Chapter 24Holomorphic Functions and the Cauchy–Riemann Equations
The definition of the derivative at a point that we learned in high school mathematics is $\displaystyle f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$. It was enough merely to check that the real number $h$ gives the same limit value whether it approaches $a$ from the right ($h\to+0$) or from the left ($h\to-0$)—for there are only two “directions” on the number line. What happens, then, if we carry the same defining formula over unchanged to a complex function $f(z)$? Here $z=x+iy$ is a point in the plane, and $h$ is also a complex number, so the “directions” of approach toward $z$ are not just from the right or from the left; there are in fact infinitely many—from directly above, from a 45° diagonal, spiraling in, and so on. For a complex function to be “differentiable,” it is required that the limit take the same value for all of these ways of approaching. In this chapter we verify that this requirement, which seems at first glance to be a matter of course, is in fact a surprisingly strong constraint, one that forces the system of partial differential equations $u_x=v_y,\ u_y=-v_x$—the Cauchy–Riemann equations (C-R equations)—to hold between the real part $u(x,y)$ and the imaginary part $v(x,y)$ of the function. A function satisfying these equations is called a holomorphic function, and it is holomorphic functions that play the leading role in all of complex analysis.
This chapter begins by establishing the vocabulary of neighborhoods and domains—the stage on which we can talk about the “differentiability” of complex functions (Section 24.1). We then define continuity and holomorphy (Section 24.2), and give a complete proof, in both directions, of the equivalence “holomorphic $\iff$ differentiable,” whose proof broke off in its second half in the source notes, with only a forward reference to the “mean value theorem for functions of two variables” (Section 24.3). Finally, using this equivalence as our main tool, we actually decide whether concrete functions such as $z^2+3z,\ \bar z,\ |z|^2,\ e^z,\ \dfrac1z$ are holomorphic, and then derive entirely by ourselves the derivatives of $e^z,\ \sin z,\ \cos z,\ \log z,\ \alpha^z$ and the linearity, product, and quotient rules of differentiation, including the parts that the original notes left as “proof omitted” (Section 24.4). The complex integration and Cauchy's integral theorem that we study from the next chapter (Chapter 25) onward are built on the foundation of holomorphy established in this chapter.
- The definitions of neighborhood ($r$-neighborhood) and domain (open domain)—openness (the boundary is not included) and connectedness (any two points can be joined by a polygonal path lying inside)
- The definition of continuity of a complex function and its essential difference from real functions—getting a feel for what it means to “converge to the same value no matter how the point approaches” through the concrete example $\bar z/z$, whose limit does not exist
- The definition of holomorphy and the Cauchy–Riemann equations $u_x=v_y,\ u_y=-v_x$
- A complete proof of the equivalence “holomorphic $\iff$ differentiable”—completing, with the mean value theorem for functions of two variables (Chapter 6), the direction “holomorphic $\Rightarrow$ differentiable” that was left unfinished in the original notes
- The computational skill to actually decide, using the C-R equations, that $f(z)=z^2+3z$ is holomorphic on the whole plane, that $f(z)=\bar z$ is not holomorphic at any point, and that $f(z)=|z|^2$ is differentiable only at the origin
- Complete proofs of the linearity of differentiation, the product rule, and the quotient rule (writing out the part that the original notes skipped as “omitted, since the proof is the same as in the real case”)
- A method for deriving the derivatives of $e^z,\ \sin z,\ \cos z,\ \alpha^z$ in a unified way from the general formula $\dfrac{d}{dz}e^{cz}=ce^{cz}$ ($c$ an arbitrary complex constant)
- The confirmation that the derivative of $\log z$ (principal value) is $1/z$, and the branch cut, which we become aware of for the first time there—a situation in which the notion of “domain” from Section 24.1 comes concretely into play
Source notes: Yasuhide Mochizuki, Mathematics Notes: Complex Analysis, pp. 21–26.
24.1 Neighborhoods and Domains — The Vocabulary of the Complex Plane
When we say that a real function $f(x)$ is “differentiable at $x=a$,” we are implicitly assuming that $f$ is defined on an interval extending to both sides of $a$. The same holds for a complex function $f(z)$: before we talk about being “differentiable at $z=z_0$” or “holomorphic at $z=z_0$,” we first need language to state precisely over how large a range around $z_0$ the function $f$ is defined. The tools for this are the neighborhood and the domain, which we introduce in this section.
24.1.1 Neighborhoods—Turning “near $z_0$” into a formula
The distance between the complex numbers $z=x+iy$ and $z_0=x_0+iy_0$ is, by the distance formula for two points, $|z-z_0|=\sqrt{(x-x_0)^2+(y-y_0)^2}$ ($|\cdot|$ is the absolute value of a complex number; we have merely applied the “distance from the origin” that we confirmed in the previous chapters to $z-z_0$, the “relative position when $z_0$ is regarded as the origin”). Using this distance, we define “the collection of all points within a certain range of $z_0$.”
Definition 24.1: Neighborhood ($r$-neighborhood)
For a point $z_0$ in the complex plane and a positive real number $r>0$, the set of all points $z$ satisfying
$$ |z-z_0| \lt r $$is called the $r$-neighborhood of $z_0$, or simply a neighborhood. As in Figure 24.1, the $r$-neighborhood consists of all the points in the interior of the circle with center $z_0$ and radius $r$ (the circle itself is not included).
Note that the inequality is $\lt$ (which does not include equality), not $\le$. That is, a neighborhood does not include the circumference; it is the interior of the circle in the strictest sense. This detail—“the boundary is not included”—will turn out to be decisive for the domains defined next as well.
24.1.2 Domains—Openness and connectedness
When we work with complex functions, the set on which $f$ is defined is not an “interval” but a “region” in the plane. That region cannot be of just any shape; we require two properties that make it easy to handle in complex analysis—openness and connectedness.
Definition 24.2: Domain (open domain)
When a set $D$ in the complex plane satisfies the following two conditions, $D$ is called a domain or an open domain.
(i) Being an open set: for every point $z$ in $D$ there exists a neighborhood contained in $D$ (that is, a sufficiently small disk centered at $z$ fits entirely inside $D$ without sticking out).
(ii) Being connected: any two points in $D$ can be joined by a polygonal path contained in $D$.
Intuition: Why must the boundary not be included? (condition (i))
Condition (i) says: “whichever point $z$ of $D$ we choose, there exists a positive real number $r$ such that the $r$-neighborhood of $z$ fits entirely inside $D$.” Let us consider the gourd-shaped domain $D$ (the shaded part) in Figure 24.2. If we choose a point in the interior (the center of the large dotted circle in the figure), even a neighborhood with a fairly large radius still fits inside $D$. Even for a point near a place where $D$ is pinched narrow (the center of the small dotted circle in the figure), the neighborhood still lies inside $D$ as long as we take the radius sufficiently small—however narrow the place, it suffices to make the radius smaller than that narrowness, so note that $r$ may be varied from place to place.
But the situation changes if we choose a point on the boundary line (the center of the dotted circle at the top of the figure). No matter how small we take $r$, its neighborhood will always contain points outside $D$ (because, by definition, the region just outside the boundary is outside $D$). That is, for a point on the boundary there is no “neighborhood contained in $D$” for any positive real number $r$. In other words, if we try to satisfy this condition at a point on the boundary line, the radius we need ends up being $r=0$—but the definition of a neighborhood (Definition 24.1) required $r>0$, so this is not a neighborhood. This is precisely why a domain does not include points on its boundary.
Intuition: Why is connectedness needed? (condition (ii))
Condition (ii) says: “any two points of $D$ can be joined by a polygonal path that does not leave $D$.” In Figure 24.3 we consider a situation in which, inside the domain $D$ (the interior of the large dotted circle), there is one small excluded obstacle (the solid egg-shaped curve). When we want to go from a point $z_1$ in $D$ to a point $z_2$, we can join them using only points of $D$ by drawing a polygonal path that detours around the obstacle. This is what it means to be “connected.”
This property is not very conspicuous in this chapter alone, but it becomes decisively important for the complex integrals (line integrals) that we study in the following chapters—for it is precisely connectedness that guarantees that a curve to be used as a path of integration can be drawn freely inside the domain. For now it suffices to keep only the plain mental picture that “there is always a path joining two points.”
Example 24.1: Determining whether points lie in a neighborhood
Consider the neighborhood $|z-z_0|\lt 1$ with center $z_0=1+i$ and radius $r=1$. Are the following two points contained in this neighborhood? (a) $z_1=1.5+1.5i$ (b) $z_2=2+2i$
Solution. We use the distance formula $|z-z_0|=\sqrt{(x-x_0)^2+(y-y_0)^2}$.
(a) Since $z_1-z_0=0.5+0.5i$,
$$ |z_1-z_0|=\sqrt{0.5^2+0.5^2}=\sqrt{0.5}=\frac{1}{\sqrt2}\approx0.707 $$Since this is smaller than $r=1$, $z_1$ is contained in this neighborhood.
(b) Since $z_2-z_0=1+i$,
$$ |z_2-z_0|=\sqrt{1^2+1^2}=\sqrt2\approx1.414 $$Since this is larger than $r=1$, $z_2$ is not contained in this neighborhood (verified with SymPy: $|z_1-z_0|=\frac{\sqrt2}{2}\lt1$, $|z_2-z_0|=\sqrt2\gt1$).
Example 24.2: Determining whether a given set is a domain
For each of the following four sets, determine whether it is a domain (a set that is both open and connected, in the sense of Definition 24.2).
$D_1=\{z: |z|\lt2\}$, $D_2=\{z: |z|\le2\}$, $D_3=\{z: |z|\lt1\}\cup\{z: |z-3|\lt1\}$, $D_4=\{z: 1\lt|z|\lt2\}$ (an annulus)
Solution. The four sets are shown in Figure 24.4.
$D_1$: This is the interior of the circle of radius 2 centered at the origin (the circle itself is not included). Around any interior point, a neighborhood fits snugly inside, provided we choose the radius suitably with respect to the distance to the boundary (it is an open set). Also, in the interior of a disk any two points can be joined by a straight line (it is connected). Hence it is a domain.
$D_2$ is $D_1$ with the boundary circle $|z|=2$ added. Consider a point on the boundary, for example $z=2$: no matter how small a radius $r$ we take for a neighborhood, it always contains points satisfying $|z|\gt2$ (points outside $D_2$). Therefore openness (i) fails, and it is not a domain.
$D_3$ consists of two disks of radius 1 placed with their centers a distance of 3 apart. Since each of the $2$ disks is open, $D_3$ itself is also open, but the distance $3$ between the centers is greater than the sum of the radii $1+1=2$, so the two disks do not intersect each other. If we try to draw a polygonal path joining a point of one disk to a point of the other, it must pass through a “gap” that does not belong to $D_3$. Therefore connectedness (ii) fails, and it is not a domain.
$D_4$ is an annulus (doughnut shape), obtained by hollowing out a disk of radius 1 from the center. Any interior point is away from both the inner boundary $|z|=1$ and the outer boundary $|z|=2$, so a sufficiently small neighborhood can be taken (it is open). Also, any two points in the annulus can be joined by a polygonal path that does not leave the annulus (for example, a polygonal path that goes around along a circle of the same radius) (it is connected). Hence it is a domain—but note that, unlike $D_1$, it is a domain that does not contain the center (the origin).
24.2 Continuity and Holomorphy of Complex Functions — The Cauchy–Riemann Equations
Now that the stage of neighborhoods and domains is set, we at last define the continuity of complex functions and the star of this chapter, holomorphy.
24.2.1 Continuity of complex functions—the condition “no matter how we approach”
Definition 24.3: Continuity of complex functions
Let a complex function $f(z)$ be defined at $z=z_0$. If
$$ \lim_{z\to z_0} f(z) = f(z_0) $$holds, we say that $f(z)$ is continuous at $z=z_0$.
Same form, different content—the essential difference from real functions
If we compare the formula of Definition 24.3 by appearance alone, it looks exactly the same as the definition of continuity of a real function that you learned in high school mathematics. But the content is entirely different. For a real function $f(x)$, the variable $x$ moves along the number line, so there are only two ways to approach $x_0$: “from the right” and “from the left.” For a complex function $f(z)$, however, $z$ moves in the complex plane. The paths approaching $z_0$ include those from the right, from the left, from above, from below, diagonally, spiraling in, …—infinitely many of them.
The formula $\displaystyle\lim_{z\to z_0}f(z)=f(z_0)$ of Definition 24.3 demands that, for all of these ways of approaching, the function converge to the same limit value $f(z_0)$. In other words, the limit of a complex function means that “it converges to the same value no matter how we approach.” Conversely, if even just two ways of approaching converge to different values, we can conclude right then that the limit does not exist (that is, the function is not continuous). Let us get a feel for this difference in the next example.
Example 24.3: An example in which the limit does not exist—$\bar z/z$
For $\displaystyle f(z)=\frac{\bar z}{z}$ ($z\ne0$), show that $\displaystyle\lim_{z\to0}f(z)$ does not exist.
Solution. We actually compute and compare the limit values for three ways of approaching: along the real axis, along the imaginary axis, and along a diagonal.
(a) Along the real axis: let $z=t$ ($t$ is real, $t\to0$). Since $\bar z=t$,
$$ f(t)=\frac{t}{t}=1 \quad\Longrightarrow\quad \lim_{t\to0}f(t)=1 $$(b) Along the imaginary axis: let $z=it$ ($t\to0$). Since $\bar z=\overline{it}=-it$,
$$ f(it)=\frac{-it}{it}=-1 \quad\Longrightarrow\quad \lim_{t\to0}f(it)=-1 $$(c) Along a diagonal (just to be extra safe): let $z=(1+i)t$ ($t\to0$). Since $\bar z=(1-i)t$,
$$ f\bigl((1+i)t\bigr)=\frac{(1-i)t}{(1+i)t}=\frac{1-i}{1+i}=\frac{(1-i)^2}{(1+i)(1-i)}=\frac{1-2i-1}{2}=-i \quad\Longrightarrow\quad \lim_{t\to0}f\bigl((1+i)t\bigr)=-i $$(To rationalize the denominator, we multiplied by the conjugate, just as in the real case, and used $(1+i)(1-i)=1-i^2=2$.)
Depending on the way of approach, we got three different values: $1$ along the real axis, $-1$ along the imaginary axis, and $-i$ along the diagonal. Definition 24.3 requires “the same value no matter how we approach,” so, since the values do not agree, $\displaystyle\lim_{z\to0}\bar z/z$ does not exist (verified with SymPy: all three directions give the same values as above).
24.2.2 The definition of holomorphy and the Cauchy–Riemann equations
Consider a complex function $f(z)=u(x,y)+iv(x,y)$ ($z=x+iy$, where $u,v$ are real-valued functions of two variables representing the real part and the imaginary part, respectively). If we carry the definition of the derivative at a point from high school mathematics over to complex numbers just as it is, we are led to consider the limit
$$ f'(z) = \lim_{h\to0}\frac{f(z+h)-f(z)}{h} $$As we saw in Section 24.2.1, for complex numbers there are infinitely many ways of letting $h\to0$, so the existence of this limit (that is, differentiability) is a much stronger condition than the differentiability of a real function. The definition of holomorphy below restates this strong condition solely in terms of the partial derivatives of $u,v$ (in Section 24.3 of this chapter we prove that this definition is in fact equivalent to differentiability).
Definition 24.4: Holomorphy and the Cauchy–Riemann equations
Suppose that, for $f(z)=u(x,y)+iv(x,y)$, the functions $u,v$ are once continuously differentiable in some neighborhood of the point $(x_0,y_0)$ (the partial derivatives $u_x,u_y,v_x,v_y$ exist and are all continuous functions—this property is called being of class $C^1$; “$C^1$” is read “C-one”), and moreover that
\begin{equation} u_x = v_y ,\qquad u_y = -v_x \label{eq:24-cr} \end{equation}holds. Then we say that $f(z)$ is holomorphic at $z_0=x_0+iy_0$. We call Eq. \eqref{eq:24-cr} the Cauchy–Riemann equations (hereafter, the C-R equations). The symbol $u_x=\pdiff ux$ means “the partial derivative of $u$ with respect to $x$” (Chapter 6, Section 6.2).
(The source notes merely say “once differentiable” and do not require continuity. However, when we prove “holomorphic $\Rightarrow$ differentiable” in Section 24.3.2, we use Theorem 6.2 of Chapter 6 (Taylor's theorem for functions of two variables), and for that the continuity of the partial derivatives is in fact essential. In order to carry the proof through rigorously to the end, this book explicitly adds this assumption at the stage of the definition. The functions that we actually treat in Section 24.4—polynomials, $e^z,\sin z,\cos z,\log z,1/z$, and so on—are all real analytic and hence automatically of class $C^1$ (in fact, much more strongly, differentiable arbitrarily many times), so this assumption, which is one step stronger, does not change the conclusions of the examples and exercises.)
Theorem 24.1: Equivalence of holomorphy and differentiability
For a complex function $f(z)$ ($z=x+iy\in\mathbb C$), the following two statements are equivalent.
“$f(z)$ is holomorphic at $z$” $\iff$ “$f(z)$ is differentiable at $z$ (in the complex sense)”
We prove this theorem in the next Section 24.3, taking our time and splitting it into two directions—“differentiable $\Rightarrow$ holomorphic” and “holomorphic $\Rightarrow$ differentiable.” Here, first, let us practice investigating functions by using the definition of holomorphy (the C-R equations) directly.
Example 24.4: Directly checking that $f(z)=e^z$ satisfies the C-R equations
For the complex exponential function $e^z=e^x(\cos y+i\sin y)$ defined in Chapter 23, verify directly that the C-R equations $u_x=v_y,\ u_y=-v_x$ hold (at every point of the whole plane).
Solution. From $e^z=e^x(\cos y+i\sin y)$, the real part and the imaginary part are, respectively,
$$ u(x,y)=e^x\cos y ,\qquad v(x,y)=e^x\sin y $$We compute the partial derivatives. When we differentiate $u$ partially with respect to $x$, we regard $y$ as a constant, so only the factor $e^x$ is differentiated, and
$$ u_x = e^x\cos y $$When we differentiate $u$ partially with respect to $y$, we regard $x$ as a constant, so $\cos y$ becomes $-\sin y$, and
$$ u_y = -e^x\sin y $$Similarly, for $v$,
$$ v_x = e^x\sin y ,\qquad v_y = e^x\cos y $$Therefore,
$$ u_x = e^x\cos y = v_y ,\qquad u_y = -e^x\sin y = -v_x $$Thus we have confirmed that the C-R equations hold (identically, regardless of the values of $x,y$) (verified with SymPy). Hence $e^z$ is holomorphic on the whole plane—we will also use this fact when we find the derivative in Section 24.4.
24.3 Holomorphy Is Equivalent to Differentiability
We now prove Theorem 24.1, “holomorphic $\iff$ differentiable,” which was announced in the previous section, by splitting it into its two directions. In the source notes, direction (i) was written out to the end, but direction (ii) was left unfinished at the point where the notes announced that they would use “the mean value theorem for functions of two variables”; the author ran out of steam there and broke off. In this section we complete the continuation ourselves, using the tools of Chapter 6, Section 6.6.
24.3.1 Differentiable $\Rightarrow$ holomorphic
Proof: Differentiable $\Rightarrow$ holomorphic
Suppose that $f(z)=u(x,y)+iv(x,y)$ is differentiable (in the complex sense) at $z=x+iy$. That is, suppose that the limit
$$ f'(z) = \lim_{h\to0}\frac{f(z+h)-f(z)}{h} $$exists, and moreover that it takes the same value no matter how $h$ approaches $0$. Letting $h=\alpha+i\beta$ ($\alpha,\beta$ real), we have
$$ f(z+h)-f(z) = \bigl[u(x+\alpha,y+\beta)-u(x,y)\bigr] + i\bigl[v(x+\alpha,y+\beta)-v(x,y)\bigr] $$and hence we obtain the limit
$$ f'(z) = \lim_{\substack{\alpha\to0\\ \beta\to0}}\left(\frac{u(x+\alpha,y+\beta)-u(x,y)}{\alpha+i\beta}+i\,\frac{v(x+\alpha,y+\beta)-v(x,y)}{\alpha+i\beta}\right) $$Since the value is “the same for any way of approaching,” the limits computed for the following two special ways of approaching—those in which $h$ is moved only in the real-axis direction or only in the imaginary-axis direction—must of course also have the same value as the original $f'(z)$. Figure 24.5 illustrates these two directions.
[1] When $\beta=0,\ \alpha\to0$ (we move $h$ only in the real-axis direction; that is, we fix $y$ and move only $x$). Setting $\beta=0$ in the formula above, we obtain
\begin{equation} f'(z) = \lim_{\alpha\to0}\left(\frac{u(x+\alpha,y)-u(x,y)}{\alpha}+i\,\frac{v(x+\alpha,y)-v(x,y)}{\alpha}\right) = \pdiff ux + i\pdiff vx = u_x + iv_x \label{eq:24-dir1} \end{equation}(Each term on the right-hand side is exactly the definition of the partial derivatives $u_x,v_x$: differentiating with respect to $x$ while holding $y$ fixed.)
[2] When $\alpha=0,\ \beta\to0$ (we move $h$ only in the imaginary-axis direction; that is, we fix $x$ and move only $y$). Setting $\alpha=0$ makes the denominator $i\beta$, so
$$ f'(z) = \lim_{\beta\to0}\left(\frac{u(x,y+\beta)-u(x,y)}{i\beta}+i\,\frac{v(x,y+\beta)-v(x,y)}{i\beta}\right) $$Here we use $\dfrac1i=-i$ (because $i\times(-i)=-i^2=1$) to remove the $i$ from the denominator, obtaining
\begin{equation} f'(z) = \lim_{\beta\to0}\left(-i\,\frac{u(x,y+\beta)-u(x,y)}{\beta}+\frac{v(x,y+\beta)-v(x,y)}{\beta}\right) = -i\pdiff uy + \pdiff vy = -iu_y + v_y \label{eq:24-dir2} \end{equation}Why is it enough to compare just these two directions?
The differentiability of Definition 24.4 demands “the same limit for every way of approaching.” What we want to find out now is what follows from that strong demand. [1] and [2] are only two special cases (the real-axis and imaginary-axis directions) among the many ways of approaching. But if $f'(z)$ really is determined as a single value, then the value obtained in computation [1] and the value obtained in computation [2] must both equal that same $f'(z)$—this is the perfectly ordinary logic that if “everyone gives the same answer,” then “the answers of two of them also agree.” So the trick of this proof is that equating the right-hand sides of Eq. \eqref{eq:24-dir1} and Eq. \eqref{eq:24-dir2} draws out relations among the partial derivatives of $u,v$.
Since the complex function is assumed to be differentiable, [1] and [2] both equal the same $f'(z)$, so
$$ u_x + iv_x = -iu_y + v_y $$Both sides are complex numbers, and two complex numbers are equal exactly when their real parts are equal and their imaginary parts are equal (covered in high school mathematics). The left-hand side has real part $u_x$ and imaginary part $v_x$; the right-hand side has real part $v_y$ and imaginary part $-u_y$. Hence, comparing real parts gives $u_x=v_y$, and comparing imaginary parts gives $v_x=-u_y$, that is, $u_y=-v_x$.
$$ \therefore\ u_x=v_y,\quad u_y=-v_x \qquad\text{(Cauchy–Riemann equations)} $$∎ (differentiable $\Rightarrow$ holomorphic)
Corollary 24.1: Formula for the derivative of a holomorphic function
By Eq. \eqref{eq:24-dir1} (or \eqref{eq:24-dir2}), when $f=u+iv$ is holomorphic, its derivative can be computed in either of the following forms.
$$ f'(z) = u_x + iv_x = v_y - iu_y $$(The equality on the right is obtained by substituting the C-R equations $u_x=v_y,\ u_y=-v_x$ into the expression on the left.) This formula is the most basic tool when we actually compute the derivatives of various complex functions in Section 24.4.
24.3.2 Holomorphic $\Rightarrow$ differentiable—completing the notes from where they broke off
Caution: What it means to be holomorphic “in a neighborhood”
Rereading Definition 24.4, we find that it says “$u,v$ are once differentiable in some neighborhood of the point $(x_0,y_0)$, and the C-R equations hold.” Please note how far this “in a neighborhood” extends—it is not enough for the C-R equations to hold at just the one point $z_0$; they must hold at every point throughout the whole neighborhood of $z_0$. The proof we are about to give uses this condition, “holds throughout the whole neighborhood,” in an essential way. Conversely, what happens when the C-R equations hold at only a single point is checked at the end of this section (Example 24.7) with $f(z)=|z|^2$ as the example—to state the conclusion first, a strange thing happens: such a function is differentiable at the origin but not holomorphic there.
Suppose that $f(z)=u(x,y)+iv(x,y)$ is holomorphic at $z_0=x_0+iy_0$. That is, suppose that within some neighborhood $N$ of $z_0$ the functions $u,v$ have continuous partial derivatives (class $C^1$), and that the C-R equations $u_x=v_y,\ u_y=-v_x$ hold at every point of $N$. We show that $f'(z_0)$ then exists (regardless of the way of approach) and that $f'(z_0)=u_x(x_0,y_0)+iv_x(x_0,y_0)$.
Proof: Holomorphic $\Rightarrow$ differentiable
Take an arbitrary $h=\alpha+i\beta\to0$ (small enough that $z_0+h$ lies in $N$).
$$ f(z_0+h)-f(z_0) = \bigl[u(x_0+\alpha,y_0+\beta)-u(x_0,y_0)\bigr] + i\bigl[v(x_0+\alpha,y_0+\beta)-v(x_0,y_0)\bigr] $$To the contents of each of the two $[\ ]$ on the right-hand side we apply Theorem 6.2 of Chapter 6 (Taylor's theorem for functions of two variables) with $n=0$—this is precisely the “mean value theorem for functions of two variables,” the point at which the notes announced it and ran out of steam. Setting $n=0$ in Theorem 6.2 leaves only the remainder term, and with $(a,b)=(x_0,y_0)$ and $(h,k)=(\alpha,\beta)$ we get
$$ u(x_0+\alpha,y_0+\beta)-u(x_0,y_0) = \alpha\,u_x(P_1) + \beta\,u_y(P_1) $$ $$ v(x_0+\alpha,y_0+\beta)-v(x_0,y_0) = \alpha\,v_x(P_2) + \beta\,v_y(P_2) $$(there exist $\theta_1,\theta_2\in(0,1)$ such that $P_1=(x_0+\theta_1\alpha,\,y_0+\theta_1\beta)$ and $P_2=(x_0+\theta_2\alpha,\,y_0+\theta_2\beta)$). Since $u$ and $v$ are different functions, the values of $\theta$ guaranteed by Theorem 6.2 may in general differ—so we have distinguished them with subscripts as $\theta_1,\theta_2$. In any case, $P_1,P_2$ lie on the line segment joining $z_0$ and $z_0+h$, so as $h\to0$ (that is, $\alpha,\beta\to0$) we have $P_1\to(x_0,y_0)$ and $P_2\to(x_0,y_0)$.
Substituting these two relations, we obtain
$$ f(z_0+h)-f(z_0) = \alpha\,u_x(P_1) + \beta\,u_y(P_1) + i\bigl[\alpha\,v_x(P_2) + \beta\,v_y(P_2)\bigr] $$Now $P_1,P_2$ are both points of the neighborhood $N$, so the C-R equations hold there as well (this is where “holds throughout the whole neighborhood” comes into play). That is, substituting $u_y(P_1)=-v_x(P_1)$ and $v_y(P_2)=u_x(P_2)$, we obtain
$$ f(z_0+h)-f(z_0) = \alpha\,u_x(P_1) - \beta\,v_x(P_1) + i\bigl[\alpha\,v_x(P_2) + \beta\,u_x(P_2)\bigr] $$Our goal is to show that this is arbitrarily close to $h\cdot\bigl(u_x(z_0)+iv_x(z_0)\bigr)$. Expanding the latter explicitly, we get
$$ h\bigl(u_x(z_0)+iv_x(z_0)\bigr) = (\alpha+i\beta)\bigl(u_x(z_0)+iv_x(z_0)\bigr) = \bigl[\alpha u_x(z_0)-\beta v_x(z_0)\bigr] + i\bigl[\alpha v_x(z_0)+\beta u_x(z_0)\bigr] $$(Note that the forms of the real and imaginary parts correspond exactly to those of the equation preceding it.) So we form the difference $E(h):=f(z_0+h)-f(z_0)-h\bigl(u_x(z_0)+iv_x(z_0)\bigr)$; subtracting corresponding terms, we get
$$ E(h) = \alpha\underbrace{\bigl[u_x(P_1)-u_x(z_0)\bigr]}_{=:\varepsilon_1} - \beta\underbrace{\bigl[v_x(P_1)-v_x(z_0)\bigr]}_{=:\varepsilon_2} + i\Bigl(\alpha\underbrace{\bigl[v_x(P_2)-v_x(z_0)\bigr]}_{=:\varepsilon_3} + \beta\underbrace{\bigl[u_x(P_2)-u_x(z_0)\bigr]}_{=:\varepsilon_4}\Bigr) $$The functions $u_x,v_x$ are continuous at $z_0$ (by the $C^1$ assumption), and $P_1,P_2\to(x_0,y_0)$ as $h\to0$, so the four quantities $\varepsilon_1,\varepsilon_2,\varepsilon_3,\varepsilon_4$ all converge to $0$ as $h\to0$. Also, $|\alpha|\le|h|$ and $|\beta|\le|h|$ (by the Pythagorean theorem, $|h|=\sqrt{\alpha^2+\beta^2}\ge|\alpha|,|\beta|$), so by the triangle inequality,
$$ |E(h)| \le |\alpha||\varepsilon_1|+|\beta||\varepsilon_2|+|\alpha||\varepsilon_3|+|\beta||\varepsilon_4| $$Now we apply $|\alpha|\le|h|,\ |\beta|\le|h|$ to each of the four terms, and further replace $|\varepsilon_1|,|\varepsilon_2|,|\varepsilon_3|,|\varepsilon_4|$ all by $\max(|\varepsilon_1|,|\varepsilon_2|,|\varepsilon_3|,|\varepsilon_4|)$ (we bound each of the four terms by the largest possible value, so the estimate is somewhat crude but certain), obtaining
$$ |E(h)| \le |h|\,|\varepsilon_1|+|h|\,|\varepsilon_2|+|h|\,|\varepsilon_3|+|h|\,|\varepsilon_4| \le 4|h|\max(|\varepsilon_1|,|\varepsilon_2|,|\varepsilon_3|,|\varepsilon_4|) $$Dividing both sides by $|h|$ ($\ne0$), we get
$$ \left|\frac{E(h)}{h}\right| \le 4\max(|\varepsilon_1|,|\varepsilon_2|,|\varepsilon_3|,|\varepsilon_4|) \ \xrightarrow[h\to0]{}\ 0 $$What is important is that this estimate does not depend at all on the direction (argument) of $h$ but comes only from the smallness of the size $|h|$—which is exactly why the same conclusion holds however $h$ approaches $0$, along any path. Therefore,
$$ \lim_{h\to0}\frac{f(z_0+h)-f(z_0)}{h} = u_x(z_0)+iv_x(z_0) + \lim_{h\to0}\frac{E(h)}{h} = u_x(z_0)+iv_x(z_0) $$That is, $f'(z_0)$ exists (regardless of the way of approach) and satisfies $f'(z_0)=u_x(z_0)+iv_x(z_0)$. This is precisely the definition of complex differentiability.
∎ (holomorphic $\Rightarrow$ differentiable)
This completes the proof of both directions of Theorem 24.1, “holomorphic $\iff$ differentiable.” The C-R equations translate the abstract condition of differentiability into equations between concretely computable quantities, the partial derivatives of $u,v$—let us put this advantage of “computability” to use right away in the next three examples.
Example 24.5: Checking that $f(z)=z^2+3z$ is holomorphic on the whole plane
Using the C-R equations, verify that $f(z)=z^2+3z$ is holomorphic for every $z\in\mathbb C$. Also find its derivative.
Solution. Substitute $z=x+iy$ and separate the real and imaginary parts.
$$ f(z) = (x+iy)^2+3(x+iy) = (x^2-y^2+2ixy)+(3x+3iy) = \underbrace{(x^2-y^2+3x)}_{u(x,y)} + i\underbrace{(2xy+3y)}_{v(x,y)} $$(We used $(x+iy)^2=x^2+2ixy+i^2y^2=x^2-y^2+2ixy$.) Computing the partial derivatives, we get
$$ u_x=2x+3,\quad u_y=-2y,\quad v_x=2y,\quad v_y=2x+3 $$Hence $u_x=2x+3=v_y$ and $u_y=-2y=-(2y)=-v_x$, so the C-R equations hold identically, at every $(x,y)$. Therefore $f(z)=z^2+3z$ is holomorphic on the whole plane (all of $\mathbb C$). By Corollary 24.1, the derivative is
$$ f'(z) = u_x+iv_x = (2x+3)+i(2y) = 2(x+iy)+3 = 2z+3 $$This is exactly the same form of result as the $\dfrac{d}{dx}(x^2+3x)=2x+3$ that you learned in high school mathematics, and this computation also bears out that the same formula can be used unchanged for complex functions (verified with SymPy).
Example 24.6: Checking that $f(z)=\bar z$ is not holomorphic at any point
Show that $f(z)=\bar z$ (the function that takes the complex conjugate) is not holomorphic at any point of $\mathbb C$.
Solution. When $z=x+iy$, we have $\bar z=x-iy$, so $u(x,y)=x$ and $v(x,y)=-y$. The partial derivatives are
$$ u_x=1,\quad u_y=0,\quad v_x=0,\quad v_y=-1 $$Looking at the first of the C-R equations, $u_x=v_y$, we get $1=-1$, which does not hold for any $(x,y)$ (it is identically contradictory). Therefore $f(z)=\bar z$ does not satisfy the C-R equations at any point of $\mathbb C$, and is not holomorphic at any point (verified with SymPy: computing $u_x-v_y$ gives $2$ identically, independent of $(x,y)$).
As we point out once more at the beginning of Section 24.4, a function containing $\bar z$ is, as here, almost never holomorphic—holomorphy is a special property possessed only by functions that are “made of $z$ alone,” such as those built from the four arithmetic operations on $z$ together with powers, exponentials, logarithms, and trigonometric functions.
Example 24.7: Checking that $f(z)=|z|^2$ is differentiable only at the origin
For $f(z)=|z|^2$, find the points at which the C-R equations hold. Also, can we say that $f$ is holomorphic at such a point?
Solution. Since $|z|^2=z\bar z=x^2+y^2$ (a real number), we have $u(x,y)=x^2+y^2$ and $v(x,y)=0$. The partial derivatives are
$$ u_x=2x,\quad u_y=2y,\quad v_x=0,\quad v_y=0 $$The C-R equation $u_x=v_y$ holds only when $2x=0$, that is, when $x=0$. The other equation, $u_y=-v_x$, holds only when $2y=0$, that is, when $y=0$. The only point that satisfies both at the same time is $(x,y)=(0,0)$, that is, just the single point $z=0$.
Now let us recall the caution at the beginning of Section 24.3.2—for a function to be holomorphic, the C-R equations must hold throughout some neighborhood. But whatever point other than the origin we take, its surroundings necessarily contain points at which the C-R equations are violated (points with $x\ne0$ or $y\ne0$). In other words, there is no neighborhood of the origin on which the C-R equations hold throughout. Therefore, even if “the defining formula of differentiability makes sense” at the single point $z=0$, the function $f(z)=|z|^2$ is not holomorphic even at the origin (because holomorphy is a property that involves not just a single point but its entire neighborhood). At every other point the C-R equations themselves do not hold, so by the contrapositive of Theorem 24.1 (not holomorphic $\Rightarrow$ not differentiable), the function is not even differentiable there.
(Verified with SymPy: solving $u_x=v_y$ gives $x=0$, solving $u_y=-v_x$ gives $y=0$, and the only common solution is the origin.) Thus $|z|^2$ is a typical example of a function that is almost never holomorphic, for the same reason as in Example 24.6, even if it does not look like an “expression in $z$ alone” (in fact $|z|^2=z\bar z$ contains $\bar z$).
24.4 Derivatives and Differentiation Formulas for Various Complex Functions
Thanks to the C-R equations, checking holomorphy has become a concrete task: “compute the partial derivatives of $u,v$ and check.” In this section we make full use of this tool to prove all of the basic formulas for differentiating complex functions—linearity and the product and quotient rules (including the parts that the original notes dismiss as “omitted, since the proof is the same as in the real case”)—and finally use them to find the derivatives of the concrete functions $e^z,\sin z,\cos z,\log z,\alpha^z$. In the source notes, the list of functions (numbered (5)) is presented first, followed by linearity and the product and quotient rules (numbered (6) and (7)); in this textbook, however, we rearrange the order so that we first establish the tools (linearity and the product and quotient rules) and then use them to derive each formula in the list (the content includes every item in the notes).
24.4.1 Linearity of differentiation
The defining formula $\displaystyle f'(z)=\lim_{\Delta z\to0}\frac{f(z+\Delta z)-f(z)}{\Delta z}$ for the derivative of a complex function has exactly the same form as the defining formula for the derivative of a real function. The properties of limits (the same as for real numbers: limits can be distributed over sums and constant multiples) also hold unchanged for limits of complex numbers, so differentiation has the same linearity.
Formula 24.1: Linearity of differentiation
$$ \frac{d}{dz}\bigl(f(z)\pm g(z)\bigr) = \frac{d}{dz}f(z)\pm\frac{d}{dz}g(z) ,\qquad\qquad \frac{d}{dz}\bigl(\alpha f(z)\bigr) = \alpha\,\frac{d}{dz}f(z)\quad(\alpha\text{ is a complex constant}) $$Proof
For the sum and difference, computing the limit directly from the definition gives
$$ \frac{d}{dz}\bigl(f(z)\pm g(z)\bigr) = \lim_{\Delta z\to0}\frac{\bigl(f(z+\Delta z)\pm g(z+\Delta z)\bigr)-\bigl(f(z)\pm g(z)\bigr)}{\Delta z} $$Rearranging the numerator into the difference of $f$ and the difference of $g$ (we merely changed the order of the additions and subtractions), we get
$$ = \lim_{\Delta z\to0}\left(\frac{f(z+\Delta z)-f(z)}{\Delta z}\pm\frac{g(z+\Delta z)-g(z)}{\Delta z}\right) $$Since limits can be distributed over sums and differences (because each of the limits for $f,g$ exists), we get
$$ = \lim_{\Delta z\to0}\frac{f(z+\Delta z)-f(z)}{\Delta z} \pm \lim_{\Delta z\to0}\frac{g(z+\Delta z)-g(z)}{\Delta z} = \frac{d}{dz}f(z)\pm\frac{d}{dz}g(z) $$Similarly, for constant multiples,
$$ \frac{d}{dz}\bigl(\alpha f(z)\bigr) = \lim_{\Delta z\to0}\frac{\alpha f(z+\Delta z)-\alpha f(z)}{\Delta z} = \alpha\lim_{\Delta z\to0}\frac{f(z+\Delta z)-f(z)}{\Delta z} = \alpha\,\frac{d}{dz}f(z) $$(Since $\alpha$ is a constant that has nothing to do with $\Delta z\to0$, it can be pulled outside the limit.)
∎
24.4.2 The product rule and the quotient rule
The original notes say only “omitted, since the proof is the same as in the real case,” but here we actually write out what calculation this “same proof” consists of.
Formula 24.2: Product rule and quotient rule
$$ \frac{d}{dz}\bigl(f(z)g(z)\bigr) = \frac{d}{dz}f(z)\cdot g(z) + f(z)\cdot\frac{d}{dz}g(z) $$ $$ \frac{d}{dz}\left(\frac{f(z)}{g(z)}\right) = \frac{\dfrac{d}{dz}f(z)\cdot g(z) - f(z)\cdot\dfrac{d}{dz}g(z)}{\{g(z)\}^2}\qquad(g(z)\ne0) $$Proof: Product rule
We form the difference quotient directly from the definition.
$$ \frac{(fg)(z+\Delta z)-(fg)(z)}{\Delta z} = \frac{f(z+\Delta z)g(z+\Delta z)-f(z)g(z)}{\Delta z} $$We insert $-f(z+\Delta z)g(z)+f(z+\Delta z)g(z)$ (terms that cancel to give $0$) into the numerator—this is exactly the same trick that is used in the proof of the product rule for real functions.
$$ f(z+\Delta z)g(z+\Delta z)-f(z)g(z) = f(z+\Delta z)\bigl[g(z+\Delta z)-g(z)\bigr] + g(z)\bigl[f(z+\Delta z)-f(z)\bigr] $$(Expanding the right-hand side gives $f(z+\Delta z)g(z+\Delta z)-f(z+\Delta z)g(z)+g(z)f(z+\Delta z)-g(z)f(z)$, a form in which the middle two terms are left standing without canceling each other—$-f(z+\Delta z)g(z)+g(z)f(z+\Delta z)=0$—and if we confirm this, we see that it agrees with the left-hand side.) Dividing both sides by $\Delta z$, we get
$$ \frac{(fg)(z+\Delta z)-(fg)(z)}{\Delta z} = f(z+\Delta z)\cdot\frac{g(z+\Delta z)-g(z)}{\Delta z} + g(z)\cdot\frac{f(z+\Delta z)-f(z)}{\Delta z} $$Since $f$ is differentiable, it is also continuous, so $f(z+\Delta z)\to f(z)$ as $\Delta z\to0$; moreover, the two difference quotients on the right-hand side converge to $g'(z),f'(z)$, respectively. Hence,
$$ \frac{d}{dz}\bigl(f(z)g(z)\bigr) = f(z)g'(z) + g(z)f'(z) = f'(z)g(z)+f(z)g'(z) $$∎
Proof: Quotient rule
Let $\varphi(z)=f(z)/g(z)$. Regarding this as $f(z)=\varphi(z)g(z)$ and differentiating both sides by the product rule (the formula we have just proved), we get
$$ f'(z) = \varphi'(z)g(z)+\varphi(z)g'(z) $$Solving for $\varphi'(z)$, we get
$$ \varphi'(z) = \frac{f'(z)-\varphi(z)g'(z)}{g(z)} = \frac{f'(z)-\dfrac{f(z)}{g(z)}g'(z)}{g(z)} = \frac{f'(z)g(z)-f(z)g'(z)}{\{g(z)\}^2} $$(In the last step, we multiplied the numerator and denominator by $g(z)$ to simplify the fraction within a fraction.) That is,
$$ \frac{d}{dz}\left(\frac{f(z)}{g(z)}\right) = \frac{f'(z)g(z)-f(z)g'(z)}{\{g(z)\}^2} $$∎ (verified numerically with SymPy: for $f(z)=z^2+1,\ g(z)=z-1$, direct differentiation and the formula agree)
24.4.3 Derivatives of various complex functions
Now that we have the tools—linearity and the product and quotient rules—we find the derivatives of concrete functions. We use the definitions of $e^z,\log z,\alpha^z,\sin z,\cos z$ given in Chapter 23.
Formula 24.3: Derivatives of various complex functions
$$ \frac{d}{dz}(e^z)=e^z \qquad\qquad \frac{d}{dz}(\alpha^z)=\alpha^z\log\alpha\quad(\alpha:\text{complex constant}) $$ $$ \frac{d}{dz}\sin z=\cos z \qquad\qquad \frac{d}{dz}\cos z=-\sin z $$ \begin{equation} \frac{d}{dz}\tan z=\frac{1}{\cos^2z} \qquad\qquad \frac{d}{dz}\log z=\frac1z \label{eq:24-derivtable} \end{equation}Remember!—Functions containing $\bar z$ are almost never holomorphic
Every function listed in Formula 24.3 is written purely as an “expression in $z$ itself” (a combination of the four arithmetic operations, powers, exponentials, logarithms, and trigonometric functions of $z$), and $\bar z$ never appears. This is no coincidence. As we confirmed in Example 24.6 ($\bar z$) and Example 24.7 ($|z|^2=z\bar z$), expressions containing $\bar z$ almost never satisfy the C-R equations and so are not holomorphic. One may even say that the property of holomorphy itself means almost the same thing as “being writable in terms of $z$ alone, without using $\bar z$” (this intuition returns in an even clearer form when, from Chapter 26 onward, we characterize holomorphic functions using power series).
In what follows, we derive each item of Formula 24.3 in turn. First, proving the following general formula, with which $e^z,\sin z,\cos z$ can be handled all at once, gives us a clear view of the whole.
Lemma: $\dfrac{d}{dz}\bigl(e^{cz}\bigr)=c\,e^{cz}$ ($c$ is an arbitrary complex constant)
Let $c=p+iq$ ($p,q$ are real numbers). Since $cz=(p+iq)(x+iy)=(px-qy)+i(qx+py)$, Euler's formula (Chapter 21, Section 21.4) gives
$$ e^{cz} = e^{(px-qy)+i(qx+py)} = e^{px-qy}\bigl[\cos(qx+py)+i\sin(qx+py)\bigr] $$Therefore $u(x,y)=e^{px-qy}\cos(qx+py)$ and $v(x,y)=e^{px-qy}\sin(qx+py)$. To differentiate $u$ partially with respect to $x$, we use the product rule (the exponent of $e^{px-qy}$ contains $x$, and the argument of $\cos(qx+py)$ also contains $x$, so we differentiate it as a product of two functions that both depend on $x$). Differentiating $e^{px-qy}$ with respect to $x$ brings out in front the factor $p$, the derivative of the exponent $px-qy$ with respect to $x$, giving $p\,e^{px-qy}$; differentiating $\cos(qx+py)$ with respect to $x$ brings out in front the factor $q$, the derivative of the inside $qx+py$ with respect to $x$, giving $-q\sin(qx+py)$ (the chain rule). Hence,
$$ u_x = p\,e^{px-qy}\cos(qx+py) + e^{px-qy}\cdot\bigl(-q\sin(qx+py)\bigr) = e^{px-qy}\bigl[p\cos(qx+py)-q\sin(qx+py)\bigr] $$Similarly, we differentiate $v$ partially with respect to $y$. Differentiating $e^{px-qy}$ with respect to $y$ brings out in front $-q$, the derivative of the inside with respect to $y$, giving $-q\,e^{px-qy}$; differentiating $\sin(qx+py)$ with respect to $y$ brings out in front $p$, the derivative of the inside with respect to $y$, giving $p\cos(qx+py)$. Therefore,
$$ v_y = -q\,e^{px-qy}\sin(qx+py) + e^{px-qy}\cdot p\cos(qx+py) = e^{px-qy}\bigl[p\cos(qx+py)-q\sin(qx+py)\bigr] $$Comparing $u_x$ and $v_y$, we see that they agree completely, so $u_x=v_y$ is confirmed. In the same way, computing $u_y,-v_x$ shows that they agree as well (verified with SymPy: we computed all four partial derivatives and confirmed the C-R equations). Hence $e^{cz}$ is holomorphic on the whole plane (whatever complex constant $c$ is). By Corollary 24.1, the derivative is obtained from
$$ \frac{d}{dz}\bigl(e^{cz}\bigr) = u_x+iv_x $$Computing $v_x$ in the same way gives $v_x=e^{px-qy}\bigl[p\sin(qx+py)+q\cos(qx+py)\bigr]$, and therefore
$$ u_x+iv_x = e^{px-qy}\Bigl\{\bigl[p\cos(qx+py)-q\sin(qx+py)\bigr] + i\bigl[p\sin(qx+py)+q\cos(qx+py)\bigr]\Bigr\} $$Grouping what is inside the curly braces into the $p$ terms and the $q$ terms, we get
$$ = e^{px-qy}\Bigl\{p\bigl[\cos(qx+py)+i\sin(qx+py)\bigr] + iq\bigl[\cos(qx+py)+i\sin(qx+py)\bigr]\Bigr\} = e^{px-qy}(p+iq)\bigl[\cos(qx+py)+i\sin(qx+py)\bigr] $$(We used the fact that terms such as $-q\sin(\cdot)+iq\cdot i\sin(\cdot)$ combine into the form $iq[\cos+i\sin]=iq\cos-q\sin$.) Finally, using $p+iq=c$ and $e^{px-qy}[\cos(qx+py)+i\sin(qx+py)]=e^{cz}$ (returning to the first formula), we obtain
$$ \frac{d}{dz}\bigl(e^{cz}\bigr) = c\,e^{cz} $$∎ (verified numerically with SymPy at six points with random complex numbers $c,z$)
Derivation: $\dfrac{d}{dz}(e^z)=e^z$
It suffices to set $c=1$ ($p=1,q=0$) in the lemma we have just proved.
$$ \frac{d}{dz}\bigl(e^{1\cdot z}\bigr) = 1\cdot e^{1\cdot z} \quad\Longrightarrow\quad \frac{d}{dz}(e^z)=e^z $$∎ (this is also consistent with the fact that we confirmed the C-R equations in Example 24.4)
Derivation: $\dfrac{d}{dz}(\alpha^z)=\alpha^z\log\alpha$
The definition of the complex power (Chapter 23) is $\alpha^z=e^{z\log\alpha}$ ($\log\alpha$ is a complex constant fixed on one particular branch). This is exactly $e^{cz}$ for the case $c=\log\alpha$ (a fixed complex constant) in the lemma, so the lemma immediately gives
$$ \frac{d}{dz}\bigl(\alpha^z\bigr) = \frac{d}{dz}\bigl(e^{z\log\alpha}\bigr) = (\log\alpha)\,e^{z\log\alpha} = \alpha^z\log\alpha $$∎
Derivation: $\dfrac{d}{dz}\sin z=\cos z$, $\dfrac{d}{dz}\cos z=-\sin z$
The definition of the complex trigonometric functions (Chapter 23; the expression in terms of exponential functions derived from Euler's formula) is
$$ \sin z = \frac{e^{iz}-e^{-iz}}{2i} ,\qquad \cos z = \frac{e^{iz}+e^{-iz}}{2} $$Here $e^{iz}$ is nothing other than the case $c=i$, and $e^{-iz}$ the case $c=-i$, of $e^{cz}$ in the lemma. Differentiating $\sin z$ term by term using Formula 24.1 (linearity), we get
$$ \frac{d}{dz}\sin z = \frac{d}{dz}\left(\frac{e^{iz}-e^{-iz}}{2i}\right) = \frac{1}{2i}\left(\frac{d}{dz}e^{iz}-\frac{d}{dz}e^{-iz}\right) = \frac{1}{2i}\Bigl(i\,e^{iz}-(-i)e^{-iz}\Bigr) $$(We applied the lemma with $c=i$ and with $c=-i$, respectively: $\frac{d}{dz}e^{iz}=i\,e^{iz}$, $\frac{d}{dz}e^{-iz}=(-i)e^{-iz}$.) Simplifying the expression inside the parentheses gives $i\,e^{iz}+i\,e^{-iz}=i(e^{iz}+e^{-iz})$, so
$$ \frac{d}{dz}\sin z = \frac{i(e^{iz}+e^{-iz})}{2i} = \frac{e^{iz}+e^{-iz}}{2} = \cos z $$Similarly, differentiating $\cos z$ term by term, we get
$$ \frac{d}{dz}\cos z = \frac{d}{dz}\left(\frac{e^{iz}+e^{-iz}}{2}\right) = \frac12\Bigl(i\,e^{iz}+(-i)e^{-iz}\Bigr) = \frac{i(e^{iz}-e^{-iz})}{2} = i\cdot\frac{e^{iz}-e^{-iz}}{2} $$Since $\dfrac{e^{iz}-e^{-iz}}{2}=i\sin z$ (this can be seen by comparing the denominators in the defining formula: multiplying both sides of $\sin z=\frac{e^{iz}-e^{-iz}}{2i}$ by $i$ gives $i\sin z=\frac{e^{iz}-e^{-iz}}{2i}\cdot i=\frac{e^{iz}-e^{-iz}}{2}$), we obtain
$$ \frac{d}{dz}\cos z = i\cdot i\sin z = i^2\sin z = -\sin z $$∎ (verified with SymPy: the definitions of the form $(e^{iz}\pm e^{-iz})$ agree numerically with SymPy's $\sin z,\cos z$, and we confirmed that their derivatives also agree with $\cos z,-\sin z$)
Math Note: Expanding directly in terms of $u,v$ gives the same result
If we expand $\sin z,\cos z$ directly into the form $u(x,y)+iv(x,y)$, then using the addition formulas and the hyperbolic functions (Chapter 21, Section 21.6, $\cosh y=\frac{e^y+e^{-y}}2,\ \sinh y=\frac{e^y-e^{-y}}2$), we obtain
$$ \sin z=\sin x\cosh y+i\cos x\sinh y ,\qquad \cos z=\cos x\cosh y-i\sin x\sinh y $$The functions take this form (verified numerically with SymPy). Even if we check the C-R equations for these functions directly (for $\sin z$, $u_x=\cos x\cosh y=v_y$ and $u_y=\sin x\sinh y=-v_x$), the formula $u_x+iv_x$ of Corollary 24.1 gives the same derivatives $\cos z$ and $-\sin z$. The derivation above using exponential functions is quicker, but confirming that this direct computation also arrives at the same answer should increase your trust in the C-R equations as a tool.
Derivation: $\dfrac{d}{dz}\tan z=\dfrac{1}{\cos^2z}$
Apply Formula 24.2 (the quotient rule) to $\tan z=\dfrac{\sin z}{\cos z}$ ($\cos z\ne0$) with $f(z)=\sin z$ and $g(z)=\cos z$:
$$ \frac{d}{dz}\tan z = \frac{(\sin z)'\cos z-\sin z(\cos z)'}{\cos^2z} = \frac{\cos z\cdot\cos z-\sin z\cdot(-\sin z)}{\cos^2z} = \frac{\cos^2z+\sin^2z}{\cos^2z} $$Here we use the fact that the same Pythagorean relation $\sin^2z+\cos^2z=1$ as for real numbers also holds for complex numbers (see the Math Note below). Hence,
$$ \frac{d}{dz}\tan z = \frac{1}{\cos^2z} $$∎ (verified with SymPy: SymPy's derivative of $\tan z$ agrees with $1/\cos^2z$)
Math Note: $\sin^2z+\cos^2z=1$ also holds for complex numbers
This can be checked directly by returning to the definitions in terms of exponential functions.
$$ \sin^2z+\cos^2z = \left(\frac{e^{iz}-e^{-iz}}{2i}\right)^2+\left(\frac{e^{iz}+e^{-iz}}{2}\right)^2 = \frac{-(e^{2iz}-2+e^{-2iz})}{4}+\frac{e^{2iz}+2+e^{-2iz}}{4} $$(We expanded the denominator of the first term, noting that $1/i^2=-1$.) When we add the numerators, the terms in $e^{2iz}$ and $e^{-2iz}$ cancel exactly, and
$$ = \frac{(-e^{2iz}+2-e^{-2iz})+(e^{2iz}+2+e^{-2iz})}{4} = \frac{4}{4} = 1 $$(Verified with SymPy.) We see that the identity $\sin^2\theta+\cos^2\theta=1$ from high school mathematics continues to hold unchanged when $\theta$ is replaced by a complex number $z$.
Derivation: $\dfrac{d}{dz}\log z=\dfrac1z$
The principal value of the complex logarithm (Chapter 23) is defined as $\log z=\log|z|+i\arg z$. Let $z=x+iy$ and $|z|=\sqrt{x^2+y^2}$, and take the principal value of the argument so that $\arg z\in(-\pi,\pi]$. In the region $x\gt0$ we can write $\arg z=\tan^{-1}\dfrac yx$, so
$$ u(x,y) = \log\sqrt{x^2+y^2} = \frac12\log(x^2+y^2) ,\qquad v(x,y) = \tan^{-1}\frac{y}{x} $$We compute the partial derivatives. For $u$ we use the chain rule for composite functions (the inside of the $\log$, $x^2+y^2$, must be differentiated further with respect to $x$):
$$ u_x = \frac12\cdot\frac{2x}{x^2+y^2} = \frac{x}{x^2+y^2} $$When we differentiate $v=\tan^{-1}(y/x)$ partially with respect to $y$, we multiply the differentiation formula $\dfrac{1}{1+u^2}$ for $\tan^{-1}u$ (with $u=y/x$) by $1/x$, the derivative of $u$ with respect to $y$, to get
$$ v_y = \frac{1}{1+(y/x)^2}\cdot\frac1x = \frac{x^2}{x^2+y^2}\cdot\frac1x = \frac{x}{x^2+y^2} $$We have confirmed that $u_x=v_y$. Similarly, $u_y=\dfrac{y}{x^2+y^2}$ and $v_x=\dfrac1{1+(y/x)^2}\cdot\left(-\dfrac{y}{x^2}\right)=-\dfrac{y}{x^2+y^2}$, so $u_y=-v_x$ also holds (verified with SymPy). Hence $\log z$ is holomorphic (in the region $x\gt0$). By Corollary 24.1,
$$ \frac{d}{dz}\log z = u_x+iv_x = \frac{x}{x^2+y^2} - i\,\frac{y}{x^2+y^2} = \frac{x-iy}{x^2+y^2} = \frac{\bar z}{|z|^2} = \frac{\bar z}{z\bar z} = \frac1z $$(In the region $x\lt0$ as well, only the way of expressing the principal value of $\arg z$ changes; exactly the same computation gives the same result $1/z$.)
∎ (verified numerically with SymPy: checked at six points in the region $x\gt0$)
Caution: The domain of $\log z$ is the region “with the origin and the negative real axis removed”
The principal value of $\arg z$ is usually taken in the range $(-\pi,\pi]$, but when $z$ crosses the negative real axis (the direction $\arg z=\pi$), the value of $\arg z$ jumps suddenly from $+\pi$ to $-\pi$ (or the other way around)—it ceases to be continuous. Consequently $\log z=\log|z|+i\arg z$ is also not continuous on the negative real axis, let alone differentiable. So the domain of $\log z$ is usually taken to be $D=\mathbb C\setminus\{x\le0,\ y=0\}$ (the complex plane with the origin and the negative real axis removed). A line that is deliberately removed in this way in order to make a function holomorphic is called a branch cut. Figure 24.6 shows this—the word “domain” that we defined in Section 24.1 is actually put to work here. Since $D$ does not contain the origin and does not touch the negative real axis, it is a perfectly good domain, satisfying both openness and connectedness.
Example 24.8: Holomorphy of $f(z)=\dfrac1z$ ($z\ne0$) and its derivative via the C-R equations
Verify that $f(z)=\dfrac1z$ ($z\ne0$) satisfies the C-R equations, and find its derivative.
Solution. From $\dfrac1z=\dfrac{\bar z}{z\bar z}=\dfrac{\bar z}{|z|^2}=\dfrac{x-iy}{x^2+y^2}$, we get
$$ u(x,y)=\frac{x}{x^2+y^2} ,\qquad v(x,y)=\frac{-y}{x^2+y^2} $$(To make the denominator real, we multiplied the complex number $z$ in the denominator by its conjugate $\bar z$; this is the same idea as rationalizing the denominator in high school mathematics.) Computing the partial derivatives in the manner of the quotient rule, we get
$$ u_x = \frac{1\cdot(x^2+y^2)-x\cdot2x}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2} ,\qquad v_y = \frac{-1\cdot(x^2+y^2)-(-y)\cdot2y}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2} $$We have confirmed that $u_x=v_y$. Similarly,
$$ u_y = \frac{-2xy}{(x^2+y^2)^2} ,\qquad v_x = \frac{2xy}{(x^2+y^2)^2} $$so $u_y=-v_x$ also holds (at every point with $z\ne0$, that is, $(x,y)\ne(0,0)$). Therefore $f(z)=1/z$ is holomorphic for $z\ne0$. By Corollary 24.1,
$$ f'(z) = u_x+iv_x = \frac{y^2-x^2}{(x^2+y^2)^2}+i\,\frac{2xy}{(x^2+y^2)^2} = \frac{(y^2-x^2)+i\cdot2xy}{(x^2+y^2)^2} $$Comparing the numerator with $(x-iy)^2=x^2-2ixy-y^2=(x^2-y^2)-2ixy$, we see that $(y^2-x^2)+i\cdot2xy=-\bigl[(x^2-y^2)-2ixy\bigr]=-(x-iy)^2$. Using $x-iy=\bar z$ and $x^2+y^2=|z|^2=z\bar z$, we obtain
$$ f'(z) = \frac{-(x-iy)^2}{(x^2+y^2)^2} = -\frac{\bar z^2}{|z|^4} = -\frac{\bar z^2}{(z\bar z)^2} = -\frac{1}{z^2} $$That is, $f'(z)=-\dfrac{1}{z^2}$, which has the same form as the formula $\left(\dfrac1x\right)'=-\dfrac{1}{x^2}$ that you learned in high school mathematics (verified numerically with SymPy: random check at five points).
Example 24.9: An application of the product rule—$\dfrac{d}{dz}\bigl(z^2\sin z\bigr)$
Use the product rule (Formula 24.2) to compute $\dfrac{d}{dz}\bigl(z^2\sin z\bigr)$.
Solution. Let $f(z)=z^2$ and $g(z)=\sin z$. Then $f'(z)=2z$ (the same form as the derivative of a power function that you already learned in high school, and consistent with the computation in Example 24.5) and $g'(z)=\cos z$ (Formula 24.3), so by the product rule,
$$ \frac{d}{dz}\bigl(z^2\sin z\bigr) = f'(z)g(z)+f(z)g'(z) = 2z\sin z + z^2\cos z $$(Verified with SymPy.)
24.5 Summary and Exercises
24.5.1 Summary
- Neighborhood: the set of all points satisfying $|z-z_0|\lt r$. Domain: a set satisfying openness (it contains no boundary points) and connectedness (any two points can be joined by a polygonal path lying inside the set).
- The continuity of a complex function demands that “the limit be the same no matter how the point is approached”—an essentially stronger condition than the continuity of a real function (the limit of $\bar z/z$ does not exist: Example 24.3).
- Holomorphy: the condition that $u,v$ are once continuously differentiable on a whole neighborhood and satisfy the Cauchy–Riemann equations (C-R equations)
- Theorem 24.1: “holomorphic $\iff$ differentiable”—we proved both directions completely in this chapter (in the proof of “holomorphic $\Rightarrow$ differentiable,” we used the mean value theorem for functions of two variables, that is, the case $n=0$ of Theorem 6.2 of Chapter 6).
- Corollary 24.1: the derivative of a holomorphic function can be computed as $f'(z)=u_x+iv_x=v_y-iu_y$—this formula was the foundation of all the computations in Section 24.4.
- Functions containing $\bar z$ (such as $\bar z,\ |z|^2$) are almost never holomorphic—holomorphy is nearly synonymous with “being expressible in terms of $z$ alone” (Examples 24.6 and 24.7). Holomorphy is a property not of a single point alone but of a whole neighborhood of that point.
- We completely proved the linearity of differentiation (Formula 24.1) and the product and quotient rules (Formula 24.2), and from the general formula $\dfrac{d}{dz}e^{cz}=ce^{cz}$ we derived the derivatives of $e^z,\sin z,\cos z,\alpha^z$ in a unified way. The derivative of $\log z$ (principal value) is $1/z$, but its domain requires a branch cut (removal of the negative real axis).
At the end of the source notes (p. 26), under the heading “(8) Cauchy–Riemann Relation (C-R equations),” a single sentence—“the function $f(z)=u(x,y)+iv(x,y)$ of $z=x+iy$ defined on a domain $D$ is holomor-”—remains, interrupted before the verb was finished. The content itself is exactly the same as the C-R equations introduced in Section 24.2, and it is probably a trace of the notes' author trying, at the close of the chapter, to confirm once more the name “Cauchy–Riemann equations.” This name is the one we have used many times throughout this chapter for the most fundamental relation in the theory of holomorphic functions, a relation that appears in every textbook on complex analysis.
24.5.2 Exercises
Exercise 24.1: Holomorphy of $f(z)=x^2-y^2+2ixy$
Determine at which points $f(z)=x^2-y^2+2ixy$ is holomorphic.
Hint: Let $u=x^2-y^2$ and $v=2xy$, and check the C-R equations. Also observe what familiar expression this $f(z)$ actually equals.
Exercise 24.2: Harmonic conjugate—constructing a holomorphic function from its real part
Construct a holomorphic function $f(z)=u+iv$ whose real part is $u(x,y)=x^3-3xy^2$, by finding the imaginary part $v(x,y)$ from the C-R equations.
Hint: Integrating $v_y=u_x$ with respect to $y$ determines the form of $v$ (in place of an integration constant, a function $h(x)$ of $x$ alone appears). Use the other equation, $v_x=-u_y$, to determine $h(x)$.
Exercise 24.3: Holomorphy and derivative of $f(z)=\cos z$
Verify the holomorphy of $f(z)=\cos z$ from the C-R equations, and find its derivative.
Hint: Use $\cos z=\cos x\cosh y-i\sin x\sinh y$ (see the “Math Note” in Section 24.4).
Exercise 24.4: Application of the quotient rule
Use the quotient rule to find the derivative of $g(z)=\dfrac{z^2+1}{z-1}$, and point out the points at which $g(z)$ is not holomorphic.
Hint: Apply Formula 24.2 with $f(z)=z^2+1$ and $h(z)=z-1$. Pay attention to the point where the denominator becomes $0$.
Exercise 24.5: A concrete check that the limits from two directions agree
For $f(z)=z^2$, actually compute, for $u=x^2-y^2$ and $v=2xy$, the limit [1] $\beta=0,\alpha\to0$ and the limit [2] $\alpha=0,\beta\to0$ used in the proof in Section 24.3.1, and confirm that they give the same value.
Hint: Simply substitute these $u,v$ into Eq. \eqref{eq:24-dir1} and Eq. \eqref{eq:24-dir2} and compute.
Exercise 24.6: The real part of a holomorphic function is a harmonic function
Show from the C-R equations that, when $f(z)=u+iv$ is holomorphic, the real part $u$ is a harmonic function, that is, satisfies $u_{xx}+u_{yy}=0$.
Hint: Differentiate both sides of $u_x=v_y$ partially with respect to $x$ and both sides of $u_y=-v_x$ partially with respect to $y$, and combine the two resulting equations using Schwarz's theorem in Chapter 6, Section 6.3 ($v_{xy}=v_{yx}$).
24.5.3 References
- Yasuhide Mochizuki, Mathematics Notes: Complex Analysis (handwritten lecture notes, in Japanese). The source of this chapter.
- L. v. Ahlfors, Complex Analysis, 3rd ed. (McGraw-Hill, 1979).