SMA 305: COMPLEX ANALYSIS I
Purpose of the Course
To introduce the learner to complex functions, their differentiation and integration.
Expected Learning Outcomes of the Course
By the end of the course, the learner will be able to:
1. Use Cauchy –Reimann equations to determine whether a given complex-valued function is
analytic or not.
2. Apply key integral theorems to perform integration of complex functions.
3. Use the Laurent series expansion to classify singularities of a given complex function.
Course Content (Revised)
Limits, Continuity and differentiability of complex functions. Analytic and elementary complex
functions. Cauchy Riemann equations, Harmonic functions. Mappings by elementary functions.
Integration of complex functions: Cauchy’s integral Theorem, Cauchy-Goursat Theorem. Taylor
series, Laurent Series, Singularities, poles, residues and Cauchy’s residue theorem.
Pre-requisite: SMA 200
Course Assessment
Continuous assessment tests: 30%
Final examination: 70%
Core Reading Materials
1. Brown,J.W, etal. (2013). Complex Variables and Applications. 9th Edition. Mc Graw-Hill,
Singapore.
2. Gamelin, T.W. (2003). Complex Analysis. Springer, New York.
3. Spiegel, M.R etal (2009) . Schaum’s Outline of Complex Variables, 2ed. Mc Graw-Hill
Professional, India.
Recommended Reading Materials
1. A.S.A. Mshimba: Basic Complex Analysis Vol. I, University of Dar es salaam, 1984.
2. J E Marsden & M. Hoffman: Basic Complex Analysis, Freeman & Co., New York, 1972.
, FUNCTIONS, LIMITS AND CONTINUITY
Functions of a complex variable
Let S be a set of complex number. A function f defined on S is a rule which assigns to each z in
S a complex number w is called the value of f at z. The number w is called the value of f at z and
denoted by f z , that is w f z .
The set S is called the domain of definition of f. When the domain of definition is not
mentioned, we agree that the longest possible set is to be taken. Thus if we speak of the function
f z 1 , the domain is the set of all non-zero complex numbers.
z
Transformation Or Mapping
Suppose that w u iv is the value of the function f at z x iy i.e. u iv f x iy . Then
each of the real numbers u and v depends on the real variables x and y.
That is
u u x, y and v v x, y 1)
Therefore, f z u x, y iv x, y
Thus given a point P ( x, y ) in the z-plane, there corresponds a point P ' (u, v) in the w-plane.
The set of equations 1) is called a transformation or a mapping.
Example
1. If f z 4 5i z 6 2i , then
f x iy 4 5i x iy 6 2i
4 x 5 y 6 i 5x 4 y 2
Since f z u iv , then equating real and imaginary parts, we get
u 4 x 5 y 6 , v 5x 4 y 2
If particular, if P 1, 2 , then x 1 , y 2 and the coordinator of P ' are given by
u 4 10 6 12 and v 5 8 2 11
, P (1, 2)
z- plane w- plane
y
v
. P(1,2)
. P (-12,11)
1
x u
Exercise
Let f z z 2 , write f z in the form u iv .
Limits
Let f z be defined and single-valued in a neighborhood of z z0 with possible exception of
z z0 itself. We say that the number L is the limit of f z as z approaches z0 and write
lim f z L if for any positive number (however small) we can find some positive number
(usually depending on ) such that f z L whenever 0 z z0
In such a case we also say that f z approaches L as z approaches z0 and write f z L as
z z0 .
Example
iz i
Consider the function f z defined on the open disk z 1 . Show that lim f z .
2 Z 1 2
Solution
1. We must show that given 0 , we can find (depending on ) such that
i
f z whenever 0 z 1 .
2
Now
i iz i
f z
2 2 2
, i z 1
2
i z 1
2
z 1
2 2
Taking 2 , the required result follows
2. Show that lim 2 x iy 2 4i.
z 2i
Solution
We must show that given 0 there exist a 0 such that
f z 4i whenever 0 z 2i that is 0 x y 2 i
Now,
f z 4i 2 x iy 2 4i
2x i y2 4
2 x i y 2 y 2
If y 2 2 , then
f z 4i 2 x i y 2 2
Therefore choose and y 2 2
2
z2 1
3. Evaluate lim .
z i z6 1
Solution
z2 1 2 z 2i 1
lim lim 5 .
z i z 1 z i 6 z
6
6i 3
Continuity
A function f is continuous at a point z0 if all the following conditions are satisfied
1) lim f z exists
z z0
Purpose of the Course
To introduce the learner to complex functions, their differentiation and integration.
Expected Learning Outcomes of the Course
By the end of the course, the learner will be able to:
1. Use Cauchy –Reimann equations to determine whether a given complex-valued function is
analytic or not.
2. Apply key integral theorems to perform integration of complex functions.
3. Use the Laurent series expansion to classify singularities of a given complex function.
Course Content (Revised)
Limits, Continuity and differentiability of complex functions. Analytic and elementary complex
functions. Cauchy Riemann equations, Harmonic functions. Mappings by elementary functions.
Integration of complex functions: Cauchy’s integral Theorem, Cauchy-Goursat Theorem. Taylor
series, Laurent Series, Singularities, poles, residues and Cauchy’s residue theorem.
Pre-requisite: SMA 200
Course Assessment
Continuous assessment tests: 30%
Final examination: 70%
Core Reading Materials
1. Brown,J.W, etal. (2013). Complex Variables and Applications. 9th Edition. Mc Graw-Hill,
Singapore.
2. Gamelin, T.W. (2003). Complex Analysis. Springer, New York.
3. Spiegel, M.R etal (2009) . Schaum’s Outline of Complex Variables, 2ed. Mc Graw-Hill
Professional, India.
Recommended Reading Materials
1. A.S.A. Mshimba: Basic Complex Analysis Vol. I, University of Dar es salaam, 1984.
2. J E Marsden & M. Hoffman: Basic Complex Analysis, Freeman & Co., New York, 1972.
, FUNCTIONS, LIMITS AND CONTINUITY
Functions of a complex variable
Let S be a set of complex number. A function f defined on S is a rule which assigns to each z in
S a complex number w is called the value of f at z. The number w is called the value of f at z and
denoted by f z , that is w f z .
The set S is called the domain of definition of f. When the domain of definition is not
mentioned, we agree that the longest possible set is to be taken. Thus if we speak of the function
f z 1 , the domain is the set of all non-zero complex numbers.
z
Transformation Or Mapping
Suppose that w u iv is the value of the function f at z x iy i.e. u iv f x iy . Then
each of the real numbers u and v depends on the real variables x and y.
That is
u u x, y and v v x, y 1)
Therefore, f z u x, y iv x, y
Thus given a point P ( x, y ) in the z-plane, there corresponds a point P ' (u, v) in the w-plane.
The set of equations 1) is called a transformation or a mapping.
Example
1. If f z 4 5i z 6 2i , then
f x iy 4 5i x iy 6 2i
4 x 5 y 6 i 5x 4 y 2
Since f z u iv , then equating real and imaginary parts, we get
u 4 x 5 y 6 , v 5x 4 y 2
If particular, if P 1, 2 , then x 1 , y 2 and the coordinator of P ' are given by
u 4 10 6 12 and v 5 8 2 11
, P (1, 2)
z- plane w- plane
y
v
. P(1,2)
. P (-12,11)
1
x u
Exercise
Let f z z 2 , write f z in the form u iv .
Limits
Let f z be defined and single-valued in a neighborhood of z z0 with possible exception of
z z0 itself. We say that the number L is the limit of f z as z approaches z0 and write
lim f z L if for any positive number (however small) we can find some positive number
(usually depending on ) such that f z L whenever 0 z z0
In such a case we also say that f z approaches L as z approaches z0 and write f z L as
z z0 .
Example
iz i
Consider the function f z defined on the open disk z 1 . Show that lim f z .
2 Z 1 2
Solution
1. We must show that given 0 , we can find (depending on ) such that
i
f z whenever 0 z 1 .
2
Now
i iz i
f z
2 2 2
, i z 1
2
i z 1
2
z 1
2 2
Taking 2 , the required result follows
2. Show that lim 2 x iy 2 4i.
z 2i
Solution
We must show that given 0 there exist a 0 such that
f z 4i whenever 0 z 2i that is 0 x y 2 i
Now,
f z 4i 2 x iy 2 4i
2x i y2 4
2 x i y 2 y 2
If y 2 2 , then
f z 4i 2 x i y 2 2
Therefore choose and y 2 2
2
z2 1
3. Evaluate lim .
z i z6 1
Solution
z2 1 2 z 2i 1
lim lim 5 .
z i z 1 z i 6 z
6
6i 3
Continuity
A function f is continuous at a point z0 if all the following conditions are satisfied
1) lim f z exists
z z0
