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Class notes MATH 126

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This class note includes how to divide complex numbers, solutions to quadratic equations, the substitution method, elimination method and special cases when there is no/infinite solutions. These notes include example problems as well as a detailed explanation of what is being covered.

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Document information

  • October 18, 2024
  • 1
  • 2024/2025
  • Class notes
  • Robert jasper
  • All classes

Subjects

  • complex numbers and powers of i

Written for

All documents for this subject (10)

Seller

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kiankido

Content preview

Week 4 Lecture 4 .




of i 1 6 2 3
1. 6 1 . .
3 Complex Numbers andPowers . . .
Dividing complex numbers

to divide the numerator + Denominator the Conjugate of denominator
#s, multiply by
-


complex we

-

Expressionof the form at bi where a and I are real's and i is an imaginary unit (exi M i
=
= - 1) -
-

, -
-


a is our real part and b is the imaginary part (i . . 3+
e 4i) Ex1) Find
#
the Real +
mag parts of tx2) 7 3i
4i
.



Oil
real imaginary 35 bi It Zi
3 iti-T :
G
·
_


add
1-2: It zi : 12-28i
To or subtract complex #'s we combine the real parts+ XX ↓
-




,
combine the Imaginary parts 16
1 + 4 = 5
3 7i R = 3/4
To complex #'s , FOIL
multiply
-
= .




3
real
"It : 4 i = T/4
powers of i i= i i 1 , = = i i= 1 =
, = i
-
=


imag =
, , ,




1. 6 3 3 solutions to Quadratic Equations Quiz 1 6
Ex1)
. . .



express in arbi form
-

Square roots of negative #'s 1)9
1) (3 + 5i) + (4 2i) -
2)(y+ 5i) (4- zi) -




9+ j
If his positive , Fr : Fift is
-


then =


(3 + 4) + (5i 2i) (3-4)r (51522)
A.
-




T zi It Ti
+ Ex )Fix : N (4) 4:
-

1 .

: :




- 23
2) E3 : F :
it
4) ;
((3+ 54 -2i) 3) F6 F4 I can't user rule if negative)

12-bitzoi
· :


20j3 ,



Bi2
4i zi :


ji45. i 3
= -
8
2 . 2 + 14 i
2 in = -
i
ComplexSolutions of Quadratic equations
-




recall : Distac
29


Ex 1) Solve X+ 4x + 5 = 0



a= 1 -

42 (5) -
42 F
b= 4 2(1) 2

C= 5
4520-42 :
= -




2

= -
21



9
. 1 1 3 substitution method
. .
. 1 2 3 Elimination Method
9 .
.
& 1 3 3 Special cases
. .
-
No/ infinite Solutions
(3)
systems of 1) manipulate equation tx1) 24x- by 15
·

equations in 8X 5 -
>
zy
= =
linear so one -




(2)
pair of variable have opposite 12 x + Ye by 14
-


2 variables 3y =
24x + :
-




coefficients 0 = 29
1) solve for one variable No solutions
False
2) And the equations
2) substitute into other equation
(9)
3) solve for variable solve
3) a Back substitute Ex 2) 3x-by = 12 >
- 12x -

24 : 48




6
2)
By 16(3)
(-




4) Back substitute Ex)4x 3y = 11 4X - = -
1x + 24y : 40
8x +
by 22
-
- -
>
- -
=




Ex) X+y 0 0
=

= 1 2x- = 1 -
2(2) &x +
y = 12 ex + 4y =
12 inf # of Solut




:
.
= It Y 4X 3) 1)
-
-
= / Ly: Let X = X

5
144
:

4X y = -

1 3x -


by = 12
-
3x - 3x
3x
+ Ex I X= 2
-




by
x:
x= y y = + x = y+ 1 2x + by = 23
!
4x 25
10 , -2)
+ 3y : X = 311 =
4 X -

5y = - 11
(2 ,
-
1)
4 (y + 1) +
3y 25 X = 4
↳ 12 214tby
= X =
: 3

4y +
45 by =25 X= y Ow5y =2

Ly: y= :
Y=3
+ y +y
3 = 11
x= y y 2x + -



= > x
- =
y+ 1 X -

y = 2 > X = 2+
- y 2x +
y= 11 + 353 3x + Yy = 193x+ yy = 19
Yx + 3y 29 14 3)
=
=
4x 3y X 2y 1 - X= 1 2x =
+ 32 29 X yy 1
= =
X = 4+ 1 =5 2+ 3 + -

=3
12 y
=
1 = -
3x +



T
-




y(2 +y) + 3y 29X = 5 2y Ly
·




yy)
=

y(y + 1) + 3y X= 5 *



By
=
32
x = 1
11
-




My + 4 + 3y =
32
8 + yy + zy = 29 2(1 -

(g) by :
X 5

I
/I =
2
yyty :
y
=
1
-




↳=
3Y
=
4

( 3)-
a 3x52y /1
3x +y =
=



102X
4X by = -
25 ->
&X-by
: -

44 2x 16 6X 15y 48
5y
-
-
= - -
+ =



22 6x + %
8x +
yy = 5) 8x +
7y 51 3x yy = 44
11
-
- = + =


3 (1) +2y =



a
-
8x + 7(5) 51 = y= 5 2x + 4(4) : 22
=e X =
1

-
8x + 35 = 51
-
35 -
35
3x + 16 = 22
y= 4 ↳
s
- X = 2

X= -
z

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