MATH135 Lecture Notes  Modulus Guitars, Parallelogram Law, Joule
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Published on 16 Oct 2011
Department
Mathematics
Course
MATH135
Professor
MATH 135 CLASSNOTE FOR MAKEUP CLASS ON NOVEMBER 16, 2010
WENTANG KUO
Chapter 8 Complex Numbers
We recall the Quadradic formula: if a, b, c ∈Rwhere a6= 0, then the solution of ax2+bx +c= 0
is
x=−b±√b2−4ac
2a.
The solutions are if b2−4ac ≥0. What if b2−4ac < 0?
8.2 Complex numbers
Deﬁnition
Let the symbol ibe a square root of −1, i.e.,i2=−1.
Acomplex number zis a number of the form z=x+iy where x, y ∈R(Cartesian form).
We call xthe real part of z, denoted by Re(z) and ythe imaginary part of z, denoted it by Im(z).
Note that x∈Rif and only if y= 0.
The set of all complex numbers is denoted by C.
The addition and multiplication on Care deﬁned as follows: let z=x+iy and w=u+iv. Then
z+w= (x+iy)+(u+iv) = (x+u) + i(y+v)
and
z·w= (x+iy)·(u+iv)
=xu +iyu +xiv +i2yv
= (xu −yv) + i(yu +xv).
Example Let z= 2 + iand w= 4 −i. Find Im(z+w) and Re(z2w).
We have
z+w= (2 + 4) + i(1 + (−1)) = 6 + 0 ·i= 6.
Thus, Im(z+w) = 0. Also,
z2w= (2 + i)2(4 −i) = (22+ 4i+i2)(4 −i) = (3 + 4i)(4 −i) = 16 + 13i.
Thus, Re(z2w) = 16.
Example Find z∈Csuch that z2= 1 + 2√2i.
Write z=x+iy with x, y ∈R. Then
z2= (x+iy)2=x2+ 2xyi +i2y2= (x2−y2)+2xyi = 1 + 2√2i.
Thus, we have x2−y2= 1 and 2xy = 2√2, i.e., xy =√2. Write y=√2
x. We have
x2−y2=x2−√2
x2=x2−2
x2= 1.
Date: November 15, 2010.
1
2 WENTANG KUO
Multiplying by x2, we get
x4−2 = x2,i.e., x4−x2−2=0,i.e., (x2−2)(x2+ 1) = 0.
Hence, x2= 2 or −1. Since x∈R,x2≥0. Hence, x2= 2, i.e., x=±√2. Also, y=√2/±√2 = ±1.
Hence, z=√2 + ior z=−√2−i.
8.3 The complex plane &8.4 Properties of complex numbers
The complex plane
A way to visualize Cand its operations.
Idea
Let xaxis be the real axis and yaxis the imaginary axis. We associate z=x+iy to the point
(x, y).

xreal
6
y
imaginary
sz=x+iy = (x, y)
1
Example Let z=x+iy and w=u+iv. Then z+w= (x+u) + i(y+v).

xu x +u
6
y
v
y+v
s
s
z
w
z+w
1
>
Addition can be visualized as the parallelogram law for adding vectors.
Given z=x+iy ∈C, we have
(x+iy)·(x−iy) = x2+y2∈R.
(Thus, by multiplying (x+iy) by (x−iy), we get back a real number)
Deﬁnition
The complex conjugate of z=x+iy is
¯z=x−iy.

x
6
y
−y
sz=x+iy
s¯z=x−iy
1
PPPPPPPP
Pq
MATH 135 CLASSNOTE FOR MAKEUP CLASS ON NOVEMBER 16, 2010 3
Conjugation is just the reﬂection with respect to the real axis.
Deﬁnition
The modulus of z=x+iy is
z=px2+y2.
Note that the distance from (0,0) to (x, y) is
px2+y2=z.

x
6
y
sz=x+iyr =px2+y2
1
Thus, zis the distance from 0 to z.
Remark (1) We have
z·¯z=z2.
(2) Note that if y= 0, i.e., z=x∈R, then
z=√x2=x,
i.e., for real numbers, modulus is just the absolute value.
Example Find the inverse of z=x+iy ∈C, where xand yare not both 0.
We have
z−1=1
z=1
x+iy =1
x+iy ·x−iy
x−iy =x−iy
x2+y2=¯z
z2.
Example Find z, w ∈Csuch that
z+w= 7 — (1)
2iz + 7w= 9 — (2).
From (1), we have z= 7 −w. Put it into (2). We get
2i(7 −w)+7w= 9,i.e., (7 −2i)w= 9 −14i.
Thus,
w=9−14i
7−2i=9−14i
7−2i·7+2i
7+2i=63 −98i+ 18i−28i2
72+ 22=(63 + 28) −80i
72+ 22=91 −80i
53 .
Also,
z= 7 −w= 7 −91 −80i
53 =371 −(91 −80i)
53 =280 + 80i
53 .
Properties of Conjugation 8.42
If z, w ∈C, we have
(1) z+w= ¯z+ ¯w.
(2) z·w= ¯z·¯w.
(3) ¯
¯z=z.
(4) z·¯z=z2.