Complex Differentiation Thursday, March 28, 2024
If $ V \subseteq \mathbb{C} $ is open, we say a function $ f \colon V \to \mathbb{C} $ is complex-differentiable
at a point $ z \in V $ if the usual difference quotient converges. That is,
$$\begin{equation}
f^\prime(z) = \lim_{h \to 0} \frac{ f(z+h)-f(z) }{ h }
\end{equation}$$
where, of course, $ h \in \mathbb{C} \setminus \left \{ 0 \right \} $.1
The incredible thing about complex differentiation is how strong it is! We’ll see more on why this is true with the rest of this post, but two fascinating facts are:
- Being one-time complex-differentiable implies you are infinitely-many times complex-differentiable;
- A function is complex-smooth if and only if it is analytic (analytic meaning locally represented by a convergent power series).
Complex numbers as matrices
Recall that the complex numbers can be seen as ordered pairs of real numbers $ (x,y) \in \mathbb{R}^2 $ written $ x+iy $.
Further recall that they have the usual pointwise addition defined on them, $ (x+iy)+(a+ib) = (x+a) + i (y+b) $, and that they have a special multiplication defined on them:
$$\begin{equation}
(x+iy) \cdot (a+ib) = (xa - yb) + i (xb + ya).
\end{equation}$$
This ends up endowing $ \mathbb{C} $ with a field structure (i.e., we have additive and multiplicative identities, and both addition and multiplication have our usual properties).
We may also represent these numbers as $ 2 \times 2 $ matrices with real entries. Indeed, we may identify2 $$\begin{equation} x+iy \longleftrightarrow \begin{pmatrix} x & -y \\ y & x \end{pmatrix}. \end{equation}$$
Differentiation in $ \mathbb{R}^2 $
Consider a function $ F \colon \mathbb{R}^2 \to \mathbb{R}^2 $. If the function is differentiable, the derivative takes the form of the Jacobian, i.e., if $ F = (F_1, F_2) $, $$\begin{equation} D F = \begin{pmatrix} \partial_x F_1 & \partial_y F_1 \\ \partial_x F_2 & \partial_y F_2 \end{pmatrix}. \end{equation}$$
Requiring derivatives are complex
Recalling our first section, it is clear that we may identify functions $ f \colon \mathbb{C} \to \mathbb{C} $ with functions $ F \colon \mathbb{R}^2 \to \mathbb{R}^2 $.
This then allots us the opportunity to study what something being complex-differentiable would mean.
Indeed, the following two subsections explore this idea and make the same conclusion: the existence of the Cauchy-Riemann Equations.
Consider a function $ f \colon \mathbb{C} \to \mathbb{C} $ given by $ f = u+iv $, where $ u,v \colon \mathbb{R}^2 \to \mathbb{R}^2 $ (obviously with us identifying $ \mathbb{C} $ with $ \mathbb{R}^2 $).
Then, taking the derivative of $ f $ in the $ \mathbb{R}^2 $-sense, we get that
$$\begin{equation}
Df =
\begin{pmatrix}
\partial_x u & \partial_y u \\
\partial_x v & \partial_y v
\end{pmatrix}.
\end{equation}$$
But if we require the result of this differentiation to always be a complex number, we immediately get two constraints, known as the Cauchy-Riemann Equations
(AKA the C-R Equations):3
$$\begin{equation}
\begin{cases}
u_x &= v_y \\
u_y &= -v_x
\end{cases}
\end{equation}$$
A quick way to recall these is by thinking about how matching indices are positive, and differing indices are negative, i.e., $ u $ is the first component and $ v $ is the second, so $ u_x = v_y $ with no negative sign (similarly for $ u_y = - v_x $).
From this we also get a nice (and hopefully clear) consequence of the C-R Equations being satisfied: equivalence to complex differentiability!
An important example
There are functions which are real-differentiable but NOT complex-differentiable.
Take for example: $ f(x+iy) = x $. We can then identify $ f $ with the function $ F(x,y) = x $. Clearly, then, $$\begin{equation} DF = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}. \end{equation}$$ Hence the C-R Equations are not satisfied and our function, although real-differentiable, is NOT complex-differentiable.
Exercises
Suppose that $ f $ is defined in a neighborhood of the complex number $ z $. Show that $ f $ is complex-differentiable at $ z $ if and only if there exists a complex number $ a $ such that $$\begin{equation} f(z+h) = f(z) + ah + o(h) \quad \text{as} \quad h \to 0 \end{equation}$$ (where we are using Landau’s Little $ o $-notation, i.e., our above equation means that $$\begin{equation} f(z+h) = f(z) + ah + E(h) \end{equation}$$ for some “error term” $ E(h) $ satisfying $ E(0) = 0 $ and $$\begin{equation} \lim_{h \to 0} \frac{ \left | E(h) \right | }{ \left | h \right | } = 0. \end{equation}$$ This is also just good to know in general.)
($\Rightarrow$)
Suppose $ f $ is complex-differentiable at $ z $.
Then there exists a complex number $ a $ such that for all $ \varepsilon > 0 $, there exists $ \delta > 0 $ such that
$$\begin{equation}
0 < |h| < \delta \implies \left | \frac{ f(z+h) - f(z) }{ h } - a \right | < \varepsilon.
\end{equation}$$
Thus taking $ h $ in this neighborhood, we see that we have an error term $ E $ such that
$$\begin{equation}
f(z+h) = f(z) + ah + E(h)
\end{equation}$$
and
$$\begin{equation}
|E(h)| < \varepsilon |h| \quad \text{and} \quad E(0) = 0
\end{equation}$$
Hence,
$$\begin{equation}
f(z+h) = f(z) + ah + o(h)
\end{equation}$$
as desired.
($\Leftarrow$) Suppose that there exists a complex number $ a $ such that $$\begin{equation} f(z+h) = f(z) + ah + o(h) \quad \text{as} \quad h \to 0. \end{equation}$$ By definition, we know then that we have an error term $ E(h) $ such that $$\begin{equation} f(z+h) = f(z) + ah + E(h) \end{equation}$$ with $ E(0) = 0 $ and $ |E(h)/h| \to 0 $ as $ h \to 0 $. It immediately follows from our properties of $ E $ that for all $ \varepsilon > 0 $, there exists a $ \delta > 0 $ such that $$\begin{equation} 0 < |h| < \delta \implies |E(h)| < \varepsilon |h|, \end{equation}$$ whereby we get $$\begin{equation} |f(z+h) - f(z) - ah| = |E(h)| < \varepsilon |h|. \end{equation}$$ Thus: $$\begin{equation} \left | \frac{ f(z+h)-f(z) }{ h } - a \right | < \varepsilon, \end{equation}$$ so $ f $ is complex-differentiable at $ z $ as desired.
Suppose that $ f $ is complex-differentiable at $ z $. Show that $ f $ is continuous at $ z $.
Suppose that $ f $ is complex-differentiable at $ z $. Let $ \varepsilon > 0 $ be smaller than $ 1 $. Then there exists some $ 0 < \delta < 1 $ such that $$\begin{equation} 0 < |w-z| < \delta \implies \left | \frac{ f(w)-f(z) }{ w-z } - f’(z) \right | < \frac{ \varepsilon }{2 |w-z|} \end{equation}$$ so $$\begin{equation} |f(w)-f(z) - (w-z)f^\prime(z)| < \frac{ \varepsilon }{ 2|w-z| }|w-z| < \frac{ \varepsilon }{ 2 } \end{equation}$$ Taking $ |w-z| < \min \left \{ \delta, \varepsilon|f^\prime(z)|^{-1}/2 \right \} $, $$\begin{align} |f(w)-f(z)| &\leq |f(w)-f(z) - (w-z)f^\prime(z)| + |w-z||f^\prime(z)| \\ &< \frac{ \varepsilon }{ 2 } + \frac{ \varepsilon }{ 2 } = \varepsilon, \end{align}$$ so $ f $ is continuous! :)
Define $ f \colon \mathbb{C} \to \mathbb{C} $ by
$$\begin{equation}
f(x+iy) =
\begin{cases}
0 \quad& x = 0 \\
0 \quad& y = 0 \\
1 \quad& \text{otherwise}.
\end{cases}
\end{equation}$$
Show that $ f $ satisfies the Cauchy-Riemann equations at the origin even though it is NOT complex-differentiable at the origin.
Hint. By our if and only if from earlier, this function must then fail to be a $ \mathbb{R}^2 $-differentiable.
It is clear that $$\begin{equation} u_x(0,0) = 0 = v_y(0,0) \quad \text{and} \quad u_y(0,0) = 0 = -v_x(0,0), \end{equation}$$ so the C-R Equations are satisfied, but it is also clear that (although the partials exist) $ f $ is not $ \mathbb{R}^2 $-differentiable.
In general we will not be specifying the set in which we are taking the limit over, but I felt it is worthwhile to do it the first time. ↩︎
I was discussing this idea with my roommate Kyle he had a good bit of intuition for this fact: complex numbers can be treated as rotations and dilations, and a rotation matrix (with some scaling factors) takes this form! ↩︎
Here we use the notation $ u_x $ for $ \partial u/\partial x $ because it looks a lot prettier. :) ↩︎