100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
Previously searched by you
Previously searched by you
logo
AS Level/ A-Level Core Pure 1 – A* Further Mathematics Pearson Edexcel Summary Notes (8FM0) (9FM0) £4.99
Add to cart
Quickly navigate to
Preview
  2.  
  3. PEARSON (PEARSON)
  4.  
  5. Further Mathematics
  6.  
  7. Core Pure 1

Summary

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

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 Num...

[Show more]

Preview 2 out of 6  pages

Document information

  • March 11, 2024
  • 6
  • 2022/2023
  • Summary

Subjects

Written for

All documents for this subject (11)

Seller

avatar-seller
celinesim9988

Content preview

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

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through credit card for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying these notes from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller celinesim9988. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for £4.99. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

69411 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy revision notes and other study material for 15 years now

Start selling
£4.99
  • (0)
Add to cart
Added