Functions, Limit and Continuity

Lecture 3 Functions, Limit and Continuity

Functions of a complex variable

Let S ⊆ C. A complex valued function f on S is a function f : S → C.

We write w = f (z). The set S is called the domain of f and the set{f (z) : z ∈ S} is called range of f .

Suppose z = x + iy , that f (z) = w = u + iv . Then

f (z) = f (x + iy) = u(x , y) + iv(x , y),

i.e., u and v are real valued functions of two real variables.

Ex. If w = z2, then

u(x , y) + iv(x , y) = (x + iy)2 = (x2 − y2) + i · 2xy ,

i.e., u(x , y) = x2 − y2, v(x , y) = 2xy .

(In polar form): Suppose z = re iθ and f (z) = w = u + iv . We can write

f (z) = f (re iθ) = u(r , θ) + iv(r , θ).

Ex. For w = z2, u(r , θ) + iv(r , θ) = z2 = r2e i2θ so that

u(r , θ) = r2 cos 2θ, v(r , θ) = r

2 sin 2θ.

Lecture 3 Functions, Limit and Continuity

Visualizing a complex function

A real valued function of a real variable is visualized with its graph.However, graph of a complex function is not a curve.

Example: Consider w = f (z) = z , defined on C. Image of each point isthe reflection about the real axis. What is the image of the set{z : |z − i | ≤ 2}?

For visualizing a complex function w = f (z) we often need two planes.

Take xy-plane as z-plane, and the domain is on this plane.Take uv-plane as w-plane, and the codomain is on this plane.

w = f (x) is visualized by the images of of sets and curves under themapping.

Example: Consider w = f (z) = z2, defined on C.

What is the image of a point z? Use z = re iθ.image of the set {z = e iθ : 0 ≤ θ ≤ π/2}?of the set {z = re iθ : 0 ≤ θ ≤ π/2}? If r < 1? r > 1?of the set {z = re iθ : 0 < r ≤ r0, 0 ≤ θ ≤ π/2}?of the set {z = re iθ : 0 < r ≤ r0, 0 ≤ θ ≤ π}?

Lecture 3 Functions, Limit and Continuity

Visualizing a complex function

Example: Again consider w = f (z) = z2, defined on C. Note that

w = u(x , y) + iv(x , y) = (x + iy)2 = (x2 − y2) + i · 2xy ,

i.e., u(x , y) = x2 − y

2, v(x , y) = 2xy .

What is the image of the hyperbola x2 − y2 = c1, c1 > 0?

O x

y

O u

v

u = c1 > 0

v = c2 > 0

It is a vertical line in w -plane given by u = c1.Note: v = ±2y

√

y2 + c1, y ∈ R.Each branch of the hyperbola maps the line in one-one manner.As you travel upward (downward) on the branch right (left) toy -axis, you travel upward on the image.

Lecture 3 Functions, Limit and Continuity

Visualizing a complex function

Example (contd.): For w = f (z) = z2,

u(x , y) = x2 − y

2, v(x , y) = 2xy .

What is the image of the hyperbola 2xy = c2, x > 0, c2 > 0?

O x

y

O u

v

u = c1 > 0

v = c2 > 0

It is the horizontal line v = c2.

Note: u = x2 −c22x2, limx→0+ u = −∞, limx→∞ u = ∞.

The hyperbola mapped to the line in one-one manner. Inwhich orientation?

What is the image of {z = x + iy : 0 < x2 − y2 ≤ c}?

What is the image of {z = x + iy : 0 < 2xy ≤ c, x > 0}?

Lecture 3 Functions, Limit and Continuity

Limit of a function

Limit of a function: Suppose f is a complex valued function defined ona deleted neighborhood of z0. We say f has a limit a as z → z0 if forevery ε > 0, there is a δ > 0 such that

|f (z)− a| < ε whenever 0 < |z − z0| < δ.

We then writelimz→z0

f (z) = a.

Example: limz→i

2i

z= 2.

Take any ε and draw diagram and see:

w = f (z) =2i

zis defined in a deleted neighborhood of i ,

|f (z) − 2| =2|z − i |

|z |< ε if |z − i | <

ε

4and |z | > 1

2 ,

If δ = min{ε

4, 12}, then |f (z)− 2| < ε when |z − i | < δ.

The limit of a function f (z) at a point z0, if exists, is unique.

Lecture 3 Functions, Limit and Continuity

Limit of a function

If f (z) = u(x , y) + iv(x , y) and z0 = x0 + iy0, pause then

limz→z0

f (z) = u0 + iv0 ⇐⇒

lim(x,y)→(x0,y0)

u(x , y) = u0 and

lim(x,y)→(x0,y0)

v(x , y) = v0.

Ex. limz→z0

z2 = z

20 .

The point z0 can be approached from any direction. If the limitlimz→z0

f (z) exists, then f (z) must approach a unique limit, no matter how

z approaches z0.

If the limit limz→z0

f (z) is different for different path of approaches then

limz→z0

f (z) does not exists.

Ex. limz→0

z

zdoes not exist.

Take z = (x , 0) → 0 and z = (0, y) → 0 separately and see.

Lecture 3 Functions, Limit and Continuity

Limit contd....

Let f , g be complex valued functions with limz→z0

f (z) = α and limz→z0

g(z) = β.

Then,

limz→z0

[f (z)± g(z)] = limz→z0

f (z)± limz→z0

g(z) = α± β.

limz→z0

[f (z) · g(z)] = limz→z0

f (z) · limz→z0

g(z) = αβ.

limz→z0

f (z)

g(z)=

limz→z0

f (z)

limz→z0

g(z)=

α

β(if β 6= 0).

limz→z0

Kf (x) = K limz→z0

f (z) = Kα ∀ K ∈ C.

Lecture 3 Functions, Limit and Continuity

Continuous functions

Continuity at a point: Let D be a domain or a region. A functionf : D → C is continuous at a point z0 ∈ D if for for every ε > 0, there isa δ > 0 such that

|f (z)− f (z0)| < ε whenever |z − z0| < δ.

In other words, f is is continuous at a point z0 in the domain if the

following conditions are satisfied.

limz→z0

f (z) exists,

limz→z0

f (z) = f (z0).

A function f is continuous on D if it is continuous at each and everypoint in D .

A function f : D → C is continuous at a point z0 ∈ D if and only ifu(x , y) = Re (f (z)) and v(x , y) = Im (f (z)) are continuous at z0.

Lecture 3 Functions, Limit and Continuity

Continuity

Let f , g : D ⊆ C → C be continuous functions at the point z0 ∈ D . Then

f ± g , fg , Kf (k ∈ C),f

g(g(z0) 6= 0) are continuous at z0.

Composition of continuous functions is continuous.

f (z), |f (z)|,Re (f (z)) and Im (f (z)) are continuous.

If a function f (z) is continuous and nonzero at a point z0, then there is aε > 0 such that f (z) 6= 0, ∀ z ∈ B(z0, ε).

Continuous image of a compact set (closed and bounded set) is compact.

Lecture 3 Functions, Limit and Continuity