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Let D be a subset of C. A function f : D Notes 1. The set D of numbers that are mapped is called the domain of f. Notice that we now have a double use of this word. Where the domain is unspecified, we assume it to be the largest subset of C for which f(z) is defined. 2. The set of image elements {w | w = f(z)} is called the range or image of the function. 3. The above definition specifies a unique image for each z
If f(z) only assumes real values, f is real-valued.
We can break w = f(z) into real and imaginary parts. Thus if w = u + iv, z = x + iy, then w = f(z) = u(x, y) + iv(x, y)
This expression in terms of real and imaginary parts may be easier said than done! In theory, it allows us to deduce properties of complex functions from our knowledge of the real numbers. A real function y = f(x) can be represented geometrically by a graph: the set of points {(x, y) | y = f(x)}. To represent the complex function w = f(z) geometrically, in general we need four dimensions or two planes: a plane for the domain, and a plane for the range. For simple functions, we can use the same plane twice.
We shall be mapping curves and regions rather than just points.
All work on functions of two variables now carries over directly. A minor difference is that because we are dealing with complex numbers, the length or norm of a vector represented by w becomes the modulus |w| of w. Thus the definition of limit becomes:
for all
Theorem 2.1 If w = f(z) = u + iv, z = x + iy, z0 = x0 + iy0, then
In brief, this theorem says: lim(u + iv) = lim u + i lim v. Proof ( By definition, given 0 < |x – x0 + i(y – y0)| < We deduce that 0 < |x – x0| < This completes this part of the proof. ( Then there exist 0 < |x – x0| < 0 < |x – x0| < Choose 0 < |(x – x0) + i(y – y0)| < |(u(x, y) + iv(x, y)) – (u0 + iv0)| Our previous theorem quickly leads to the well-known and useful Limit Theorems. We use an easy to remember abbreviated notation. Theorem 2.2. (Limit Theorems) If lim f, lim g exist, then lim (f Proof (a) Set f = u + iv, lim f = u0 + iv0, g = U + iV, lim g = U0 + iV0. Now lim (f + g) = lim (u + U + i(v + V)) (substitute and rearrange) The other proofs are similar. Definition Function f is said to be continuous at z0 if f satisfies the following three conditions. (a) f(z0) exists ; (b) lim z Notes 1. Just writing statement (c) assumes the truth of (a) and (b). 2. Expressing (c) in terms of the limit definition, we obtain: |z – z0| < This is slightly different from the usual definition of limit, in that we allow the possibility z = z0 (omitting 0 < |z – z0| ... ). Function f = f(z) is said to be continuous in a region if it is continuous at each point of the region. The sum f + g, difference f – g, product f.g and quotient f/g of two continuous functions is continuous at a point z = z0, with the proviso that in the last case g(z0)
We can also compose complex functions f, g to obtain the new function f o g defined by (f o g)(z) = f(g(z)). If f, g are continuous, will f o g be continuous also?
Alternatively, a continuous function of a continuous function is a continuous function. Formally, the proof of this theorem is exactly as for the real case, and is omitted here.
By Theorem 2.1, f(z) = u + iv is continuous
Definition Say f is bounded in region R if |f(z)| If f is continuous in R, then f is bounded because of the corresponding properties of u, v. Show this! Formally, the definition of the derivative f '(z) for functions of a complex variable is the same as for real functions. Let f be a function whose domain contains a neighbourhood of point z0. Then
if the limit exists. In this case the function f is said to be differentiable at z0. It is sometimes preferable to use the alternative form of the derivative obtained by setting z = z0 +
Note Since z0 lies in an (open) neighbourhood of the domain of f, f(z0 + We can evaluate simple derivatives by using the basic definition.
The usual differentiation formulae hold as for real variables. For example, However, care is required for more unusual functions.
This last statement is proved using the basic definition. Show it! As in the real case, f is differentiable The same rules apply for the sum, product, quotient and composite of two differentiable functions (where defined).
Again, with more unusual functions, we may have to use the limit definition of differentiation.
Theorem 2.1 gives us conditions for continuity for a function of a complex variable in terms of the continuity of the real and imaginary parts. We now ask: Is there any test for differentiability? Theorem 2.4 The derivative f '(z) of f = u + iv exists at z ux = vy, uy = –vx (the Cauchy-Riemann equations). Further, f '(z) = ux + ivx = vy – iuy. Proof Since the derivative of f exists,
By Theorem 2.1, lim Re(*) = a, lim Im (*) = b. Set y Set x Hence all the first partial derivatives exist, ux = vy, uy = –vx, and 1. Set f(z) = z2 = (x + iy)2 = (x2 – y2) + 2ixy. Now f'(z) = 2z exists for all z. So the Cauchy-Riemann equations are satisfied. 2. Set f(z) = |z|2. We show f '(z) does not exist for z Now f(z) = x2 + y2, i.e. u = x2 + y2, v = 0, ux = 2x, uy = 2y, vx = 0 = vy. Hence f '(z) can only exist at (0, 0). Does f '(0) exist? Yes; as suggested earlier; we must use a first principles argument. Theorem 2.4 gives necessary conditions for f to be differentiable. We now seek sufficient conditions for f ' to exist: that is, a similar statement, but using Theorem 2.5 Let f = u + iv as before. Suppose that That is, f is differentiable at z0 Proof We omit this proof. It is not hard, but rather messy. The real functions u = ex cos y, v = ex sin y are defined and continuous everywhere. So are ux, vx, uy, vy and you can easily check that the Cauchy-Riemann equations are satisfied. Hence the function f(z) = ex cos y + i ex sin y is differentiable everywhere. Since ux = u, vx = v, Definitions f(z) is analytic at z0 if f '(z) exists not only at z0 but for all z in some neighbourhood of z0. f(z) is analytic in a domain of the z-plane if it is analytic at every point of the domain. f(z) is entire if it is analytic everywhere.
Question How can we tell if a function is analytic? We can use Theorem 2.5, or Theorem 2.6: If f = f(z) is analytic, then in any formula for f, x and y can only occur in the combination x + iy. Proof We note that x = (1/2)(z +
equating real, imaginary parts to zero. Hence f analytic
Derivative Theorems The theorems on derivatives quickly transfer to analytic functions. Let f = u + iv be analytic in some domain of z-plane. Then the Cauchy-Riemann equations hold: ux = vy, uy = –vx It can be shown that for analytic function, the partial derivatives of all orders exist uxx = vyx and uyy = –vxy. Assuming continuity of vyx, vxy, we have vyx = vxy, and hence uxx + uyy = 0. This is Laplace's equation, and u is called an harmonic function. In the same way we get vxx + vyy = 0; i.e. v is an harmonic function. If f = u + iv, u and v are conjugate harmonic functions. In applied mathematics (partial differential equations) we often seek an harmonic function on a given domain which satisfies given boundary conditions. If we are given one of two conjugate harmonic functions, it is a simple matter to find the other. We use the Cauchy-Riemann equations.
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C is a rule that associates with each z in D a unique complex number w. We write w = f(z).
D. Later we shall extend this definition to include multivalued functions.
i}.
Example 1. w = f(z) = z + 2.
Example 2. w = f(z) = 
.
(x2 + y2) – iy.
c.
Thus the image of this circle is the line segment u = c, –c =w.gif){kind=small}
means 
> 0, there exists 
> 0 : for all z, 0 < |z – z0| < 
|f(z) – w0| < 
) Now let us suppose that 
0)
zo f(z) exists ; (c) lim z
ai z i is continuous forall z since it is constructed as sums and products of the continuous constant functions (ai) and the continuous function f(z) = z.
u(x, y), v(x, y) are continuous. Thus:
z :
(2z2 + i)5 = 5(2z2 + i)4.4z = 20z(2z2 + i)4.
y0 to get 
(z = 1, 2 are singularities).
and
(x) and now