# complex 4-5 module 2 ```MATH F112 (MATHEMATICS-II)
Gaurav Dwivedi
Department of Mathematics
BITS Pilani, Pilani Campus.
Module-9(Functions of Complex Variable)
Function of a Complex Variable
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
3 / 22
Function of a Complex Variable
Function of a complex variable
Let S be a set of complex numbers. Then function f
defined on S is a rule that assigns to each z ∈ S, a
complex number w, and we write
f (z) = w.
The set S is called domain of definition of f .
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
4 / 22
Suppose that w = u + iv is the value of the function f
at z = x + iy, so that u + iv = f (x + iy). Thus each of
real number u and v depends on the real variables x and
y i.e.,
f (z) = u(x, y) + iv(x, y).
Similarly, in the polar form
f (z) = u(r, θ) + iv(r, θ).
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
5 / 22
Examples
f (z) = z 2 = (x2 − y 2 ) + i(2xy). Here
u(x, y) = x2 − y 2 and v(x, y) = 2xy.
f (z) = ez = ex (cos y + i sin y). Here
u(x, y) = ex cos y and v(x, y) = ex sin y.
In polar form f (z) = z 2 gives
f (z) = (reiθ )2 = r2 e2iθ = r2 (cos 2θ + i sin 2θ).
Here u(r, θ) = r2 cos 2θ and v(r, θ) = r2 sin 2θ.
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
6 / 22
Polynomial function:
f (z) = a0 + a1 z + a2 z 2 + &middot; &middot; &middot; + an z n ,
where n is zero or a positive integer and
a0 , a1 , . . . , an are complex constants and an 6= 0.
The domain of definition is the entire z-plane.
Rational function: the quotients P (z)/Q(z) of
polynomials. The domain of definition is
{z ∈ C : Q(z) 6= 0}.
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
7 / 22
Limit
Limit
Let f be a function defined at all points of z in some
deleted nbd of z0 , then
lim f (z) = w0 ,
z→z0
if given ε &gt; 0, there exists a δ &gt; 0 such that
|f (z) − w0 | &lt; whenever 0 &lt; |z − z0 | &lt; δ.
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
8 / 22
That means the point w = f (z) can be made arbitrarily
close to w0 if we choose the point z close enough to z0
but distinct from it.
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
9 / 22
Exercise Show that if f (z) =
|z| &lt; 1, then
iz̄
in the open disk
2
i
lim f (z) = .
z→1
2
2
Exercise Show that lim zz does not exist.
z→0
Sol. Use two paths test.
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
10 / 22
Some Theorems on Limits
Theorem 1
Let f (z) = u(x, y) + iv(x, y),
z0 = x0 + iy0 ,
(z = x + iy), and
w0 = u0 + iv0 ,
then
lim f (z) = w0 ,
z→z0
if and only if
lim
(x,y)→(x0 ,y0 )
u(x, y) = u0
Gaurav Dwivedi (BITS Pilani)
and
lim
(x,y)→(x0 ,y0 )
MATH F112 (MATHEMATICS-II)
v(x, y) = v0 .
March 14, 2020
11 / 22
Some Theorems on Limits
Theorem 2
If lim f (z) and lim g(z) both exist, then
z→z0
z→z0
lim [f (z) &plusmn; g(z)] = lim f (z) &plusmn; lim g(z).
z→z0
z→z0
z→z0
lim [f (z)g(z)] = lim f (z) lim g(z).
z→z0
lim
z→z0
z→z0
h
f (z)
g(z)
i
Gaurav Dwivedi (BITS Pilani)
z→z0
lim f (z)
=
z→z0
lim g(z) ,
provided lim g(z) 6= 0.
z→z0
MATH F112 (MATHEMATICS-II)
z→z0
March 14, 2020
12 / 22
Stereographic Projection and
the Riemann Sphere
Let us think of the complex plane as passing through the
equator of the unit sphere S centred at origin. To each
point z in the plane there corresponds exactly one point
P on the surface of the sphere. The point P is the point
where the line through z and the north pole N intersects
the sphere. In like manner, to each point P on the
surface of the sphere, other than the north pole N, there
corresponds exactly one point z in the plane. By letting
the point N of the sphere correspond to the point at
infinity, we obtain one to one correspondence between S
and the extended complex plane.
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
13 / 22
Stereographic Projection and
the Riemann Sphere
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
14 / 22
Point at Infinity
Point at Infinity
The point at infinity is denoted by ∞, and the complex
plane together with ∞ is called the extended complex
plane.
For each small &gt; 0, the set
1
,
S = z : |z| &gt;
is called an neighborhood of ∞.
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
15 / 22
Definition of lim f (z) = f0
z→∞
Let f (z) be a complex function of the complex variable
z, and f0 be a complex constant. If for every real
number ε, there exists a real number r such that
|f (z) − f0 | &lt; ε for every |z| &gt; r, then we say that
lim f (z) = f0 .
z→∞
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
16 / 22
Definition of lim f (z) = ∞
z→z0
Let f (z) be a complex function of the complex variable
z, and z0 be a complex constant. If for every real
number ε, there exists a δ &gt; 0 such that |f (z)| &gt; ε for
every 0 &lt; |z − z0 | &lt; δ, then we say that
lim f (z) = ∞.
z→z0
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
17 / 22
Theorem Related to ∞
1
f
(z)
z→z0
lim f (z) = ∞ ⇔ lim
z→z0
1
z
lim f (z) = w0 ⇔ lim f
z→∞
z→0
lim f (z) = ∞ ⇔ lim f
z→∞
z→0
1
( z1 )
= 0.
= w0 .
= 0.
Proof. Proofs are exercise.
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
18 / 22
Continuity
Continuous Function
A function f (z) is said to be continuous at a point z0 if
f (z0 ) is defined.
lim f (z) exists.
z→z0
lim f (z) = f (z0 ).
z→z0
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
19 / 22
Continuity
Continuous Function
In terms of ε, δ notation, a function f (z) is said to be
continuous at a point z0 if for a given &gt; 0, there is a
δ &gt; 0, such that
|f (z) − f (z0 )| &lt; whenever |z − z0 | &lt; δ.
A function f (z) is said to be continuous in a region R if
it is continuous at all the points of the region R.
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
20 / 22
Some Results on Continuity
Theorem 1.
A composition of continuous functions is itself
continuous.
Theorem 2.
f (z) = u(x, y) + iv(x, y) is continuous iff both u(x, y)
and v(x, y) are continuous.
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
21 / 22
Some Results on Continuity
Theorem 3.
If f (z) and g(z) are continuous, then
f (z) &plusmn; g(z),
f (z)g(z), and
f (z)
g(z) ,
g(z) 6= 0,
all are continuous.
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
22 / 22
Some Results on Continuity
Theorem 4.
Let f (z) is continuous at z0 and f (z0 ) 6= 0. Then
f (z) 6= 0 throughout in some nbd of z0 .
Proof. Since f (z) is continuous at z0 , therefore for each
&gt; 0, there is a δ &gt; 0 such that
|f (z) − f (z0 )| &lt; whenever |z − z0 | &lt; δ.
(1)
Equation (??) is valid for all , so in particular it holds
for = |f (z2 0 )| .
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
23 / 22
Some Results on Continuity
Then there exists δ &gt; 0 such that when |z − z0 | &lt; δ
|f (z) − f (z0 )| &lt;
|f (z0 )|
.
2
If f (z) = 0 in some nbd of z0 , then |f (z0 )| &lt; |f (z2 0 )| ,
which is a contradiction. Therefore, we must have
f (z) 6= 0 for all z in Nδ (z0 ).
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
24 / 22
Some Results on Continuity
Theorem 5.
Every continuous function in a closed and bounded
region R is bounded i.e.,
|f (z)| ≤ M for all z ∈ R.
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
25 / 22
Examples
Discuss the continuity of f (z) =
Sol. We have f (0) = 0. Now
lim f (z) =
z→0
Re z
1+|z|
at z = 0.
x
p
= 0 = f (0).
(x,y)→(0,0) 1 +
x2 + y 2
lim
Therefore f (z) is continuous at 0.
Discuss the continuity of f (z) =
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
Re z
z
at z = 0.
March 14, 2020
26 / 22
Derivatives
Differentiable Function
Let f be a function defined on S containing Nρ (z0 ). If
f (z) − f (z0 )
,
z→z0
z − z0
lim
or equivalently
f (z0 + ∆z) − f (z0 )
,
∆z→0
∆z
lim
exists then we say f is differentiable at z0 and the value
of the limit denoted by f 0 (z0 ) is the derivative of f at z0 .
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
27 / 22
Thus
f (z) − f (z0 )
,
z→z0
z − z0
f 0 (z0 ) = lim
or
f (z0 + ∆z) − f (z0 )
.
∆z→0
∆z
f 0 (z0 ) = lim
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
28 / 22
Example
If f (z) = z 2 + 2, then show that f 0 (z0 ) = 2z0 .
Sol. We have
f (z) − f (z0 )
(z 2 + 2) − (z02 + 2)
= lim
z→z0
z→z0
z − z0
z − z0
= lim (z + z0 )
lim
z→z0
= 2z0 .
Thus f 0 (z0 ) = 2z0 .
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
29 / 22
Differentiability implies continuity
If f is differentiable at z0 then f is continuous at z0 .
Proof.
f (z) − f (z0 )
.(z − z0 )
lim [f (z) − f (z0 )] = lim
z→z0
z→z0
z − z0
f (z) − f (z0 )
= lim
. lim (z − z0 )
z→z0
z→z0
z − z0
= f 0 (z0 ).0 = 0.
Thus limz→z0 f (z) = f (z0 )(Why?).
Hence f is continuous at z0 .
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
30 / 22
Remark
The converse of above result is not true i.e., a continuous
function may not be differentiable.
Example
The function f (z) = |z|2 is continuous everywhere but
differentiable only at 0.
Sol. We have f (z) = |z|2 = x2 + y 2 = u(x, y) + iv(x, y).
Since both u(x, y) = x2 + y 2 and v(x, y) = 0 are
continuous everywhere, therefore f (z) is continuous
everywhere. Now, for z 6= z0 , we have
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
31 / 22
Example
f (z) − f (z0 ) |z|2 − |z0 |2
=
z − z0
z − z0
z z̄ − z0 z0
=
z − z0
z z̄ − z̄z0 + z̄z0 − z0 z0
=
z − z0
z̄(z − z0 ) + z0 (z̄ − z0 )
=
z − z0
∆z
= z̄ + z0
, ∆z = z − z0
∆z
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
32 / 22
f (z) − f (z0 )
∆x − i∆y
= z + z0
z − z0
∆x + i∆y
(
z + z0 when ∆z = (∆x, 0)
=
z − z0 when ∆z = (0, ∆y)
f (z) − f (z0 )
exists, then, z + z0 = z − z0 ,
z→z0
z − z0
(z0 )
i.e., z0 = 0. Thus, if z0 6= 0, limz→z0 f (z)−f
does not
z−z0
exist and so f is not differentiable at z0 6= 0.
Thus if lim
Remark: Note that, this does not prove that f (z) is
differentiable at z0 = 0.
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
33 / 22
Now for z0 = 0:
|z|2
f (z) − f (0)
= lim
z→0 z
z→0
z−0
= lim z̄
lim
z→0
= 0.
Thus f is differentiable at 0 and f 0 (0) = 0.
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
34 / 22
Differentiation Rules
d
dz (c)
= 0 for a constant c.
d
n
dz (z )
= nz n−1 .
[f (z) &plusmn; g(z)]0 = f 0 (z) &plusmn; g 0 (z).
[f (z)g(z)]0 = f (z)g 0 (z) + f 0 (z)g(z).
h
f (z)
g(z)
i0
=
g(z)f 0 (z)−f (z)g 0 (z)
,
(g(z))2
Gaurav Dwivedi (BITS Pilani)
g(z) 6= 0.
MATH F112 (MATHEMATICS-II)
March 14, 2020
35 / 22
Chain Rule for Differentiation
Let F (z) = g(f (z)), and assume that f is differentiable
at z0 and g is differentiable at f (z0 ), then F (z) is
differentiable at z0 and
F 0 (z0 ) = g 0 (f (z0 ))f 0 (z0 ).
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
36 / 22
Q. 9, Page - 63
Let f (z) be a function defined by
( 2
(z̄)
z , if z 6= 0,
f (z) =
0,
if z = 0.
Show that f 0 (0) does not exist.
Sol. We have
2
(z̄)
−0
f (z) − f (0)
= lim z
lim
z→0
z→0 z − 0
z−0
2
z̄ 2
x − iy
= lim
= lim
.
z→0 z
(x,y)→(0,0) x + iy
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
37 / 22
Now, along x-axis
lim
(x,y)→(0,0)
x − iy
x + iy
2
= 1,
while along the line y = x
2
x − iy
lim
= −1.
(x,y)→(0,0) x + iy
(0)
Hence limz→0 f (z)−f
does not exist and hence f 0 (0)
z−0
does not exist.
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
38 / 22
Homework If f (z) = z̄, show that f 0 (z) does not exist
at any point z.
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
39 / 22
Cauchy Riemann Equations (C-R
Equations)
Cauchy Riemann Equations (C-R Equations)
Let f (z) = u(x, y) + iv(x, y), then the equations
ux = vy
and uy = −vx ,
are called Cauchy Riemann Equations (C-R Equations).
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
40 / 22
Necessary Condition for
Differentiability
Necessary Condition for Differentiability
Suppose f (z) = u(x, y) + iv(x, y) is differentiable at a
point z0 = x0 + iy0 . Then the first order partial
derivatives of u and v exist and satisfy C-R equations at
(x0 , y0 ).
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
41 / 22
Proof
Since f (z) = u(x, y) + iv(x, y) is differentiable at
z0 = x0 + iy0 , therefore
f (z0 + 4z) − f (z0 )
∆z→0
4z
u(x0 + 4x, y0 + 4y) − u(x0 , y0 )
=
lim
4x + i4y
(∆x,∆y)→(0,0)
v(x0 + 4x, y0 + 4y) − v(x0 , y0 )
+i
.
4x + i4y
f 0 (z0 ) = lim
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
42 / 22
u(x0 + 4x, y0 + 4y) − u(x0 , y0 )
4x + i4y
(∆x,∆y)→(0,0)
v(x0 + 4x, y0 + 4y) − v(x0 , y0 )
.
+i
lim
4x + i4y
(∆x,∆y)→(0,0)
=
lim
Now along the path where ∆y = 0, it gives
f 0 (z0 ) = ux (x0 , y0 ) + ivx (x0 , y0 ),
and along the path where ∆x = 0, it gives
f 0 (z0 ) = −iuy (x0 , y0 ) + vy (x0 , y0 ).
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
43 / 22
This shows the existence of the first order partial
derivatives of u and v.
Now since f is differentiable at z0 therefore f 0 (z0 ) is
unique (limits along every path unique) and so, on
comparing:
ux (x0 , y0 ) = vy (x0 , y0 ),
Gaurav Dwivedi (BITS Pilani)
uy (x0 , y0 ) = −vx (x0 , y0 ).
MATH F112 (MATHEMATICS-II)
March 14, 2020
44 / 22
Sufficient Condition
Sufficient Condition for Differentiability
Let f (z) = u(x, y) + iv(x, y) be any function defined
throughout in N (z0 ) such that
1
The first order partial derivatives of u and v
(ux , uy , vx , vy ) exist in N (z0 ),
The first order partial derivatives of u and v are
continuous and satisfy C-R equations at (x0 , y0 ).
Then f is differentiable at z0 and
2
f 0 (z0 ) = ux (x0 , y0 ) + ivx (x0 , y0 ).
Gaurav Dwivedi (BITS Pilani)
MATH F112 (MATHEMATICS-II)
March 14, 2020
45 / 22
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