Math144Notes - Advanced Engineering Math II Math 144...

Advanced
Engineering Math II
Math 144
Lecture Notes
by
Stefan Waner
(First printing: 2003)
Department of Mathematics, Hofstra University

2
1. Algebra and Geometry of Complex Numbers (based on §§17.1–17.3 of Zill)
Definition 1.1 A complex number has the form z = (x, y), where x and y are real
numbers. x is referred to as the real part of z, and y is referred to as the imaginary part
of z . We write
Re (z) = x , Im (z) = y.
Denote the set of complex numbers by C I . Think of the set of real numbers as a subset of
C I by writing the real number x as (x, 0) . The complex number (0, 1) is called i.
Examples
3 = (3, 0), (0, 5), (-1, - π ), i = (0, 1).
Geometric Representation of a Complex Number- in class.
Definition 1.2 Addition and multiplication of complex numbers, and also multiplication
by reals are given by:
(x, y) + (x ' , y ' ) = ((x+x ' ), (y+y ' ))
(x, y)(x ' , y ' ) = ((xx ' -yy ' ), (xy ' +x ' y))
(x, y) = ( x, y).
Geometric Representation of Addition- in class. (Multiplication later)
Examples 1.3
(a) 3+4 = (3, 0)+(4, 0) = (7, 0) = 7 (b) 3 4 = (3, 0)(4, 0) = (12-0, 0) = (12,
0) = 12
(c) (0, y) = y(0, 1) = yi (which we also write as iy).
(d) In general, z = (x, y) = (x, 0) + (0, y) = x + iy. z = x + iy
(e) Also, i
2
= (0, 1)(0, 1) = (-1, 0) = -1. i
2
= -1
(g) 4 - 3i = (4, -3) .
Note In view of (d) above, from now on we shall write the complex number (x, y) as
x+iy .
Definitions 1.4 The complex conjugate , z , of the complex number z = x+iy given by
z = x - iy.
The magnitude , |z| of z = x+iy is given by
|z| = x
2
+y
2
.
Examples and Geometric Representation of Conjugation and Magnitude - in class.
Notes
1. z + z = (x+iy) + (x-iy) = 2x = 2 Re (z) . Therefore, Re (z) =
1
2
(z+z)

3
z - z = (x+iy) - (x-iy) = 2iy = 2i Im (z) . Therefore, Im (z) =
1
2i
(z-z)
2. Note that zz = (x+iy)(x-iy) = x
2
-i
2
y
2
= x
2
+y
2
= |z|
2
zz = |z|
2
3 . If z ≠ 0 , then z has a multiplicative inverse. Why? because:
z ·
z
|z| 2
=
zz
|z|
2
=
|z|
2
|z|
2
= 1. Hence, z
-1
=
z
|z| 2
Examples
(a)
1
i
= -i (b)
1
3+4i
=
3-4i
25
(c)
1
1
2
(1+i)
=
1
2
(1-i) (d)
1
cos ø + i sin ø
= cos (-ø) + i sin (-ø)
4. There is also the Triangle Inequality:
|z
1
+ z
2
| ≤ |z
1
| + |z
2
|.
Proof We square both sides and compare them. Write z
1
= x
1
+ iy
1
and z
2
= x
2
+ iy
2
.
Then
|z
1
+ z
2
|
2
= (x
1
+x
2
)
2
+ (y
1
+y
2
)
2
= x
1
2
+ x
2
2
+ 2x
1
x
2
+ y
1
2
+ y
2
2
+ 2y
1
y
2
.
On the other hand,
(|z
1
| + |z
2
|)
2
= |z
1
|
2
+ 2|z
1
||z
2
| + |z
2
|
2
= x
1
2
+ x
2
2
+ y
1
2
+ y
2
2
+ 2|z
1
||z
2
|.
Subtracting,
(|z
1
| + |z
2
|) 2 - |z
1
+ z
2
| 2 = 2|z
1
||z
2
| - 2(x
1
x
2
+ y
1
y
2
)
= 2
[
|(x
1
,y
1
)||(x
2
,y
2
)| - (x
1
,y
1
).(x
2
,y
2
)
]
(in vector form)
= 2
[
|(x
1
,y
1
)||(x
2
,y
2
)| - |(x
1
,y
1
)||(x
2
,y
2
)| cos
]
= 2 |(x
1
,y
1
)||(x
2
,y
2
)| (1 - cos )
≥ 0,
giving the result.
Note The triangle inequality can also be seen by drawing a picture of z
1
+ z
2
.
5. We now consider the polar form of these things: If z = x+iy, we can write x = r cos ø
and y = r sin ø, getting z = r cos ø + ir sin ø, so
This is called the polar form of z . It is important to draw pictures in order to feel
comfortable with the polar representation. Here r is the magnitude of z, r = |z| , and ø is
called the argument of z , denoted arg (z). To calculate ø , we can use the fact that tan ø =
y/x. Thus ø is not arctan (y/x) as claimed in the book, but by:
since the arctan function takes values between - π /2 and π /2. The principal value of
arg (z) is the unique choice of ø such that - π < ø ≤ π . We write this as Arg (z)
- π < Arg (z) ≤ π - π ≤ A

4
Examples
(a) Express z = 1+i in polar form, using the principal value
(b) Same for 3 + 3 3 i
(c) 6 = 6( cos 0 + i sin 0)
6.