Chapter 13. Complex Numbers and Function & Complex Differentiation
2018학년도 2학기
반도체 공학과 공학수학2 (GEDB005) 강의교안 (2018학년도용)
주교재: Erwin Kreyszig, Engineering Mathematics, 10th Edition
부교재: 최신공학수학 I 과 II, 한빛출판사 및 (영문) 다변량 Calculus (이상구, 김응기 et al) (http://www.hanbit.co.kr/EM/sage/)
강의시간: E12 (금 12:00 - 14:45), 반도체관 400126호 담당교수: 김응기 박사
주차 |
주교재 |
부교재 |
9 |
13.1: Complex Numbers. Complex Plane 13.2: Polar Form of Complex Numbers. Powers and Roots 13.3: Derivative. Analytic Function |
13.1 복소수 13.2 복소함수 |
web |
Week 9
Chapter 13. Complex Numbers and Function
Complex Differentiation
13.1 Complex Numbers and Their Geometric Representation
A complex number is an ordered pair of and .
.
is the real part of
is the imaginary part of
Two complex numbers and are equal.
(1) is the imaginary unit.
Addition, Multiplication. Notation
Addition of two complex numbers and is
(2) .
Multiplication of two complex numbers and is
(3) .
For ,
and .
The complex numbers “extend” the real numbers.
We can thus write
. Similarly
because by and the definition of multiplication we have
.
Complex numbers are
(4) .
Electrical engineers often write instead if because they need for the current.
is pure imaginary.
Also, and give
(5) . the definition of multiplication.
Addition of two complex numbers and is
.
Multiplication of two complex numbers and is
.
Example 1 Real Part, Imaginary, Sum and Product of Complex Numbers
Let and .
, , , .
,
.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
8
3
9
-2
I + 17
11*I + 78
Subtraction, Division
Subtraction is defined as the inverse operations of addition.
Division is defined as the inverse operations of multiplication..
Difference of two complex numbers and is
(6) .
Quotient of two complex numbers and is
(7)
By multiplying numerator and denominator of by
Example 2 Difference and Quotient of Complex Numbers
and .
and
.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
5*I - 1
43/85*I + 66/85
Complex Plane
Two perpendicular coordinate axes,
The horizontal axis is the real axis.
The vertical axis is the imaginary axis.
The real axis and imaginary axis have the same unit of length.
This is a Cartesian coordinate system.
A given complex number as the point with coordinate .
The -plane in which the complex numbers are represented is the complex plane.
The complex plane
The point represented by in the complex plane The point in the complex plane
Visualization of addition and subtraction
Addition of complex numbers Subtraction of complex numbers
Complex Conjugate Numbers
is the complex conjugate of a complex number .
Obtained by reflecting the point in the real axis.
Ex.
Fig. Complex conjugate numbers
Complex conjugate is important because it permits us to switch complex to real.
, .
,
. real part of
. imaginary part of
is real, , by the definition of .
(9) .
Example 3 Illustration of and
Let and . Then by ,
.
The multiplication formula in is verified by
,
.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
4
3
-3*I + 4
-5*I + 2
-26*I - 7
-26*I - 7
True
13.2 Polar From of Complex Numbers. Powers and Roots
The polar coordinates of a point are related to the rectangular coordinates by the equations
To convert from polar coordinates to rectangular coordinates ,
(1) , .
Polar form of is
(2) .
is the absolute value or modulus of .
Hence
(3) .
Complex plane, polar form Distance between two
of a complex number points in the complex plane
is the distance of the point from the origin.
is the distance between and .
is the argument of .
(4) .
is the directed angle from the positive -axis to .
All angles are measured in radians and positive in the counterclockwise sense.
Define the principal value of by the double inequality
(5) .
Positive real
Negative real
For a given , the other values of are ().
Example 1 Polar Form of Complex Numbers. Principal Value
.
, ,
(the principle value).
, and .
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
sqrt(2)
1/4*pi
sqrt(2)*e^(1/4*I*pi)
I + 1
Triangle Inequality
The triangle inequality of the complex
(6) .
Fig. 326 Triangle inequality
Generalized triangle inequality
(6*)
Absolute value of a sum cannot exceed the sum of the absolute values of the terms.
Example 2 Triangle Inequality
If and then .
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
True
Multiplication and Division in Polar Form
Let and .
Multiplication
,
Absolute value of a product equals the product of the absolute values of the factors,
.
Argument of a product equals the sum of the arguments of the factors,
(9) (up to multiples of )
Division
(12)
Absolute value of a division equals the product of the absolute values of the factors,
(10) .
Argument of a division equals the difference of the arguments of the factors,
(11) (up to multiples of )
Example 3 Illustration of Formulas
Let and .
,
.
Then ,
, .
For the arguments we obtain
, ,
.
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
z1*z2 = -6*I - 6
z1/z2 = 2/3*I + 2/3
|z1| = 2*sqrt(2)
|z2| = 3
|z1*z2| = 6*sqrt(2)
|z1/z2| = 2/3*sqrt(2)
arg(z1) = 3/4*pi
arg(z2) = 1/2*pi
arg(z1*z2) = -3/4*pi
arg(z1/z2) = 1/4*pi
Example 4 Integer Powers of . De Moivre’s Formula
Form and with we obtain by induction for
(13) .
Similarly, with and gives for .
For , formula becomes De Moivre’s Formula
(13*) .
We can use this to express and in terms of and .
For ,
, .
.
Roots
Each value of there corresponds one value of .
(14) an root of
Hence this symbol is multivalued, namely, -valued.
The values of can be obtained as follows. We write and in polar from
and .
Then the equation becomes, by De Moivre’s formula (with instead of )
.
where is positive real.
where is an integer.
For we get distinct values of .
Further integers of would give values already obtained.
For instance, gives , hence the corresponding to , etc.
Consequently, , for , has the distinct values
(15) , where .
These values lie on a circle of radius with center at the origin and constitute the vertices of a regular polygon of sides.
The values of obtained by taking the principal value of and in is called the principal value of .
Taking in , we have and . Then gives
(16) , .
These values are called the th roots of unity. They lie on the circle of radius and center , called the unit circle.
Show that .
We write and in polar from
and .
,
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
[w == 1/2*I*sqrt(3) - 1/2, w == -1/2*I*sqrt(3) - 1/2, w == 1]
Evaluate
1
1/2*I*sqrt(3) - 1/2
-1/2*I*sqrt(3) - 1/2
Show that .
We write and in polar from
and .
,
Show that .
We write and in polar from
and .
,
,
Hence
.
If denotes the value corresponding to in , then the values of can be written
.
More generally, if is any root of an arbitrary complex number , then the values of in are
(17) ,
because multiplying by corresponds to increasing the argument of , by .
Formula motivates the introduction of roots of unity and shows their usefulness.
Example
Calculate the square root of 5 of .
Solution
Taking , the square root of of is
.
Hence
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
[x == (-sqrt(3) + I)^(1/5)*e^(2/5*I*pi), x == (-sqrt(3) + I)^(1/5)*e^(4/5*I*pi), x == (-sqrt(3) + I)^(1/5)*e^(-4/5*I*pi), x == (-sqrt(3) + I)^(1/5)*e^(-2/5*I*pi), x == (-sqrt(3) + I)^(1/5)]
Evaluate
1/2*(sqrt(3) + I)*2^(1/5)
(I*sin(17/30*pi) + cos(17/30*pi))*2^(1/5)
(I*sin(29/30*pi) + cos(29/30*pi))*2^(1/5)
(I*sin(41/30*pi) + cos(41/30*pi))*2^(1/5)
(I*sin(53/30*pi) + cos(53/30*pi))*2^(1/5)
Example
Prove that
.
Solution
13.3. Derivative. Analytic Function
Unit Circle :
General circle of radius and center
.
The set of all whose distance from the center equals .
is interior (“open circular disk”)
is interior plus the circle itself (“closed circular disk”)
is exterior.
Open circular disk is a neighborhood of .
is a -neighborhood of .
Any set containing -neighborhood of is a neighborhood of .
Closed circular disk :
Exterior : .
Neighborhood of : An open circular disk
Open annulus (circular ring) : .
This is the set of all whose distance from a is greater than but less than .
Closed annulus : .
Open circular disk is a neighborhood of or, -neighborhood of .
Any set containing -neighborhood of is a neighborhood of .
Half-Planes
The set of all points .
Upper half-plane :
Lower half-plane :
Right half-plane :
Left half-plane :
Point set
Collection of finitely many of infinitely many points.
Examples : the solution of a quadratic equation, the points of a line, the [point in the interior of a circle.
Set is open
Every point of has a neighborhood consisting entirely of points that belong to .
Example, the points in the interior of a circle or a square from an open set, and so do the points of the right half-plane .
Set is connected
Any two of its points can be joined by a broken line of points that belong to .
Domain
Open connected set is a domain.
Complement of a set in the complex plane
Set of all points of the complex plane that do not belong to .
is closed
Its complement is open.
Example, the points on and inside the unit circle from a closed set (“closed unit disk”) since its complement is open.
Boundary point of a set
A point every neighborhood of which contains both points that belong to and points that do not belong to .
Example, the boundary points of an annulus are the points on the two bounding circles.
Clearly, if a set is open, then no boundary point belongs to if is closed, then every boundary point belongs to .
Boundary of
Set of all boundary points of a set .
Region
Set consisting of a domain plus some or all of its boundary points.
Complex function
is a set of complex numbers. And a function defined on is a rule that assigns to every a complex number (the value of at ).
.
Here, varies in and is a complex variable.
Set is the domain of definition of or, briefly, the domain of .
Set of all values of a function is the range of .
Example 1 Function of a Complex Variable
Let . Find and and calculate the value of at .
Solution
and .
and .
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
x^2 + 2*I*x*y - y^2 + 3*x + 3*I*y
15*I - 5
Example 2 Function of a Complex Variable
Let . Find and and calculate the value of at .
Solution
and .
.
and .
Sage Coding
http://math3.skku.ac.kr/ http://sage.skku.edu/ http://mathlab.knou.ac.kr:8080/
Evaluate
2*I*x - 2*y + 6*conjugate(x) - 6*I*conjugate(y)
-23*I - 5
Limit, Continuity
Limit
(1) .
have the limit as approaches a point
(2) .
Fig. Limit
Continuity
(3) and is defined.
Function is continuous at
Derivative
The derivative of a complex function at a point is written and is defined by
(4)
derivative of at a point
Then is differentiable at .
(4') . ,
Example 3 Differentiability, Derivative
Function is differentiable for all derivative
.
The differentiation rules
Analytic function and and constant ,
Chain rules and the power rule ().
is differentiable at is continuous at .
Example 4 not Differentiable
It may come as a surprise that there are many complex functions that do not have a derivative at any point.
For instance, is not differentiable.
To see this, we write and obtain
(5)
do not have a derivative at any point .
Fig. Paths in
Definition : Analyticity
A function is said to be analytic in a domain if is defined and differentiable at all points of .
The function is analytic at a point in if is analytic in a neighborhood of .
Also, by an analytic function we mean a function that is analytic in some domain.
A more modern term for analytic in is holomorphic in .
Example 5 Polynomials, Rational Functions
The nonnegative integer powers are analytic in the entire complex plane, Polynomials are
where are complex constants.
Quotient two polynomials and ,
is a rational function.
Rational function is analytic except at the points .
http://matrix.skku.ac.kr/sglee/
[한빛 아카데미] Engineering Mathematics with Sage:
[저자] 이상 구, 김영 록, 박준 현, 김응 기, 이재 화
Contents
A. 공학수학 1 – 선형대수, 상미분방정식+ Lab
Chapter 01 벡터와 선형대수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-1.html
Chapter 02 미분방정식의 이해 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-2.html
Chapter 03 1계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-3.html
Chapter 04 2계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-4.html
Chapter 05 고계 상미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-5.html
Chapter 06 연립미분방정식, 비선형미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-6.html
Chapter 07 상미분방정식의 급수해법, 특수함수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-7.html
Chapter 08 라플라스 변환 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-8.html
B. 공학수학 2 - 벡터미적분, 복소해석 + Lab
Chapter 09 벡터미분, 기울기, 발산, 회전 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-9.html
Chapter 10 벡터적분, 적분정리 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-10.html
Chapter 11 푸리에 급수, 적분 및 변환 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-11.html
Chapter 12 편미분방정식 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-12.html
Chapter 13 복소수와 복소함수, 복소미분 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-13.html
Chapter 14 복소적분 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-14.html
Chapter 15 급수, 유수 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-15.html
Chapter 16 등각사상 http://matrix.skku.ac.kr/EM-Sage/E-Math-Chapter-16.html
Copyright @ 2018 SKKU Matrix Lab. All rights reserved.
Made by Manager: Prof. Sang-Gu Lee and Dr. Jae Hwa Lee
*This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2017R1D1A1B03035865).