1. Week 4 Complex numbers: analytic functions
1. Differentiation of complex functions
2. Cauchy’s Integral Theorem

2. 1. Differentiation of complex functions
The notion of differentiability of complex functions is very different from that of their real counterparts.
For example, if a real function is differentiable in a domain D, it’s not necessarily twice differentiable in D.
Comment:
Can you come up with an example of such a function?
A differentiable complex function, on the other hand, always has infinitely many derivatives.
Here’s another difference: real functions defined through non-singular algebraic expressions are differentiable – which is, however, not true for their complex counterparts.

3. ۞ A complex function f(z) is said to be continuous at z = z0 if f(z) exists in a certain neighbourhood of z0 and
regardless of the path along which z approaches z0.

4. ۞ The derivative f '(z) of a complex function f(z) at z = z0 is defined by
(1)
where the limit has the same value regardless of the path along which z approaches z0.
The rules of complex differentiation are the same as in real calculus, and their proofs are literally the same.

5. Example 1:
Calculate ‘from the first principles’ the derivative of f(z) = z2.
Example 2:
Show that the derivative of f(z) = z* does not exist for any z.
۞ A function f(z) is said to be analytic in a domain D if f(z) is differentiable at all points of D.
۞ A function f(z) is said to be analytic at a point z0 if f(z) is analytic in a neighbourhood of z0.

6. Theorem 1: the Cauchy–Riemann Equations
Let f(z) = u(x, y) + i v(x, y) be defined and continuous in some neighborhood of a point z and be differentiable at z itself.
Then, at z, the first-order partial derivatives of u and v exist and satisfy the so-called Cauchy–Riemann equations:
(2)
(3)
and the derivative of f(z) at z = z0 is given by either
(4) or
(5)

7. Proof:
Introduce Δz = z – z0 and rewrite definition (1) of f '(z) as
Let’s re-arrange it further, in the form
(6)
where Δx = Re Δz and Δy = Im Δz.

8. Recall that the limit in (6) can be calculated along any path we want – thus, consider the following two paths:
For P1, expression (6) yields
uy(x0, y0)
vy(x0, y0)
(7)
Do the two limits in this expression remind you of something?...

9. Thus, (7) reduces to (5).
Similarly, for path P2, (7) reduces to (4).
Finally, comparing (4) and (5), we shall obtain the desired Cauchy–Riemann equations. █
Theorem 2:
If two real-valued continuous functions u(x, y) and v(x, y) of two real variables x and y have continuous partial derivatives satisfying the Cauchy–Riemann equations in a domain D, then the complex function f(z) = u(x, y) + i v(x, y) is analytic in D.
Proof: See Kreyszig, Appendix 4 (non-examinable).

10. Theorem 3: Laplace’s Equation
If f(z) = u(x, y) + i v(x, y) is analytic in a domain D, then u and v both satisfy in D the Laplace’s equation:
or, in a different notation,
Proof: This theorem follows from the Cauchy–Riemann equations.
۞ Real solutions of Laplace’s equation, with continuous second-order partial derivatives, are called harmonic functions.

11. ۞ If real-valued functions u(x, y) and v(x, y) satisfy the Cauchy–Riemann equations in a domain D, they are said to be harmonically conjugated in D.
Comments:
(1) “Harmonically conjugated functions” have absolutely nothing to do with “conjugate complex numbers” (the use of “conjugate(d)” is coincidental).
(2) Harmonically conjugated functions both satisfy Laplace’s equation (due to Theorem 3).

12. Example 3:
Verify that u = x2 – y2 – y satisfies Laplace’s equation and find all functions v harmonically conjugated to u.
Soln to the 2nd part of the question:
Harmonically conjugated functions satisfy the Cauchy – Riemann equations (2)-(3):
hence, substituting for u the expression given,
hence...

13. (8)
hence,
hence,
hence, (8) yields

14. 2. Cauchy’s Integral Theorem
۞ Consider a continuous complex function f(z) and a path P on the complex plane, parameterised by
Let P be smooth, i.e. dZ/dt exists for all t  (a, b) and is continuous.
Then
(9)

15. A complex integral can be reduced to two real line integrals on the real plane. To do so, let
and rearranged (9) in the form
Then, expanding the expressions in brackets and separating the real and imaginary parts, we obtain...

16. The two integrals on the r. h. s
The two integrals on the r.h.s. are the real line integrals of the vector fields
Example 4:
Calculate
where P is the ¾ of the unit circle centred at the origin, starting from z = 1 and finishing at z = −i.