Presentation on theme: "1 Week 4 Complex numbers: analytic functions 1. Differentiation of complex functions 2. Cauchy’s Integral Theorem."— Presentation transcript:
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. A differentiable complex function, on the other hand, always has infinitely many derivatives. For example, if a real function is differentiable in a domain D, it’s not necessarily twice differentiable in D. Here’s another difference: real functions defined through non- singular algebraic expressions are differentiable – which is, however, not true for their complex counterparts. Comment: Can you come up with an example of such a function?
3 ۞ A complex function f(z) is said to be continuous at z = z 0 if f(z) exists in a certain neighbourhood of z 0 and regardless of the path along which z approaches z 0.
4 ۞ The derivative f '(z) of a complex function f(z) at z = z 0 is defined by where the limit has the same value regardless of the path along which z approaches z 0. (1) The rules of complex differentiation are the same as in real calculus, and their proofs are literally the same.
5 Example 2: Show that the derivative of f(z) = z* does not exist for any z. Example 1: Calculate ‘from the first principles’ the derivative of f(z) = z 2. ۞ A function f(z) is said to be analytic at a point z 0 if f(z) is analytic in a neighbourhood of z 0. ۞ A function f(z) is said to be analytic in a domain D if f(z) is differentiable at all points of D.
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: or and the derivative of f(z) at z = z 0 is given by either (2) (3) (4) (5)
7 Proof: Introduce Δz = z – z 0 and rewrite definition (1) of f '(z) as where Δx = Re z and Δy = Im z. (6) Let’s re-arrange it further, in the form
8 Recall that the limit in (6) can be calculated along any path we want – thus, consider the following two paths: Do the two limits in this expression remind you of something?... For P 1, expression (6) yields u y (x 0, y 0 )v y (x 0, y 0 ) (7)
9 Thus, (7) reduces to (5). 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). Similarly, for path P 2, (7) reduces to (4).
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 Example 3: Verify that u = x 2 – y 2 – y satisfies Laplace’s equation and find all functions v harmonically conjugated to u. ۞ 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 2. Cauchy’s Integral Theorem Then ۞ Consider a continuous complex function f(z) and a path P on the complex plane, parameterised by (8) Let P be smooth, i.e. dZ/dt exists for all t (a, b) and is continuous.
13 A complex integral can be reduced to two real line integrals on the real plane. To do so, let and rearrange (8) in the form Then, expanding the expressions in brackets and separating the real and imaginary parts, we obtain...
14 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. The two integrals on the r.h.s. are the real line integrals of the vector fields