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2. Limits and differentiation in the complex plane and the
Cauchy-Riemann equations
§2.1 Open sets and domains
A major difference between real and complex analysis is that the geometry of the complex plane is far richer than that of the real line. For example, the only connected subsets of R are intervals, whereas there are far more complicated connected sets in C (‘connected’ has a rigorous meaning, but for now you can assume that a subset is connected if it ‘looks’ connected: i.e. any two points in the subset can be joined by a line that does not leave the subset). We need to make precise what we mean by convergence, open sets (generalising open intervals), etc.
Remark. Throughout, we use ⊂ (rather than ⊆) to denote ‘is a subset of’. Thus A ⊂ B means that A is a subset (indeed, possibly equal to) B.
Definition. Let z 0 ∈ C and let ε > 0. We write
Nε(z 0 ) = {z ∈ C | |z − z 0 | < ε}
to denote the disc in C of complex numbers that are distance at most ε from z 0. We call Nε(z 0 ) the ε-neighbourhood of z 0.
Definition. Let D ⊂ C. We say that D is an open set if for every z 0 ∈ D there exists ε > 0 such that Nε(z 0 ) ⊂ D.
Definition. We call a set D closed if its complement C \ D is a open
Remark. Note that a set is closed precisely when the complement is open. A very common mistake is to think that ‘closed’ means ‘not open’: this is not the case, and it is easy to write down examples of sets that are neither open nor closed.
In our setting, one can often decide whether a set is open or not by looking at it and thinking carefully. (A more rigorous treatment of open sets is given in the MATH Metric Spaces course.) For example, any open disc {z ∈ C | |z − z 0 | < r} is an open set; see Figure 2.1. Open subsets of C may be very complicated. We will only be interested in ‘nice’ open sets called domains.
Definition. Let D ⊂ C be a non-empty set. Then we say that D is a domain if
(i) D is open;
(ii) every pair of points in D can be connected by a polygonal arc (i.e. one built up of line segments) lying entirely in D.
z
z 0
r
Figure 2.1.1: The open disc Nr(z 0 ) with centre z 0 and radius r > 0. This is an open set as, given any point z ∈ C, we can find another open disc centred at z that is contained in Nr(z 0 ).
Examples.
(i) The open disc {z ∈ C | |z − z 0 | < r} centred at z 0 ∈ C and of radius r is a domain.
(ii) An annulus {z ∈ C | r 1 < |z − z 0 | < r 2 } is a domain.
(iii) A half-plane such as {z ∈ C | Re(z) > a} is a domain.
(iv) A closed disc {z ∈ C | |z − z 0 | ≤ r} or a closed half-plane {z ∈ C | Re(z) ≥ a} are not domains as they are not open sets.
(v) The set D = {z ∈ C | Im(z) 6 = 0}, corresponding to the complex plane with the real axis deleted, is not a domain. Although it is an open set, there are points (such as i, −i) that cannot be connected by a polygonal arc lying entirely in D.
See Figure 2.1.2 for examples of domains.
§2.2 Limits of complex sequences
Let zn ∈ C be a sequence of complex numbers. We say that zn → z as n → ∞ if: for all ε > 0 there exists N ∈ N such that if n ≥ N then |zn − z| < ε. (Equivalently, in the language of MATH10242 Sequences and Series, zn → z if |zn − z| is a null sequence.)
Lemma 2.2. Let zn ∈ C and write zn = xn + iyn, xn, yn ∈ R. Then zn converges if and only if xn and yn converge.
Proof. Suppose that zn → z and write z = x + iy. Then
|xn − x| ≤
|xn − x|^2 + |yn − y|^2 = |zn − z| → 0
Continuity obeys the same rules as in MATH20101 Real Analysis. In particular, if f, g : D → C are complex functions which are continuous at z 0 then
f (z) + g(z), f (z)g(z), cf (z) (c ∈ C)
are all continuous, as is f (z)/g(z) provided that g(z 0 ) 6 = 0.
§2.4 Differentiable functions
Let us first consider how one differentiates real valued functions defined on R. You will cover this properly in the Real Analysis course, and some of you will have seen ‘differentiation from first principles’ at A-level. Let (a, b) ⊂ R be an open interval and letf : (a, b) → R be a function. Let x 0 ∈ (a, b). The idea is that f ′(x 0 ) is the slope of the graph of f at the point x 0. Heuristically, one takes a point x that is near x 0 and looks at the gradient of the straight line drawn between the points (x 0 , f (x 0 )) and (x, f (x)) on the graph of f ; this is an approximation to the slope at x 0 , and becomes more accurate as x approaches x 0. We then say that f is differentiable at x 0 if this limit exists, and define the derivative of f at x 0 to be the value of this limit.
Definition. Let (a, b) ⊂ R be an interval and let x 0 ∈ (a, b). A function f : (a, b) → R is differentiable at x 0 if
f ′(x 0 ) = lim x→x 0
f (x) − f (x 0 ) x − x 0
exists. We call f ′(x 0 ) the derivative of f at x 0. We say that f is differentiable if it is differentiable at all points x 0 ∈ (a, b).
Remark. Notice that there are two ways that x can approach x 0 : x can either approach x 0 from the left or from the right. The definition of the derivative in (2.4.1) requires the limit to exist from both the left and the right and for the value of these limits to be the same. (As an aside, one could instead look at left-handed and right-handed derivatives. For example, consider f (x) = |x| defined on R. The left-handed derivative at 0 is
lim x→ 0 −
f (x) − f (0) x − 0 = lim x→ 0 −
−x x
and the right-handed derivative at 0 is
xlim→0+
f (x) − f (0) x − 0 = (^) xlim→ 0 −
x x
(Here x → 0 − (x → 0+) means x tends to 0 from the left-hand side (right-hand side, respectively).) Thus the left-handed and right-handed derivatives are not equal, so f is not differentiable at the origin. This corresponds to our intuition, as the graph of the function f (x) = |x| has a corner at the origin and so there is no well-defined tangent.)
Remark. The above remark illustrates why we are interested in functions defined on open sets: we want to approach the point x 0 from either side. If f was defined on the closed interval [a, b] then we could only consider right-handed derivatives at a (and left-handed derivatives at b).
The generalisation to complex functions is as one would expect.
Definition. Let D ⊂ C be an open set and let f : D → C be a function. Let z 0 ∈ D. We say that f is differentiable at z 0 if
f ′(z 0 ) = lim z→z 0
f (z) − f (z 0 ) z − z 0
exists. (Note that in (2.4.2) we are allowing z to converge to z 0 from any direction.) We call f ′(z 0 ) the derivative of f at z 0. If f is differentiable at every point z 0 ∈ D then we say that f is differentiable on D.
Remark. Sometimes we use the notation
df dz (z 0 )
to denote the derivative of f at z 0.
As we shall see, differentiability is a very strong property for a complex function to possess; it is much stronger than the real case. For example (as we shall see) there are many functions that are differentiable when restricted to the real axis but that are not differentiable as a function defined on C. For this reason, we shall often use the following alternative terminology.
Definition. Suppose that f : D → C is differentiable on a domain D. Then we say that f is holomorphic on D.
The higher derivatives are defined similarly, and we denote them by
f ′′(z 0 ), f ′′′(z 0 ),... , f (n)(z 0 ).
Example. Let f (z) = z^2 , defined on C. Let z 0 ∈ C be any point. Then
lim z→z 0
f (z) − f (z 0 ) z − z 0 = lim z→z 0
z^2 − z^20 z − z 0 = lim z→z 0
(z + z 0 )(z − z 0 ) z − z 0 = lim z→z 0 z + z 0 = 2z 0.
Hence f ′(z 0 ) = 2z 0 for all z 0 ∈ C. Thus f is differentiable at every point in C and so is a holomorphic function on C.
All of the standard rules of differentiable functions continue to hold in the complex case:
Proposition 2.4. Let f, g be holomorphic on D. Let c ∈ C. Then the following hold:
(i) sum rule: (f + g)′^ = f ′^ + g′,
(ii) scalar rule: (cf )′^ = cf ′,
(iii) product rule: (f g)′^ = f ′g + f g′,
(iv) quotient rule:
f g
= f g
′−f ′g g^2 , (v) chain rule: (f ◦ g)′^ = f ′^ ◦ g · g′.
Proof. Recall from (2.4.2) that to calculate f ′(z) we look at points that are close to z and then let these points tend to z. The trick is to calculate f ′(z) in two different ways: by looking at points that converge to z horizontally, and by looking at points that converge to z vertically. Let h be real and consider z + h = (x + h) + iy. Then as h → 0 we have z + h → z. Hence
f ′(z) = lim h→ 0
f (z + h) − f (z) h = lim h→ 0
u(x + h, y) + iv(x + h, y) − u(x, y) − iv(x, y) h = lim h→ 0
u(x + h, y) − u(x, y) h
- i v(x + h, y) − v(x, y) h =
∂u ∂x
∂v ∂x
Now take k to be real and consider z + ik = x + i(y + k). Then as k → 0 we have z + ik → z. Hence
f ′(z) = lim k→ 0
f (z + ik) − f (z) ik = lim k→ 0
u(x, y + k) + iv(x, y + k) − u(x, y) − iv(x, y) ik = lim k→ 0
u(x, y + k) − u(x, y) ik
- i v(x, y + k) − v(x, y) ik = −i
∂u ∂y
∂v ∂y
recalling that 1/i = −i. Comparing the real and imaginary parts of (2.5.1) and (2.5.2) gives the result. 2
Example. We can use the Cauchy-Riemann equations to examine whether the function f (z) = ¯z might be differentiable on C. Note that writing z = x + iy allows us to write f (z) = ¯z = x − iy. Hence f (z) = u(x, y) + iv(x, y) with u(x, y) = x and v(x, y) = −y. Now
∂u ∂x
∂u ∂y
∂v ∂x
∂v ∂y
Hence there are no points at which ∂u ∂x
∂v ∂y
so that f (z) = ¯z is not differentiable at any point in C.
Remark. Notice however that f (z) = ¯z is continuous at every point in C. Hence f (z) = z¯ is an example of an everywhere continuous but nowhere differentiable func- tion. Such functions also exist in real analysis, but they are much harder to write down and considerably harder to study (one of the simplest is known as Weierstrass’ function w(x) =
n=0 2 −nα (^) cos 2πbnx where α ∈ (0, 1), b ≥ 2; such functions are still of interest in
current research).
We have seen that if f is differentiable then the Cauchy-Riemann equations are satisfied. One could ask whether the converse is true: if the Cauchy-Riemann equations are satisfied then is f differentiable? The answer is no, as the following example shows. Define
f (x + iy) =
0 if one or both of x, y is 0, 1 if neither x, y is 0.
Thus f takes the value 0 on the x- and y-axes, but takes the value 1 everywhere else. Then at the origin ∂u ∂x
= lim h→ 0
u(x + h, y) − u(x, y) h
= lim h→ 0
h
and similarly ∂u/∂y, ∂v/∂x, ∂v/∂y are all equal to zero at the origin. Hence the Cauchy- Riemann equations hold. However, f is not even continuous at the origin, so it cannot be differentiable. The problem with the above example is that in the definition of differentiability (2.4.2) we need to let z tend to z 0 in an arbitrary way. In calculating the partial derivatives we only know what happens at z tends to z 0 either horizontally or vertically. Hence we need some extra hypotheses on u, v at z 0 ; the correct hypotheses are continuity of the partial derivatives.
Proposition 2.5.2 (Converse of the Cauchy-Riemann equations) Let f : D → C be a continuous function and let z 0 = x 0 + iy 0 ∈ D. Write f (x + iy) = u(x, y) + iv(x, y). Suppose that ∂u ∂x
∂u ∂y
∂v ∂x
∂v ∂y exist and are continuous at z 0 , and further suppose that the Cauchy-Riemann equations hold. Then f is differentiable at z 0.
Proof. The proof is based on the following lemma, whose proof we omit.
Lemma 2.5. Suppose that ∂w/∂x, ∂w/∂y exist at (x, y) and ∂w/∂x is continuous at (x, y). Then there exist functions ε(h, k) and η(h, k) such that
w(x + h, y + k) − w(x, y) = h
∂w ∂x
(x, y) + ε(h, k)
∂w ∂y
(x, y) + η(h, k)
and ε(h, k), η(h, k) → 0 as h, k → 0.
Now consider z = z 0 + h + ik. Applying the above lemma to both u and v we can write f (z) − f (z 0 ) = u(x 0 + h, y 0 + k) + iv(x 0 + h, y 0 + k) − u(x 0 , y 0 ) − iv(x 0 , y 0 )
= h
∂u ∂x
∂u ∂y
∂v ∂x
∂v ∂y
where ε 1 , ε 2 , η 1 , η 2 → 0 as h, k → 0. Using the Cauchy-Riemann equations, we can write the above expression as
f (z) − f (z 0 ) = (h + ik)
∂u ∂x
∂v ∂x
- hε 1 + kη 1 + ihε 2 + ikη 2
= (z − z 0 )
∂u ∂x
Exercise 2. Let f (z) =
|xy| where z = x + iy.
(i) Show from the definition (2.4.2) that f is not differentiable at the origin.
(ii) Show however that the Cauchy-Riemann equations are satisfied at the origin. Why does this not contradict Proposition 2.5.2?
Exercise 2. Suppose that f (z) = u(x, y) + iv(x, y) is holomorphic. Use the Cauchy-Riemann equations to show that both u and v satisfy Laplace’s equation:
∂^2 u ∂x^2
∂^2 u ∂y^2
∂^2 v ∂x^2
∂^2 v ∂y^2
(you may assume that the second partial derivatives exist and are continuous). (Functions which satisfy Laplace’s equation are called harmonic functions.)
Exercise 2. For f (z) = z^3 calculate u, v so that f (z) = u(x, y) + iv(x, y) (where z = x + iy). Verify that both u and v satisfy Laplace’s equation.
Exercise 2. Suppose f (z) = u(x, y) + iv(x, y). Suppose we know that u(x, y) = x^5 − 10 x^3 y^2 + 5xy^4. By using the Cauchy-Riemann equations, find all the possible forms of v(x, y). (Thus Cauchy Riemann equations have the following remarkable implication: suppose f (z) = u(x, y) + iv(x, y) is differentiable and that we know a formula for u, then we can recover v (up to a constant); conversely, if we know v then we can recover u (up to a constant). Hence for complex differentiable functions, the real part of a function determines the imaginary part (up to constants), and conversely.)
Exercise 2. Suppose that u(x, y) = x^3 − kxy^2 + 12xy − 12 x
for some constant k ∈ C. Find all values of k for which u is the real part of a holomorphic function f.
Exercise 2. Show that if f is holomorphic and f has a constant real part then f is constant.
Exercise 2. Show that the only holomorphic function f of the form f (x + iy) = u(x) + iv(y) is given by f (z) = λz + a for some λ ∈ R and a ∈ C.
Exercise 2. Suppose that f (z) = u(x, y) + iv(x, y) is a holomorphic function and that
2 u(x, y) + v(x, y) = 5 for all z = x + iy ∈ C.
Show that f is constant.
