dividing by the vector $x - x_0$ is undefined, which prevents us from defining the derivative of $g$ in that manner. – littleO Apr 15 '17 at 14:07

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 $h : \mathbb{R}^2 \to \mathbb{R}$.