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ENGG2420B Complex functions and complex differentiation

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1 ENGG2420B Complex functions and complex differentiation
Kenneth Shum kshum ENGG2420B

2 What is a function? Domain Range a  = f(a)  b  = f(b) c  = f(c) 
kshum ENGG2420B 2

3 Real-valued function f(x)=x2
Domain Range –2 f 4 2 f -1 f nothing is mapped to -1 kshum kshum 3

4 Complex function Domain is the set of all complex numbers
Range is the set of all complex numbers z = x+i y w = f(z) w = u + i v z-plane w-plane kshum kshum 4

5 Notations in this lecture
z-plane Domain of a complex function z = x + i y x is the real part of z. y is the imaginary part of z. w-plane Image of a complex function w = u + i v, w is a function of z u is the real part of w. v is the imaginary part of w. u and v are functions of x and y. kshum kshum

6 Extension from real to complex
z-plane f(z) = z2 w-plane -1-i 1+i 2i –2 4 2 i -1 -i u = x2 – y2 v = 2 x y Now, every complex number, including -1, has at least one pre-image. kshum kshum

7 f(z) = i z u = –y v = x z-plane w-plane w = i z kshum kshum

8 f(z) = (1+i) z+ 2 u = x–y+2 v = x+y z-plane w-plane w = f(z) kshum

9 f(z) = conj(z) u = –x v = y z-plane w = conj(z) w-plane kshum kshum

10 f(z) = z2 (lower left corner is 0)
u = x2 – y2 v = 2 x y z-plane w-plane w = z2 kshum kshum ENGG2420B ENGG2012B

11 f(z) = z2 (lower left corner is 1)
u = x2 – y2 v = 2 x y z-plane w-plane w = z2 kshum kshum ENGG2420B ENGG2012B

12 f(z) = 1/z u = x/(x2+y2) v = -y/(x2+y2) z-plane w-plane w = 1/z kshum

13 f(z) = z + (1+i) conj(z) u = 2x + y v = x z-plane
w = z + (1+i) conj(z) w-plane kshum kshum

14 Question On the left-hand side in each of the previous slides, the thick red line and the thick blue line are perpendicular. The angle between the black line and the red line is arctan(3) = degrees. On the right-hand sides, these angles at the intersection point are sometime preserved, and sometime not. Is there any explanation? kshum kshum

15 Review: Differentiation of real-valued function
y = x2 y = 2x-1 kshum kshum

16 Behaviour of f(x)=x2 near x=1
An infinitesimal arrow in the domain starting at x=1 is transformed to an infinitesimal arrow in the image starting at y=1, with scaling factor = 2 Domain Image f(x) = x2 y x 1 1 kshum kshum

17 Behaviour of f(x)=x2 near x=0
An infinitesimal arrow in the domain starting at x=1 is transformed to an arrow of length 0 in the image Domain Image f(x) = x2 y x kshum kshum

18 Behaviour of f(x)=x2 near x=-0.5
An infinitesimal arrow in the domain starting at x=1 is transformed to an infinitesimal arrow in the image starting at y=1, with scaling factor = -1 Domain Image f(x) = x2 y x -0.5 -0.5 kshum kshum

19 Derivative for real function
Let f(x) be a real-valued function. Domain and image are the real numbers The function f is said to be differentiable at x0 if we can find a real number s, such that for any small real number h, we have f(x0+ h)  f(x0)+ s h. The scaling factor s may vary with the reference point x0. If we change the reference point x0, the value of s may change. The scaling factor s is called the derivative of f(z) at x0, and is usually denoted by f’(x0). Formally, f’(x0) is expressed as the limit kshum ENGG2420B

20 Complex differentiation
(We draw a complex number as a vector here) Domain z-plane Image w-plane z0 is mapped to w0. f(z0) = w0 w = f(z) A complex function f is said to be differentiable at z0, if whenever we draw a short vector h from z0, (the angle is arbitrary), the corresponding displacement in the image is well approximated by a complex multiple of h, i.e., f(z0+ h) – f(z0)  c h. The complex scaling factor c is called the derivative of f at z0 z0 w0 kshum kshum

21 An example of a non-differentiable function
Domain z-plane Image w-plane z0 is mapped to w0. f(z0) = w0 w = f(z) z0 w0 kshum kshum

22 Complex derivative f(z0+ h)  f(z0)+ c  h.
Let f(z) be a complex function. Domain and image are complex planes The function f is said to be differentiable at a point z0 if we can find a complex number c, such that for any small complex number h, we have the approximation f(z0+ h)  f(z0)+ c  h. The complex scaling factor c may vary with the point z0. If we change the reference point z0, the value of c may change. The complex scaling factor c is called the derivative of f(z) at z0, and is usually denoted by f’(z0). Formally, the derivative of f(z) at z0 is defined as a limit Here, h is a complex number. Addition, subtraction and division are all complex arithmetic. kshum ENGG2420B

23 Analytic function A complex function is said to be analytic if it is differentiable at all points in the domain of the function. An analytic function is sometime called a holomorphic function. If f(z) is differentiable at all points in the complex plane, than f(z) is called an entire function. For example, f(z) = z2 is an entire function, and the derivative is given by f’(z)=2z. kshum ENGG2420B

24 Answer to the question Suppose that there are two curves intersecting at z0 in the domain z-plane, making an angle  at the intersection point. If f(z) is differentiable at z0 with non-zero derivative, then image of the two curves in the w-plane have angle  at the intersection point w0 as well. In complex analysis, an angle-preserving mapping is called a conformal mapping. kshum kshum

25 Summary of the examples given in this lecture
Function Differentiable ? Derivative = ? f(z) = j z Differentiable everywhere f’(z) = j f(z) = (1+j) z+ 2 f’(z) = 1+j f(z) = conj(z) Not differentiable N/A f(z) = z2 f’(z) = 2 z f(z) = 1/z Differentiable everywhere except the origin f’(z) = - 1/z2 f(z) = cos(z) f’(z) = - sin(z) f(z) = z + (1+j) conj(z) kshum ENGG2420B

26 A necessary condition for being differentiable
Cauchy-Riemann condition (cartesian form) Consider a complex function f(z) = u(x,y) + i v(x,y) Suppose f(z) is defined at a point z0 = x0+i y0 and its neighborhood. Then f(z) is differentiable at z0 if \begin{align*} u_x(x_0,y_0) &= v_y(x_0,y_0) \\ u_y(x_0,y_0) &= -v_x(x_0,y_0) \end{align*} kshum ENGG2420B

27 Application We can show that the complex conjugate function is not differentiable anywhere by verifying that the Cauchy-Riemann condition fails. conj(z) = x – i y u(x,y) = x v(x,y) = –y Check that ux(x,y)=1, vy(x,y) = –1. They are not equal for any x and any y. kshum ENGG2420B


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