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Material Type: Notes; Class: Junior Seminar; Subject: Mathematics; University: Seton Hall University; Term: Spring 2004;
Typology: Study notes
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1 What is a Complex Function? 1
2 Limits and Continuity of Complex Functions 2
3 Complex Differentiation 3
4 Analyticity 5
5 Entire 7
We shall begin with a review of the basics.
Definition 1.1 (Function) A function f is a rule that assigns to each element in a set A one and only one element in a set B.
If f assigns the value b to the element a in A, we write
b = f (a)
and call b the image of a under f. As a result of this definition of function we sometimes refer to f as a mapping of A onto B. We are not concerned with just any functions. We are concerned with com- plex valued funtions of a complex variable, that is z = x + ıy, where x is the real part and y is the imaginary part (both real-valued). For complex valued functions, if w denotes the value of the function f at the point z, we then write
w = f (z).
Also, just as the complex number z decomposes into real and imaginary parts, the complex function w also decomposes into real and imaginary parts, again both real-valued. This is customarily written
w = u(x, y) + ıv(x, y),
with u and v denoting the real and imaginary parts respectively. Thus, a complex-valued function of a complex variable is, in essence, a pair of real functions of two real variables.
Example 1.1 Write the function w = z^2 +2z in the form w = u(x, y)+ıv(x, y).
Solution: By setting z = x + ıy we obtain
w = (x + ıy)^2 + 2(x + ıy) = x^2 − y^2 + ı 2 xy + 2x + ı 2 y. Which then can be rewritten as
w = (x^2 − y^2 + 2x) + ı(2xy + 2y).
The concepts of limits and continuity for complex functions are similar to those for real functions. Let’s first examine the concept of the limit of a complex- valued function.
Definition 2.1 (Limit) Let f be a function defined in some neighborhood of z 0 , with the possible exception of the point z 0 itself. We say that the limit of f (z) as z approaches z 0 is the number w 0 and write
lim z→z 0 f (z) = w 0 ,
or equivalently,
f (z) → w 0 as z → z 0 , if for any > 0 there exists a positive number δ such that |f (z) − w 0 | < whenever 0 < |z − z 0 | < δ.
This is esentially the same definition the we learned in calculus I when we were first learning limits. The condition of continuity is as follows.
Definition 2.2 (Continuous) Let f be a function defined in a neighborhood of z 0. Then f is continuous at z 0 if
lim z→z 0 f (z) = f (z 0 ).
In general, a complex function of a complex variable, f (z), is an arbitrary mapping from the xy-plane to the uv -plane. A complex function is split into real and imaginary parts, u and v, and any pair u(x, y) and v(x, y) of two- variable functions gives us a complex function u + ıv. However, notice there is something special about the pair
u 1 (x, y) = x^2 − y^2 , and v 1 (x, y) = 2xy, as opposed to u 2 (x, y) = x^2 − y^2 , and v 2 (x, y) = 3xy. The difference is that the complex function u 1 + ıv 1 treats z = x + ıy as a single ”unit”, because x^2 − y^2 + ı 2 xy = (x + ıy)^2. These are the types of functions that are complex differentiable.
Definition 3.1 Let f be a complex-valued function defined in a neighborhood of z 0. Then the derivative of f at z 0 is given by
df dz
(z 0 ) ≡ f ′(z 0 ) := lim ∆z→ 0
f (z 0 + ∆z) − f (z 0 ) ∆z
provided this limit exists. (Such an f is said to be differentiable at z 0 ).
The catch in this definition is that ∆z is a complex number, so it can ap- proach zero in many different ways. Even though this catch may make things seem slightly more difficult, the rules for differentiating real functions apply in the same way for complex-valued functions (as long as the complex-valued function is in a form where z = x + ıy is treated as a single unit).
Example 3.1 Show that, for any positive integer n,
d dz
zn^ = nzn−^1.
Solution: Using Definition 3.1 we have
(z + ∆z)n^ − zn ∆z
nzn−^1 ∆z + n(n 2 − 1)zn−^2 (∆z)^2 + · · · + (∆z)n ∆z
Thus d dz
zn^ = lim ∆z→ 0
nzn−^1 +
n(n − 1) 2
zn−^2 ∆z + · · · + (∆z)n−^1
= nzn−^1.
Theorem 3.1 If f and g are differentiable at z, then
(f ± g)′(z) = f ′(z) ± g′(z), (1) (cf )′(z) = cf ′(z) for any constant c, (2) (f g)′(z) = f (z)g′(z) + f ′(z)g(z), (3) ( f g
(z) =
g(z)f ′(z) − f (z)g′(z) g(z)^2
if g(z) 6 = 0. (4)
(5)
If g is differentiable at z and f is differentiable at g(z), then the chain rule holds: d dz
f (g(z)) = f ′(g(z))g′(z).
Now that we have a secure background we are ready to look at the theory of analytic functions.
Definition 4.1 A complex-valued function f (z) is said to be analytic on an open set G if it has a derivative at every point of G.
Analyticity is a property defined over open sets, while differentiability could hold at one point only. If the phrase ”f (z) is analytic at the point z 0 ” is used it means that f (z) is analytic in some neighborhood of z 0. To show that a function is analytic we use the following equations which must hold at z 0 = x 0 + ıy 0 :
∂u ∂x
∂v ∂y
∂u ∂y
∂v ∂x known as the Cauchy-Riemann equations.
Theorem 4.1 A necessary condition for a function f (z) = u(x, y) + ıv(x, y) to be differentiable at a point z 0 is that the Cauchy-Riemann equations hold at z 0. Consequently, if f is analytic in an open set G, then the Cauchy-Riemann equations must hold at every point of G.
Example 4.1 Show that the function f (z) = (x^2 + y) + ı(y^2 − x) is not analytic at any point.
Solution: Since u(x, y) = x^2 + y and v(x, y) = y^2 − x we have ∂u ∂x = 2x,^
∂v ∂y = 2y, ∂u ∂y = 1,^
∂v ∂x =^ −1.
Example 4.4 Find
C
ez (z+5)^3 (z−ı) dz, where^ C^ is the circle centered at the origin of radius 2.
Solution: Again in this example we must rewrite the problem in the form f (z) z−z 0.
C
ez (z + 5)^3 (z − ı)
dz
C
ez (z+5)^3 (z − ı)
dz
In this case it is easy to see that f (z) = e
z (z+5)^3 and^ z^0 =^ ı.^ In this case z 0 = ı is included in C so Theorem 4.3 can be applied.
C
ez (z+5)^3 (z − ı)
dz
= 2 πıf (ı)
= 2 πı
eı (ı + 5)^3
If f (z) is analytic on the whole complex plane, then it is said to be entire.
Definition 5.1 A function f (z) is called entire if it has a representation of the form
f (z) =
k=
akzk^ valid for |z| < ∞.
This class of functions is designated by E. E is a linear space.
Example 5.1 Some examples of entire functions are
sin(z) z
, 2 z^ ,
∫ (^) z
0
et
2 dt,
Γ(z)
Theorem 5.1 The function f (z) =
k=0 akz k (^) is entire if and only if
lim n→∞ |an| n^1 = 0.