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AS Level/ A-Level Core Pure 1 – A* Further Mathematics Pearson Edexcel Summary Notes (8FM0) (9FM0)

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AS Level/ A-Level Core Pure 1 Further Mathematics Pearson Edexcel Summary Notes (8FM0) (9FM0) All key points and example questions (including step-by-step workings) are included! Notes were designed based on the Edexcel syllabus in preparation for the 2023 Summer Exam. Chapter 1: Complex Numbers Chapter 2: Argand Diagrams Chapter 3: Series Chapter 4: Roots of Polynomials Chapter 5: Volume of Revolution Chapter 6: Matrices Chapter 7: Linear Transformation Chapter 8: Proof by Induction Chapter 9: Vectors

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CORE PURE 1


Chapter 1 :
Complex numbers

i F1




O
=




*
- =
a + bi (a DE(R)
, ,
(zEK) · if z = a + bi ,
z = a-bi (complex conjugate
iP = 1 i = -
1
Re(z) =
'Real part of z' = a · if roots of a quadratic equation are a and B ,
then 1z-x)(z B) - =
0



is = -
i
1m(z) =
'Imaginary part of z = b z2 -
(x B)z + +
xB =
0




Chapter 2 :
Argand diagram
Im
A

- =
x +
iy (cartesian form / Z z, = r, (C030 , + iSiNO , ) ,
Ec =
Uc(10902 + iSinOc)
x


| z| 6x
· U2(109(0 (0 , (
+ z , Ec
=
y modulus, r = U, 82) i sin +
O2)
·
+ +
,

N O


arg(z)
=
tan"(π argument , O
> Re ④



= (10610 ,
-
01) + i sin(8 ,
+ O

range : ↑ < arg(z) => ↑
principal argument
-



always sketch !



z , z2) =
/z //Zal
,
arg(z , zc) =
arg(z ,
) +
arg(ze)
z r (c0S0 isinO) (modulus-argument form / 1z , 1
arg/E)
=


Ei
+

= =
arg1z , ) -


arg (zz)
↓ Zal
r (cos0- i sino) =
(cos1 0) -
+
isin) -

Oll

3 3
z ,
= = ,
arg(z , 3) =
3 arg(z , )

loci =
a set of points

e. g
. (i) Iz -
z, ) =
P circle (ii) Iz-z , 1 =
1z-Ec1 perpendicular bisector (iii) arg(z-z , ) =
0 half-line

1z -
5 -
3i) =
3 1z -

31 =
/z + il
arg(z + 3 +
2i) =
arg(z -
( 3 2i))
- -

=
) the
/

midpoint (70,
Im =

1z -
15 +
3i)1 = 3 :
Im
-
X

3
IM
-




> Re
(x 5) (y 3) q
/ -
+ - = O

3 3
..
MH = -



Re X --
2
O 1 3 -

.
-

2)

3
-



y (- 2 )
- = -
3(x 2) -




3
-
·




3( 4
y
= -
+

> Re
O j




>
determining minimum and maximum of arglz) and 121 :
in argand diagram
regions
1z -

(4 + 3i)) = 3 z -
4- 21/2 ,
1z -
4K/z -
61
, 01 arg(z -
2- 2i) !
Im Im
/
·
min arg(z) =
0




h
sin (3) 1 29 rad 02arg(z-2-2i)
arg(z) 2
=
·

max = x .
1z -


4/4/z -
6
3




·
X

5
d =




O 3 z -
4- 2i22

Min
·
(z) =
5 -
3 =
2 (d -
r)
> Re -
2
i
X X




·
max(z) = 5 + 3 =
0(d +
r)


Re
d ↓




Chapter 3 : Series Chapter 5 : Volumes of Revolution
"

1 = n
V= 1
Around 11-axis :




π/ay Arth
· =
N
V =
d cylinder
r =
(n(n + 1)
V= 1


Around y-axis :
·

cone
=
5 TU'h
, =
5n(n +
1)(2n 1) +




V =
πfa) dy
2



P
N
3
=
+ (n + = ↑
V= 1

, Chapter 4 : Roots of Polynomials

Polynomials Ex =
- [: xB =
f & BU =
-


* BUS =
a) + bl + C x +
B xB

all + b( + Cx +
d x +
B +
f B +
BU +
UX XBG

a)" +
b) + C11" + dx +
2 x +
B +
8 + f B +
Bu +
08 + 8x +
B8 +
xuaB8 +
BUS +
USx +
XBS NBUS




Identities :
Finding equation of graph / transformed roots :




x B + = (Ex) -


22 xB e new roots of 1-1 B-1
.

g. ,




x+ B +
0 =
(Ex)" 2xB -
let W = ) -
1


x +
B3 = (2x)" -

3 BEx x =
W + 1


x B3 83 + + =
(Ex)3 -
3[XBEx +
3 XBU then substitute back into original equation
x B + 4 02 +
B38
:
:
(EXB)" -


2 B8 Ex




Chapter 6 :
Matrices
Matrix multiplication -2x) 2x3
multiplication is commutative.
+
: -
NOT


(i)(38 -Y ) =
(ii) AB # BA except Al =
IA =
A

(6 % ) or
-
3 x 0 +
4 x2

associative.
<
Identity matrix I =
2 rows ,
2 columns x 2 rows ,
3 columns -

multiplication is ,




no .
of columns in matrix 1 =
no .
of rows in Matrix 2 (AB)C =
A(BC)



Determinant i Inverse :




M 1,I
=
2 x 2 matrix A =
(9) 2 x 2 Matrix (AB)" =
B A-

det(m) =
=, & = ad-bc A
-
=
defcas ( 9 - ( _
(AB)(AB)" =
1

if det(M) = 0
, singular Matrix, no inverse

if det (M) # 0
, non-singular Matrix ,
has inverse A =· a 3 x 3 Matrix 4 steps !

1) calculate det(A) =
((0 -1-1) -

3/0 -2) +
1(0-8) = -1


P of
9
M 3 x 3 Matrix 2) matrix of minors
,
(a
=
13 I




i
0 4
9 h i e. g. minor of 2 in
- O


M -


-
det (M)
a Ye if
=
= A
ef
n i
-
b
df
+ C
aC
9 i gh
h i
9 - - -
minor of a
minor of b of
minor C 3) changing of signs cofactor , c = I

Simultaneous equations :
4) transpose ,
CT =
i switch rows and columns




&
x 6y 22 21
e g
-
+ -
=
. .




60 i
-
I

6x -



2y
-
z = -
16
Y
=


24
5) A" =
det(A) CT

-
2x +
3y + 52 =
24




if A 4 = V
,
then ( = Av




Consistency :


line
i




. e. same plane
Step 1 :
Find det/M).

if det(M) O
,
then one single solution
.


> if det (M) =
0, then follow Step 2 .




Step 2 :
compare the equations of the planes.




*
If 2 of parallel
i




> (i) or more them are , they
. e. not same line plane



look like these .


g )
5)
le . .

3x-24 +
2 =




parallel
6x -




44
+
22 =
9
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