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© 2005 Baylor UniversitySlide 1
Fundamentals of Engineering AnalysisEGR 1302 - Introduction to Complex Numbers, Standard Form
© 2005 Baylor UniversitySlide 2
0 5 1010
0
108.847
6.88
g x( )
100 x
Tout
T
Sinusoidal Response
Complex Numbers
Definitions and Formats Complex Numbers mathematically represent actual physical systems
Tin
SYSTEMTout
Feedback
Exponential Decay
© 2005 Baylor UniversitySlide 3
The General Quadratic Equation
02 cbxax
2
22
44)
2(
aacb
abx
aacb
abx
24
2
2
Take the Square Root
2
2
2
22
44 ab
ac
abx
abx
Complete the Square
acx
abx
2
aacbbx
242
The Solution to theGeneral Quadratic Equation
© 2005 Baylor UniversitySlide 4
Solutions of the Quadratic Equation
By solution, we mean“roots”, or where x=0
5 0 510
0
1010
10
f x( )
55 x
2nd Order
5 0 510
0
1010
10
f x( )
55 x
3rd Order
abacbx
241 2
acb 42 If there is one real roota
acbbx2
42
5 0 510
0
1010
10
f x( )
55 x
acb 42 If there are no real roots, as shown
acb 42 If there are two real roots, as above
© 2005 Baylor UniversitySlide 5
The Imaginary Number
0222 xxConsider:
12122 xxComplete the square:
11
1)1( 2
x
xTake the square root:
11 xThe solution:
1Because does not exist,we call this an “imaginary” number,and we give it the symbol “ ”or “ ”.ji
becomes ix 1
© 2005 Baylor UniversitySlide 6
Complex Numbers
Substitute into 0222 xxix 1
02)1(2)1)(1( iii
022221 2 iii
11*1* 2 iii
0)1(1 Checks!
aibacbx
24 2
A general solution for is24 bac
© 2005 Baylor UniversitySlide 7
z=x+iy
Complex Numbers
abaci
abx
24
2
2
Definitions
The “Standard Form” Im(z)=yRe(z)=x
z=x+iy and if y=0, then z=x, a real number
i3=-i i4=1i2=-1 i5=i i6=-1
z1=x1+iy1Given z2=x2+iy2and Then ifz1=z2
y1 = y2
x1 = x2
© 2005 Baylor UniversitySlide 8
Algebra of Complex Numbers
z1=x1+iy1Definitions: Given z2=x2+iy2
z1+z2=z3z3= (x1 + x2) + i(y1 + y2)
z1-z2=z3z3= (x1 - x2) + i(y1 - y2)
z1 * z2=z3 z3= (x1 + iy1)(x2 + iy2) = x1x2+ix1y2+ix2y1+i2y1y2
z3= x1x2+i2y1y2 +i(x1y2+x2y1)
Im(z3)= x2y1-x1y2
Re(z3)= x1x2-y1y2
© 2005 Baylor UniversitySlide 9
Dividing Complex Numbers
32
1 zyixbia
zz
To divide, must eliminate the “i” from the denominator
We do this with the “Complex Conjugate” - by CHANGING THE SIGN OF i
22
22
2112212122
222222
2
221212121
22
22
22
11 )()(*yx
yxyxiyyxxiyiyxiyxxiyyiyxixyxx
iyxiyx
iyxiyx
22
22
21213
)()Re(yxyyxxz
22
22
21213
)()Im(yxyxxyz
© 2005 Baylor UniversitySlide 10
Reciprocals of Complex Numbers
jjjjj
jj 133
132
1332
9432
3232*
321
321
132)Re( z
133)Im( z
Multiply by the Complex Conjugate to put in Standard Form
jjj
jj
jj
2*11
© 2005 Baylor UniversitySlide 11
Questions?