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by Indian Institute of Technology Madras (IITM)
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January-May 2011
IIT Madras M. Thamban Nair June 2011
(^1) Second Editin, Hindustan Book Agency (‘trim’ series), New Delhi, 2008.
v
After having the real field R, it is natural to look for a bigger field in which algebraic equations such as
x^2 + 1 = 0 (∗)
has a solution. Of course, the + sign here must be the symbol for addition in the bigger field. Since two fields can be considered to be identical if there is a surjective isomorphism between then, it is enough to have a field which contains an isomorphic image of R and having required properties such as solution to algebraic equations. We shall define such a field with the intention of having a solution to the equation (∗).
Definition 1.1.1 The set C of complex numbers is the set of all ordered pairs (x, y) of real numbers with the following operations of addition and multiplication:
(x 1 , y 1 ) + (x 2 , y 2 ) = (x 1 + x 2 , y 1 + y 2 ),
(x 1 , y 1 ).(x 2 , y 2 ) = (x 1 x 2 − y 1 y 2 , x 1 y 2 + x 2 y 1 ). ♦ The proof of the following theorem is left to the reader.
Theorem 1.1.1 The following hold. (i) C is a field with additive identity (0, 0) and multiplicative identity (1, 0). (ii) The map ϕ : R → C defined by ϕ(x) = (x, 0), x ∈ R,
is a field isomorphism.
1
2 Complex Plane
We observe that the multiplicative inverse of a nonzero complex number z = (x, y) is given by ( x x^2 + y^2
, − y x^2 + y^2
Writing i = (0, 1)
and for x ∈ R, x˜ = (x, 0),
we observe that i^2 = −˜ 1 ,
and C = {x˜ + i˜y : x, y ∈ R}.
With the above notations, the addition and multiplication in C can be written as
(˜x 1 + iy˜ 1 ) + (˜x 2 + iy˜ 2 ) = (˜x 1 + ˜y 2 ) + i(˜y 1 + ˜y 2 ) = (x˜ 1 + y 2 ) + i (y˜ 1 + y 2 ),
(˜x 1 + iy˜ 1 ).(˜x 2 + iy˜ 2 ) = (˜x 1 x˜ 2 − ˜y 1 ˜y 2 ) + i(˜x 1 y˜ 2 + ˜x 2 ˜y 1 ) = (x 1 x˜ 2 − y 1 y 2 ) + i (x 1 y˜ 2 + x 2 y 1 ).
Throughout this course, we identify ˜x with x for every x ∈ R, so that C is the set of all numbers of the form
a + ib with a, b ∈ R.
Thus, a + i0 = a, 0 + i1 = i and i^2 = − 1 ,
and for nonzero z = x + iy,
1 z
:= z−^1 = x x^2 + y^2
− i y x^2 + y^2
One of the important properties of this field is that, not only equa- tion (∗) has a solution in C, but every algebraic equation also has a solution. This is the so called fundamental theorem of algebra which we shall prove in the due course.
4 Complex Plane
zn → z =⇒ z ∈ S.
z ∈ A, |z − z 0 | < δ =⇒ |f (z) − f (z 0 )| < ε.
Equivalently, f is continuous at a point z 0 ∈ A if and only if for every ε > 0, there exists δ > 0 such that
z ∈ A ∩ B(z 0 , δ) =⇒ f (z) ∈ B(f (z 0 ), ε).
z ∈ A, 0 < |z − z 0 | < δ =⇒ |f (z) − ζ| < ε,
and in that case we write
zlim→z 0
f (z) = ζ.
Some Definitions and Properties 5
Theorem 1.2.2 The set C is a complete metric space.
Proof. Let (zn) be a Cauchy sequence in C. Writing zn = xn +iyn with xn, yn ∈ R, we have
|xn − xm| ≤ |zn − zm|, |yn − ym| ≤ |zn − zm|
for all n, m ∈ N. Hence, (xn) and yn) are Cauchy sequences in R. Since R is a complete metric space with respect to the absolute-value metric, there exist x, y ∈ R such that xn → x and yn → y. Then, writing z = x + iy, we have
|zn − z|^2 = (xn − x)^2 + (yn − y)^2 → 0.
Thus, (zn) converges to z.
Let z be a nonzero complex number and let θ be the angle which the line segment joining 0 to z makes with the positive real axis, and r = |z|, the length of the line segment. Then it is clear from the geometry that z = r(cos θ + i sin θ). (∗)
Definition 1.2.2 The representation (∗) of a nonzero z ∈ C is called its polar representation, and an angle θ for which (∗) holds is called an argumet of z, denoted by arg(z). ♦
Note that each (r, θ) with r > 0 and θ ∈ R represents a unique nonzero z ∈ C with the representation (∗), but a non zero z ∈ C has many polar representations, namely,
z = r(cos θ + i sin θ), θ ∈ {θ 0 + 2πk : k ∈ Z},
where θ 0 is one of the angles for which (∗) holds. We note that if z 1 = r 1 (cos θ 1 + i sin θ 1 ) and z 2 = r 2 (cos θ 2 + i sin θ 2 ), then
z 1 z 2 = r 1 r 2 [cos(θ 1 + θ 2 ) + i sin(θ 1 + θ 2 )].
Thus, if z = r(cos θ + i sin θ) and n ∈ N, then
zn^ = rn(cos nθ+i sin nθ) = rn[cos(nθ+2kπ)+i sin(nθ+2kπ)], k ∈ Z.
Problems 7
i.e., uλ = (λx, λy, 1 − λ), λ ∈ R.
Clearly, for every u = (x, y, 0) ∈ X there exists one and only λ := λu such that uλ ∈ S^2. We consider the map
u := (x, y, 0) 7 → (λux, λuy, 1 − λ)
from X to S^2 \ {(0, 0 , 1)}. Note that
uλ ∈ S^2 ⇐⇒ λ^2 x^2 + λ^2 y^2 + (1 − λ)^2 = 1
if and only if λ = 0 or λ(x^2 + y^2 + 1) − 2 = 0. The point λ = 0 correspond to u 0. Hence,
λu =
x^2 + y^2 + 1.
Thus the map
z := x + iy 7 →
2 x 1 + |z|^2 ,^
2 x 1 + |z|^2 ,^
|z|^2 − 1 1 + |z|^2
is a bijective continuous function from C onto S^2 \ {(1, 0 , 0)} with its inverse (α, β, γ) 7 → α^ +^ iβ 1 − γ
which is also continuous.
8 Complex Plane
β
γ
1 + z 1 − z =^ ib^ for some^ b^ ∈^ R.
z := x + iy 7 →
2 x 1 + |z|^2
, 2 x 1 + |z|^2
, |z|
1 + |z|^2
is a bijective continuous function from C onto S^2 \ {(1, 0 , 0)} with its inverse (α, β, γ) 7 → α^ +^ iβ 1 − γ
which is also continuous.
does not exist.
10 Analytic Functions
whenever z is in some neighbourhood of z 0.
The following theorem can be proved (exercise) using arguments similar to the real case of real valued functions of a real variable.
Theorem 2.1.1 Let z 0 be an interior point of Ω ⊆ C. Then the following holds.
(i) If f differentiable at z 0 ∈ Ω, then f is continuous at z 0. (ii) If f and g are differentiable at z 0 ∈ Ω, then f + g and f g are differentiable at z 0 , and (f +g)′(z 0 ) = f ′(z 0 )+g(z 0 ), (f g)′(z 0 ) = f ′(z 0 )g(z 0 )+f (z 0 )g′(z 0 ).
(iii) If f and g are differentiable at z 0 ∈ Ω and if g(z 0 ) 6 = 0, then f /g is differentiable at z 0 , and ( f g
(z 0 ) =
g(z 0 )f ′(z 0 ) − g′(z 0 )f (z 0 ) [g(z 0 )]^2.
(iv) If f is differentiable at z 0 ∈ Ω and g is differentiable in a neighbourhood of f (z 0 ), then g ◦ f is differentiable at z 0 and (g ◦ f )′(z 0 ) = g′(f (z 0 )f ′(z 0 ).
Now, let us write f (z) as u(z) + iv(z), where u(z) = Ref (z) and v(z) = Imf (z). Recall that f is differentiable at z 0 ∈ Ω if and only if there exists c ∈ C such that
R(z) |z − z 0 |
→ 0 as z → z 0 ,
where R(z) = f (z) − f (z 0 ) − c(z − z 0 ). Writing
z = x + iy, z 0 = x 0 + iy 0 , c = a + ib,
we have
R(z) = f (z) − f (z 0 ) − c(z − z 0 ) = [u(z) − u(z 0 )] + i[v(z) − v(z 0 )] − (a + ib)[(x − x 0 ) + i(y − y 0 )] = [u(z) − u(z 0 ) − a(x − x 0 ) + b(y − y 0 )] +i[v(z) − v(z 0 ) − b(x − x 0 ) − a(y − y 0 )] = R 1 (z) + iR 2 (z),
Differentiation 11
where R 1 (z) = u(z) − u(z 0 ) − [a(x − x 0 ) − b(y − y 0 )], R 2 (z) = v(z) − v(z 0 ) − [b(x − x 0 ) + a(y − y 0 )].
Thus,
R(z) |z − z 0 | →^0 ⇐⇒^
R 1 (z) |z − z 0 | →^0 &^
R 2 (z) |z − z 0 | →^0
if and only if u and v are differentiable as a functions of two real variables at (x 0 , y 0 ), and
a = ∂u ∂x
(x 0 , y 0 ), −b = ∂u ∂y
(x 0 , y 0 ),
b = ∂v ∂x
(x 0 , y 0 ), a = ∂v ∂y
(x 0 , y 0 ),
i.e.,
∂u ∂x (x^0 , y^0 ) =^
∂v ∂y (x^0 , y^0 )^ &^
∂u ∂y (x^0 , y^0 ) =^ −^
∂v ∂x (x^0 , y^0 ),
and in that case
f ′(z 0 ) = a + ib =
∂u ∂x (x^0 , y^0 ) +^ i
∂v ∂x (x^0 , y^0 ) = ∂v∂y (x 0 , y 0 ) − i ∂u∂y (x 0 , y 0 ).
Thus, we have proved the following theorem.
Theorem 2.1.2 The function f is differentiable at z 0 ∈ Ω if and only if its real part u and imaginary part v are differentiable at (x 0 , y 0 ) and ux, uy, vx, vy satisfy the equations
ux(z 0 ) = vy(z 0 ), uy(z 0 ) = −vx(z 0 ), (∗)
and in that case
f ′(z 0 ) = ux(z 0 ) + ivx(z 0 ) = vy(z 0 ) − iuy(z 0 ). Equations in (∗) are called the Cauchy-Riemann equations, or in short CR-equations.
Now, recalling from a sufficient condition for differentiability of a real valued function of two variables, we have the following sufficient condition of differentiability of f at z 0 ∈ Ω.
Differentiation 13
Now, r^2 = x^2 + y^2 and tan θ = y/x so that
2 r ∂x∂r = 2x, 2 r ∂r∂y = 2y,
sec^2 θ ∂x∂θ = − (^) xy 2 , secθ^ ∂θ∂y =^1 x ,
i.e., ∂r ∂x =^
x r ,^
∂r ∂y =^
y r , x^2 + y^2 x^2
∂θ ∂x
= − y x^2
, x
(^2) + y 2 x^2
∂θ ∂y
x
i.e.,
∂r ∂x =^
x r ,^
∂r ∂y =^
y r ,^
∂θ ∂x =^ −^
y r^2 ,^
∂θ ∂y =^
x r^2.
Now,
∂u ∂x =^
∂u ∂r
∂r ∂x +^
∂u ∂θ
∂θ ∂x ,^
∂u ∂y =^
∂u ∂r
∂r ∂y +^
∂u ∂θ
∂θ ∂y.
Thus, ∂u ∂x =^
x r
∂u ∂r −^
y r^2
∂u ∂θ ,^
∂u ∂y =^
y r
∂u ∂r +^
x r^2
∂u ∂θ.^ (1) Similarly,
∂v ∂x
= x r
∂v ∂r
− y r^2
∂v ∂θ
, ∂v ∂y
= y r
∂v ∂r
∂v ∂θ
Recall that the CR-equations in Cartesian coordinates are
∂u ∂x
= ∂v ∂y
, ∂u ∂y
= − ∂v ∂x
Hence, (1) − (2) give
x r
∂u ∂r −^
y r^2
∂u ∂θ =^
y r
∂v ∂r +^
x r^2
∂v ∂θ ,^ (5) y r
∂u ∂r +^
x r^2
∂u ∂θ =^
x r
∂v ∂r −^
y r^2
∂v ∂θ.^ (6) The equations (3) − (4) imply that
r ∂u ∂r
= ∂v ∂θ
, ∂u ∂θ
= −r ∂v ∂r
These are the CR-equations in polar coordinates.
14 Analytic Functions
Definition 2.2.1 Let Ω ⊆ C.
(i) A function f : Ω → C is said to be analytic at a point z 0 ∈ Ω if there exists r > 0 such that B(z 0 , r) ⊆ Ω and f is differentiable at every point in B(z 0 , r).
(ii) A function f : Ω → C is said to be holomorphic or analytic on Ω 0 ⊆ Ω if f is analytic at every point in Ω 0.
♦
Definition 2.2.2 A complex valued function defined and analytic on the entire complex plane is called an entire function. ♦
Remark 2.2.1 In the subject of complex analysis, it is very common to say a function
f is analytic at a point z 0 ∈ C
to mean that f is defined in an open neigbourhood of z 0 and f is analytic at z 0. Usually, a function is given in terms of certain expression, and in that case, the domain of definition of f is taken to be the largest sub- set of C in which the expression makes sense. For example, consider the expression
f (z) =
z.