Intuition with Complex Derivatives of a Complex Variable.

Question

I'm having some trouble wrapping my mind around derivatives of a complex function of a complex variable.This is a function of two real inputs and two real outputs, correct?

What confuses me is that in multivariable calculus, the functions were of multivariable input, and had one real output. So when calculating the derivative, you'd have to know what vector you're calculating it on. In the complex space however, with two inputs and two outputs, I feel like this would also be the case? But it seems like we can use all the standard rules from single variable calc. What am I misunderstanding? Thanks!

Answer 1

You're comparing inequivalent concepts.

Even in single variable calculus, if you want to know the directional derivative of $f$ at $x_0$ in the direction of a unit vector $u$ (which in this case can only be $\pm 1$), you would have to compute $\frac{df}{dx}\bigg|_{x_0}\cdot u$. The dot here is the multiplication of two real numbers. The concept of directional derivative is hardly used in single variable calculus because it's so simple.

For a real valued function of several variables, at a point $\vec{a}\in\mathbb{R}^n$ you calculate $(\frac{\partial f}{\partial x_1}\bigg|_\vec{a},\dots,\frac{\partial f}{\partial x_n}\bigg|_\vec{a})\cdot \vec{u}$, where $\vec{u}$ is a unit vector in $\mathbb{R}^n$. The dot here is the dot product of two vectors.

For a holomorphic function you compute $\frac{df}{dz}\bigg|_{z_0}\cdot u$, where $u$ is now a complex number of unit length. The dot here is complex multiplication. Again, this concept is not used very much.

Answer 2

You should think in terms of a single (complex) input and a single (complex) output. Take, for instance, the map $z\mapsto z^3$. It's much easier to work with it under this form rather than looking at it as$$x+yi\mapsto x^3-3xy^2+(3x^2y-y^3)i.$$

Answer 3

For $f:\mathbb{C}\to\mathbb{C}$ to be complex-differentiable, two things have to happen.

1. $f:\mathbb{R}^2\to\mathbb{R}^2$ must be differentiable, i.e , $\exists$ $T:\mathbb{R}^2\to\mathbb{R}^2$ linear transformation that: $$f(a+v)-f(a)=T\cdot v+ \rho(h)\hspace{1cm}; \lim \rho(h)=0$$

2. The Jacobian matrix of T is of the form: $$\begin{pmatrix} r_1 & -r_2 \\ r_2 & r_1 \end{pmatrix}$$

Consider the following example $f:\mathbb{R}^2\to\mathbb{R}^2$ given by $f(x,y)=(x,-y)$, this is a linear transformation so it is differentiable, but the function $f(x+yi)=x-yi$ is not complex-differentiable.

You can use calculation techniques, but you also have to take into account the second condition. That is in a sense, they are almost the same, but complex differentiable it is a stronger condition.