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Math 382-0: Kursobjekt

Aaron Greicius

Section 1.10 Cauchy-Riemann equations

We come at last to our first important result, the Cauchy-Riemann equations, which express the differentiability of a complex function \(f\) in terms of its real and imaginary parts.

Subsection Cauchy-Riemann equations

The Cauchy-Riemann equations involve the partial derivatives of the real and imaginary parts of a complex function. We give a brief reminder of the definitions and notation around this concept.

Convention 1.10.1. Partial derivatives.

Given a real-valued function \(u(x,y)\) of two real variables and an interior point \((x_0,y_0)\) of its domain, the partial derivatives of \(u\) at \((x_0,y_0)\) with respect to \(x\) and \(y\) are defined as
\begin{align*} u_x(x_0,y_0) \amp = \lim_{h\to 0}\frac{u(x_0+h,y_0)-u(x_0,y_0)}{h} \amp \\ u_y(x_0,y_0) \amp = \lim_{h\to 0}\frac{u(x_0,y_0+h)-u(x_0,y_0)}{h} \amp \text{,} \end{align*}
provided of course these limits exist. We denote the resulting partial derivative functions as \(u_x\) and \(u_y\text{,}\) or alternatively as \(\frac{\partial u}{\partial x}\) and \(\frac{\partial u}{\partial y}\text{.}\)

Proof.

Example 1.10.3. Cauchy-Riemann verification.

Verify that the real and imaginary parts of the holomorphic function \(f(z)=z^2\) satisfy the Cauchy-Riemann equations for all \(z\in \C\text{,}\) and that \(f'(z)\) satisfies (1.40).
Solution.
We have \(f=u+iv\text{,}\) where \(u(x,y)=x^2-y^2\text{,}\) \(v(x,y)=2xy\text{,}\) and thus
\begin{align*} u_x \amp = 2x \amp u_y\amp = -2y\\ v_x \amp= 2y \amp v_y\amp = 2x \text{.} \end{align*}
It is clear from these computations that
\begin{align*} u_x \amp= v_y \amp u_y \amp = -v_x\text{.} \end{align*}
Furthermore, we saw previously that \(f'(z)=2z\text{,}\) and hence that
\begin{align*} f'(z) \amp =2x+2yi\\ \amp = u_x+iv_x\text{.} \end{align*}

Example 1.10.4. Complex conjugation.

Define \(f\colon \C\rightarrow \C\) as \(f(z)=\overline{z}\text{.}\) Prove that \(f\) is not differentiable anywhere on its domain.
Solution.
We have \(f=u+iv\text{,}\) where \(u(x,y)=x\) and \(v(x,y)=-y\text{.}\) Thus for any \(z=x+iy\) we have
\begin{align*} u_x(x,y) \amp = 1 \amp u_y(x,y)\amp= 0\\ v_x(x,y) \amp = 0 \amp v_y(x,y)\amp =-1\text{.} \end{align*}
It follow that for all \(z_0=x_0+iy_0\) we have
\begin{equation*} u_x(x_0,y_0)=1\ne -1 =v_y(x_0,y_0)\text{.} \end{equation*}
We conclude that for all \(z_0\in \C\text{,}\) \(f\) is not differentiable at \(z_0\text{,}\) as claimed.
As stated, Theorem 1.10.2 provides only a necessary condition for \(f'\) to exist at \(z_0\text{;}\) it makes no claim as to whether this condition is sufficient. The next theorem improves on this slightly, by establishing something close, but not quite equal to the converse of Theorem 1.10.2. Roughly speaking, it asserts that satisfaction of the Cauchy-Riemann equations suffices as long as we throw in a certain continuity condition on the first-order partial derivatives. We will see later that when considering differentiability on open sets this extra condition is in fact automatically satisfied, yielding a true equivalence.

Proof.

In lieu of a proof, we indicate the ideas at play. As it turns out, the condition we actually need, in addition to the satisfaction of the Cauchy-Riemann equations, is that the function
\begin{align*} f\colon \amp \R^2\rightarrow \R^2\\ (x,y) \amp \mapsto (u(x,y),v(x,y)) \end{align*}
is differentiable in the sense of multivariable calculus. I will not include that definition here (look it up in your calculus text!), but suffice it to say that it is not equivalent to the partial derivatives of \(u\) and \(v\) existing. The additional continuity condition given in the theorem implies that \(f\) is differentiable in the multivariable calculus sense, and hence, that \(f\) is differentiable in the complex sense, assuming the Cauchy-Riemann equations are satisfied.
As an immediate corollary of Theorem 1.10.5 we have the following corollary, which will be our main tool for investigating differentiability.

Example 1.10.7. Exponential and trigonometric functions.

Show that \(\exp\text{,}\) \(\cos\text{,}\) and \(\sin\) are holomorphic functions on \(\C\) and compute formulas for their derivatives.
Solution.
We have \(e^z=u+iv\text{,}\) where \(u(x,y)=e^x\cos y\) and \(v(x,y)=e^x\sin y\text{.}\) We then have
\begin{align*} u_x \amp= e^x\cos y \amp u_y\amp =-e^x\sin y \\ v_x \amp=e^x\sin y \amp e^x\cos y \text{,} \end{align*}
whence we see that (a) these partials exist and are continuous, and (b) satisfy the Cauchy-Riemann equations (1.41) for all \(z=x+iy\text{.}\) We conclude that \(f(z)=e^z\) is differentiable everywhere and satisfies
\begin{equation*} \frac{d}{dz}(e^z)=u_x+i v_x=e^x\cos y+e^x\sin y i=e^z\text{.} \end{equation*}
That
\begin{align*} \cos z \amp = \frac{e^{iz}+e^{-iz}}{2} \amp \sin(z)\amp =\frac{e^{iz}-e^{-iz}}{2i} \end{align*}
are differentiable everywhere now follows from the fact that the polynomial functions \(g(z)=iz\) and \(h(z)=-iz\) are differentiable, from which it follows by the chain rule that \(e^{iz}\) and \(e^{-iz}\) are differentiable, and then from the linear combination rule of Theorem 1.9.8 that \(\cos z\) and \(\sin z\) are differentiable everywhere. Furthermore we can compute the derivatives of \(\cos\) and \(\sin\) using these same rules and our formula for \(e^z\) to conclude that
\begin{align*} \frac{d}{dz}(\cos z) \amp =-\sin z\\ \frac{d}{dz}(\sin z) \amp = \cos z\text{.} \end{align*}
As with good old-fashioned calculus, a complex function whose derivative is zero everywhere (on a connected open set) is constant. Let’s see why this is so. First some definitions and conventions.

Convention 1.10.8. Restrictions and constant functions.

Given a function \(f\colon D\rightarrow \C\) and subset \(D'\subseteq D\text{,}\) we denote by \(f\vert_{D'}\) the restriction of \(f\) to \(D'\text{:}\) that is
\begin{align*} f\vert_{D'}\colon D' \amp\rightarrow \C \\ z \amp \longmapsto f(z)\text{.} \end{align*}
Note that although \(f\vert_{D'}\) is defined via the same rule as \(f\text{,}\) if \(D'\ne D\text{,}\) it is technically a different function.
Given \(D'\subseteq D\) we say that \(f\) is on constant on \(D'\) if there is a complex number \(w_0\in \C\) such that \(f(z)=w_0\) for all \(z\in D'\text{.}\) We will write \(f\vert_{D'}=w_0\) in this case.

Definition 1.10.9. Locally constant.

A function \(f\) is locally constant on a set \(D\) if for all \(z_0\in U\) there is an open ball \(B_r(z_0)\subseteq U\) such that \(f\) is constant on \(B_r(z_0)\cap D\text{.}\)

Proof.

Implication: (i)\(\implies\)(ii).
Assume \(f\) is locally constant. Given any \(z_0\in U\) we can find an open ball \(V=B_r(z_0)\subseteq U\) such that the restriction \(f\vert_V=w_0\text{.}\) for some \(w_0\in \C\text{.}\) It is easy to see that \(f\) is differentiable at \(z_0\) if and only if its restriction \(f_V\) is differentiable at \(z_0\text{,}\) and that
\begin{equation*} f'(z_0)=(f\vert_V)'(z_0)=0\text{,} \end{equation*}
since \(f\vert_V\) is constant. (See Example 1.12.7.) It follows that \(f'(z)=0\) for all \(z\in U\text{.}\)
Implication: (ii)\(\implies\)(i).
Assume \(f'(z)=0\) for all \(z\in U\text{.}\) We will show that \(f\) is locally constant. Given \(z_0\in U\text{,}\) there is an open ball \(V=B_{r}(z_0)\subseteq U\text{.}\) Writing \(f=u+iv\) and using Theorem 1.10.2, we have
\begin{equation*} 0=f'(z)=u_x(z)+iv_x(z)=v_y(z)-iu_y(z), \end{equation*}
and thus
\begin{equation*} u_x(z)=u_y(z)=v_x(z)=v_y(z)=0 \end{equation*}
for all \(z\in V\text{.}\) Since \(V=B_r(z_0)\) is connected, it follows from multivariable calculus facts that \(u(x,y)=C\) and \(v(x,y)=D\) for some constants \(C,D\in \R\) for all \(z=x+iy\in V\text{.}\) Thus the restriction \(f\vert_V=C+iD\) is constant on \(V\text{.}\) This proves \(f\) is locally constant.
The next corollary follows directly from Theorem 1.10.10 and Theorem 1.10.13. The latter was left as a homework exercise, but is important enough to enshrine here as an official result.

Definition 1.10.12. Discrete subset.

A set \(D\subseteq \C\) is discrete if for all \(z_0\in D\) there is an open ball \(B_r(z_0)\) such that \(B_r(z_0)\cap D=\{z_0\}\text{.}\)

Proof.

This is left as a homework exercise.

Subsection Polar Cauchy-Riemann equations

As many of our familiar functions have a nice description in polar coordinates, it helps to have a corresponding polar version of the Cauchy-Riemann equations. Technically speaking, when we “represent” a function \(u(x,y)\)in polar coordinates we are really looking at the function
\begin{align*} g(r,\theta) \amp =u(r\cos \theta, r\sin\theta)=u(x(r,\theta) y(r,\theta))\text{,} \end{align*}
obtained by pre-composing \(u\) with the function \(F\colon \R^2\rightarrow \R^2\) defined as
\begin{equation*} F(r,\theta)=(r\cos\theta, r\sin\theta)\text{.} \end{equation*}
In an effort to pare down notation, we will adopt the following abuses of notation:
\begin{align*} u(r,\theta) \amp = u(r\cos\theta, r\sin\theta)=g(r,\theta)\\ u_r(r,\theta) \amp = g_r(r,\theta)\\ u_\theta(r,\theta) \amp = g_\theta(r,\theta)\text{.} \end{align*}
As a result, we can express the chain rule applied to \(g\) as follows:
\begin{align*} u_r \amp =u_x\, x_r+u_y\, y_r=\cos\theta\, u_x+\sin\theta\, u_v\\ u_\theta \amp = u_x\, x_\theta+u_y\, y_\theta=-r\sin\theta\, u_x+r\cos\theta\, u_y\text{.} \end{align*}

Proof.

This is left as a homework exercise.

Example 1.10.15. Derivative of \(\Log\).

Show that \(\Log\) is differentiable on its domain \(\C-(-\infty, 0]\) and compute a formula for its derivative.
Solution.
Fix any \(z_0\in \C-(-\infty, 0]\text{,}\) and write \(z_0=r_0e^{i\theta_0}\text{.}\) Picking a small enough neighborhood in the \(r\theta\)-plane, we can assume that for \((r,\theta)\) close enough to \((r_0,\theta_0)\) we have \(z=re^{i\theta}\) if and only if \(r=\abs{z}\) and \(\Arg z=\theta+2\pi k\) for some fixed \(k\in \Z\text{.}\) (This is because the map \((r,\theta)\mapsto \Arg(e^{i\theta})\) is continuous.) It follows that
\begin{equation*} \Log(z)=\ln\abs{z}+i(\theta+2\pi k) \end{equation*}
and thus \(u(r,\theta)=\ln r\) and \(v(r,\theta)=\theta+2\pi k\text{.}\) Computing partial derivatives, we see have
\begin{align*} u_r(r_0,\theta_0) \amp = \frac{1}{r_0} \amp u_\theta \amp = 0\\ v_r \amp = 0 \amp v_\theta\amp =1\text{,} \end{align*}
from which we see clearly that
\begin{align*} u_r \amp = \frac{1}{r}v_\theta \amp u_\theta\amp =-r\v_r\text{.} \end{align*}
It follows that \(\Log z\) is differentiable on \(\C-(-\infty,0]\) and that given \(z=re^{i\theta}\text{,}\) we have
\begin{equation*} \frac{d}{dz}(\Log z)=e^{-i\theta}(\frac{1}{r}+0i)=\frac{1}{r}e^{-i\theta}=\frac{1}{z}\text{.} \end{equation*}
We conclude that
\begin{equation*} \frac{d}{dz}\Log z=\frac{1}{z} \end{equation*}
for all \(z\in \C-(-\infty, 0]\text{,}\) as desired.

Example 1.10.16. Reciprocal function.

Define \(f\colon \C-\{0\}\rightarrow \C-\{0\}\) as \(f(z)=1/z\text{.}\) Re-derive the derivative formula \(f'(z)=-1/z^2\) using the polar Cauchy-Riemann equations.
Solution.
For \(z=re^{i\theta}\text{,}\) we have
\begin{equation*} f(z)=\frac{1}{r}e^{-i\theta}=\frac{\cos(-\theta)}{r}+i\,\frac{\sin(-\theta)}{r}\text{,} \end{equation*}
and thus \(u(r,\theta)=\cos(-\theta)/r\) and \(v(r,\theta)=\sin(-\theta)/r\text{.}\) The relevant partial derivatives are then
\begin{align*} u_r \amp = -\frac{\cos(-\theta)}{r^2} \amp u_\theta\amp =\frac{\sin(-\theta)}{r}\\ v_r \amp = -\frac{\sin(-\theta)}{r^2} \amp v_\theta\amp =-\frac{\cos(-\theta)}{r} \text{.} \end{align*}
We then easily see that
\begin{align*} u_r \amp = \frac{1}{r}v_\theta\\ u_\theta \amp = -rv_r\text{.} \end{align*}
We conclude that \(f\) is differentiable on \(\C-\{0\}\) and
\begin{align*} f'(z) \amp = e^{-i\theta}(u_r(r,\theta)+iv_r(r,\theta))\\ \amp = e^{-i\theta}\left(-\frac{\cos(-\theta)}{r^2}-i\frac{\sin(-\theta)}{r^2}\right)\\ \amp = -\frac{1}{r^2}\, e^{-i\theta}e^{-i\theta}\\ \amp = -\frac{1}{r^2}\, e^{-2i\theta}\\ \amp = -\frac{1}{z^2}\text{.} \end{align*}

Example 1.10.17. Derivative of power function.

Fix a complex number \(w\) and define \(f(z)\) to be the principle branch of \(z^w\text{.}\) Show that \(f\) is differentiable on its domain and compute a formula for its derivative.
Solution.
The function \(f\) has formula \(f(z)=e^{w\Log z}=\exp(w\Log z)\text{.}\) The derivative of \(f\) is easily computed using the the chain rule:
\begin{align*} f'(z) \amp = \exp'(w\Log z)\cdot (w\Log z)'\\ \amp = w\exp(w\Log z)\cdot \frac{1}{z} \amp (\exp'=\exp, (Log z)'=1/z\\ \amp = \frac{f(z)}{z}\text{.} \end{align*}
You were expecting \(f'(z)=z^{w-1}\text{?}\) Since \(z^w\) and \(z^{w-1}\) are not complex functions in themselves, the answer was never going to be this easy. However, writing
\begin{equation*} z^{w-1}=\frac{z^w}{z} \end{equation*}
and swapping \(z^{w}\) out for \(f\text{,}\) we see that our formula is equivalent to the familiar power rule.
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