When considering a real-valued function
\(f: \R^2 \to \R\) of two variables, there is no notion of
the derivative of a function. For such a function, we instead only have partial derivatives
\(\fderiv fx(x_0,y_0)\) and
\(\fderiv fy(x_0,y_0)\) (and also directional derivatives) which depend on the way in which we approach a point
\((x_0,y_0)
\in \R^2\text{.}\)
For a complex-valued function \(f(z)\text{,}\) we now have a new concept of the derivative \(f'(z_0)\text{,}\) which by definition cannot depend on the way in which we approach a point \(z_0 = (x_0,y_0) \in \C\text{.}\) It is logical, then, that there should be a relationship between the complex derivative \(f'(z_0)\) and the partial derivatives
\begin{align*}
\fderiv fx(z_0) \amp := \lim_{ x \to x_0 } \frac{ f(x,y_0) -
f(x_0,y_0) }{ x - x_0 }\\
\fderiv fy(z_0) \amp := \lim_{ y \to y_0 } \frac{ f(x_0,y) -
f(x_0,y_0) }{ y - y_0 } \,
\end{align*}
(so this definition is exactly as in the real-valued case).
This relationship between the complex derivative and partial derivatives is very strong, and it is a powerful computational tool. It is described by the
Cauchy–Riemann equations, named after Augustin Louis Cauchy (1789–1857) and Georg Friedrich Bernhard Riemann (1826–1866), even though the equations appeared already in the works of Jean le Rond d’Alembert (1717–1783) and Leonhard Euler (1707–1783).
Before proving
Theorem 2.3.1, we note several comments and give two applications. It is traditional, and often convenient, to write the function
\(f\) in terms of its real and imaginary parts. That is, we write
\(f(z) = f(x,y) = u(x,y) + i \,
v(x,y)\) where
\(u\) is the real part of
\(f\) and
\(v\) is the imaginary part. Then, using the usual shorthand
\(f_x = \fderiv fx\) and
\(f_y = \fderiv fy\text{,}\)
\begin{align*}
f_x \amp \ = \ u_x+i \, v_x\\
-if_y \amp \ = \ -i(u_y+i \, v_y) \ = \ v_y-i \, u_y \,\text{.}
\end{align*}
With this terminology we can rewrite
(2.2) as the pair of equations
\begin{equation}
\begin{array}{ccc}
u_x(x_0,y_0) \amp = \amp v_y(x_0,y_0) \\
u_y(x_0,y_0) \amp = \amp -v_x(x_0,y_0) \, .
\end{array}\tag{2.3}
\end{equation}
As stated, parts (a) and (b) in
Theorem 2.3.1 are not quite converse statements. However, we will show in
Corollary 5.1.6 that if
\(f\) is
holomorphic at
\(z_0 = x_0 + i y_0\) then
\(u\) and
\(v\) have continuous partials (of any order) at
\(z_0\text{.}\) That is, in
Chapter 5 we will see that
\(f = u + i v\) is holomorphic in an open set
\(G\) if and only if
\(u\) and
\(v\) have continuous partials that satisfy
(2.3) in
\(G\text{.}\)
If
\(u\) and
\(v\) satisfy
(2.3) and their second partials are also continuous, then
\begin{equation}
u_{xx}(x_0,y_0) \ = \ v_{yx}(x_0,y_0) \ = \ v_{xy}(x_0,y_0) \ = \ - u_{yy}(x_0,y_0) \,\text{,}\tag{2.4}
\end{equation}
that is,
\begin{equation*}
u_{xx}(x_0,y_0) + u_{yy}(x_0,y_0) \ = \ 0
\end{equation*}
(and an analogous identity for
\(v\)). Functions with continuous second partials satisfying this partial differential equation on a region
\(G \subset \C\) (though not necessarily
(2.3)) are called
harmonic on
\(G\text{;}\) we will study such functions in
Chapter 6. Again, as we will see later, if
\(f\) is holomorphic in an open set
\(G\) then the partials of any order of
\(u\) and
\(v\) exist; hence we will show that the real and imaginary parts of a function that is holomorphic in an open set are harmonic on that set.
Proof.
(a) If \(f\) is differentiable at \(z_0 = (x_0, y_0)\) then
\begin{equation*}
f'(z_0) \ = \ \lim_{\D z \to 0}\frac{f(z_0+\D z) - f(z_0)}{\D z} \,\text{.}
\end{equation*}
As we know by now, we must get the same result if we restrict \(\D z\) to be on the real axis and if we restrict it to be on the imaginary axis. In the first case, \(\D z=\D x\) and
\begin{align*}
f'(z_0) \amp \ = \ \lim_{\D x \to 0}\frac{f(z_0+\D x) -
f(z_0)}{\D x}\\
\amp \ = \ \lim_{\D x \to 0}\frac{f(x_0+\D
x,y_0)-f(x_0,y_0)}{\D x}\\
\amp \ = \ \fderiv fx(x_0,y_0) \,\text{.}
\end{align*}
In the second case, \(\D z = i \, \D y\) and
\begin{align*}
f'(z_0) \amp \ = \ \lim_{i \, \D y \to 0}\frac{f(z_0+i\D y) -
f(z_0)}{i \, \D y}\\
\amp \ = \ \lim_{\D y \to
0}\frac1i\frac{f(x_0,y_0+\D y)-f(x_0,y_0)}{\D y}\\
\amp \ = \ -i \,
\fderiv fy(x_0,y_0) \,\text{.}
\end{align*}
Thus we have shown that \(f'(z_0) = f_x(z_0) = -if_y(z_0)\text{.}\)
(b) Suppose the Cauchy–Riemann equation
(2.2) holds and the partial derivatives
\(f_x\) and
\(f_y\) are continuous in an open disk centered at
\(z_0\text{.}\) Our goal is to prove that
\(f'(z_0) = f_x(z_0)\text{.}\) By
(2.2),
\begin{align*}
f_x(z_0) \amp \ = \ \frac{\D x + i \, \D y}{\D z}\, f_x(z_0)\\
\amp \ = \ \frac{\D x}{\D z}\, f_x(z_0) + \frac{\D y}{\D z}\, i \,
f_x(z_0)\\
\amp \ = \ \frac{\D x}{\D z}\, f_x(z_0) + \frac{\D y}{\D z}\,
f_y(z_0) \,\text{.}
\end{align*}
On the other hand, we can rewrite the difference quotient for \(f'(z_0)\) as
\begin{align*}
\amp\frac{f(z_0+\D z)-f(z_0)}{\D z}\amp\\
\amp \ = \ \frac{f(z_0+\D
z)-f(z_0+\D x) + f(z_0+\D x) - f(z_0)}{\D z}\\
\amp \ = \ \frac{f(z_0+\D x+i\D y)-f(z_0+\D x)}{\D z}
+ \frac{f(z_0+\D x) - f(z_0)}{\D z} \,\text{.}
\end{align*}
Thus
\begin{align}
\amp\lim_{\D z\to0}\frac{f(z_0+\D z)-f(z_0)}{\D z} - f_x(z_0)\notag\\
\amp = \lim_{\D z\to0} \frac{\D y}{\D
z}\left(\frac{f(z_0+\D x+i\D y)-f(z_0+\D x)}{\D y} -
f_y(z_0)\right)\notag\\
\amp \qquad + \lim_{\D z\to0} \frac{\D x}{\D
z}\left(\frac{f(z_0+\D x) - f(z_0)}{\D x} - f_x(z_0)\right)\text{.}\tag{2.5}
\end{align}
We claim that both limits on the right-hand side are
\(0\text{,}\) so we have achieved our set goal. The fractions
\(\frac{ \D x}{ \D z }\) and
\(\frac{ \D y }{ \D z }\) are bounded in absolute value by
\(1\text{,}\) so we just need to see that the limits of the expressions in parentheses are
\(0\text{.}\) The second term on the right-hand side of
(2.5) has a limit of
\(0\) since, by definition,
\begin{equation*}
f_x(z_0) \ = \ \lim_{\D x\to0}\frac{f(z_0+\D x) - f(z_0)}{\D x}
\end{equation*}
and taking the limit here as \(\D z\to0\) is the same as taking the limit as \(\D x\to0\text{.}\)
We cannot do something equivalent for the first term in
(2.5), since now both
\(\D x\) and
\(\D y\) are involved, and both change as
\(\D z\to0\text{.}\) Instead we apply the Mean-Value
Theorem A.0.2 for real functions, to the real and imaginary parts
\(u(z)\) and
\(v(z)\) of
\(f(z)\text{.}\) Theorem A.0.2 gives real numbers
\(0\lt a,b\lt 1\) such that
\begin{align*}
\frac{u(x_0+\D x, \, y_0+\D y)-u(x_0+\D x, \, y_0)}{\D y} \amp
\ = \ u_y(x_0+\D x, \, y_0+a \, \D y)\\
\frac{v(x_0+\D x, \, y_0+\D y)-v(x_0+\D x, \, y_0)}{\D y} \amp
\ = \ v_y(x_0+\D x, \, y_0+b \, \D y) \,\text{.}
\end{align*}
Thus
\begin{align}
\amp \frac{f(z_0+\D x+ i \D y)-f(z_0+\D x)}{\D y} -
f_y(z_0)\notag\\
\amp = \left( \frac{ u(x_0 + \D x, y_0 + \D y) - u(x_0
+ \D x, y_0) }{ \D y } - u_y(z_0) \right) \notag\\
\amp \quad + i \left(
\frac{ v(x_0 + \D x, y_0 + \D y) - v(x_0 + \D x, y_0) }{
\D y } - v_y(z_0) \right)\notag\\
\amp = \left(u_y(x_0+\D x, y_0+a \D y)-u_y(x_0,
y_0)\right)\tag{2.6}\\
\amp \quad + i\left(v_y(x_0+\D x, y_0+b \D y)-v_y(x_0,
y_0)\right) \text{.}\notag
\end{align}
Because \(u_y\) and \(v_y\) are continuous at \((x_0,y_0)\text{,}\)
\begin{align*}
\lim_{ \D z \to 0 } u_y(x_0+\D x, \, y_0+a \, \D y) \amp \ = \
u_y(x_0, \, y_0)\\
\lim_{ \D z \to 0 } v_y(x_0+\D x, \, y_0+b \, \D y) \amp \ = \
v_y(x_0, \, y_0) \, \text{,}
\end{align*}
and so
(2.6) goes to 0 as
\(\D z \to 0\text{,}\) which we set out to prove.
