AS Level/ A-Level Core Pure 2 – A* Further Mathematics Pearson Edexcel Summary Notes (8FM0) (9FM0)

CORE PURE 2

Chapter 1 :
Complex numbers
Exponential form : z =
reio pin = -1 (COST +
isini / · reio and re-"O are conjugates
pio fil0 Oc
+

22Ti
,

= COSO + isino =
1 (COSCH + iSin2π( ·
Zizz = 0 , Va


wherer = 121 and O =
arg = eit = i (cos +
isint ·

E =
5
i 10
, e




De Moivre's Theorem .g
e . (i) Express Cos 30 in terms of powers of cos O
.


·
if z = r(1090 + isinO) ICOs0 + isinG)" =
COs30 + isin3O


z =
"(Cosn0 + isinnol Cos30 + 3icOS'OsinO + 3 i cosO sinO + i sin'O =
COS 38 + isin3O
(IM) (Re)
ino
z =
n cos"0 + Sicos'OsinO -31050 sin'o - isinO =
COSSO + i sin 30


=
COS30 =
COS30 -
31050 /1 -
COS' Ol

·
if z =
COSO +
isinO
, = 4 COS 0 -3 COS O


=> +
5 =
2 COSO ....
z + En = 2 COSNO


= gi0 + -iO = fino + e-ino (ii) Express cos"0 in terms of Cos no
.


= - =
Lisino ....
z -

In =
Lisinno let z =
COSO + i sinO


= pio -iO - = pino e-ino-
12(050)" = (z +
5)5

3210958 = z5 + 523 + 107 + 4 +
s
+
Es

32(0950 = (z5 25) + +
5(z) 3) + + 10(z E +




Sum of series : cos50 = 56/10850 +
510330 +
10 COsO

Given z = Cos + isint ,
SHON 1 + z + =C ... + z" = / + icot(c)
9(8N 1) -
EP -
I
Sp =
1
splitting summations
=
a :
,
= = ..
r -
1 - -
I
:




z -
1 (e)" -
- eπi -
1 1 -
- -
2
e C 1 5 1030 +
910920 #COS 30 +...
g.
-

= I = = -




=
.



z
-
1
e
* -
1
ei ei -
1


Se
S =
5 sino -

J sin20 +, Sin30 +...




Tierige =
2
-
-




Lisin By considering C-is ,
show that C = to 0
and find an expression for


i sin (- l C-iS 1 -5 (1050 isin8) +
(10920 +
i sin20) -

-(10930 + i sin 30)
ie-Eni
= +


-
zie-
i
i(cos(-2) +


=

qfi20 + -13. ...
10
sin 5
=
sin En
:n I
=
2isin = - + -




= i(cos()-isin() =
icos(i) -
i sin (E) a = 1
,
r =
-

5 pi0
3(3 10)
-




/En (
gicio
sin sin I + 2
So = =

It po
= =

(3 +
fi8)(3 f +
-




10)
sin(ticosten) = 1 +
icot( 9 + 3 - 10 9 + 3/1050 - isinOl
sin() -
:
=

9 +
3/210 - p- 0) go+ 10 +
6 COS O
(real)
(imaginary (
9 + 3C0S O 3 sino
an expression of the form e"II can be factorised by taking
.. C = and =

10 +
6 COSO 10 + 6 (0SO


out pie gi0 + = pie(pie-ei8) =
pie(2c0s8


"
nth roots of complex number Solutions To E of
a 1 are roots
unity
= :




e .
g.
23 = 1 + j repeats every 2Ki -n = gidKR
2kπ) (ei) 2Kπ) i 2Kπ
+


=
62(cos( # 2 KM) + +
i sin( # +
OR = z = en gives corners of regular n-gon

2(π)]5 * 4
;

- = [5(cos( +
2Kn) +
isin(# +
E ,
= 1 , z = e = N
,
Es = 2 = G
2




+, KM
25 gi)
+

25(cos/4 8Kn) i sin)π 8KT) )
+ +

W
&
- = +
OR = : I + W + W +
...
= O


k = 0 ,
zo =
25 (cos/5) +
isin(t)) = 2 git

D = 1
,
z, = 25((08(3) isin() + = 2 ·
if E ,
is one root of z" =
S and 1 ,
W ,
C" ...
"
"are the nth roots of
unity ,
then


k =
-1
,
z = 2
*
((89) 4) -
+ isin (4) = 2 the roots of z" = s are given by E ., z ,
W ,
E ,
Wh, ...,
z, Wh .


Im


.g
e n 6 Vertex at 18 8) if centre of n-gon is at (a b)
.
=
z, W
·

zz =
,
. ,




↳·
·




W = pic it
I turn i To get next Vertex ( z ,
=
(9) + rpio

Es
O
·

.
Re z ,
= 852C
i
zc =
(b) + reiox pi
gi
0
o

Ec = 852 gitt Es =
85 eit Es =
(3) + PeiO ,

q
"Te
z zx =
852 pi *, Es =
852


zo = 852e"

, Chapter
2 Series :


Method of differences :
Maclaurin Series :




Ur = (f(r) -
f( )
1) Find f(x) ,
f'll) ,
f"(l) ,
f''ll) , ...




"c
n
I

I I
4r! ,
I
2(r + a of flo) f'(0) f 10) f' (0) ,
2 .
.
g =
ar art e .
g. V= 1
r(r + 2)
V= 1 2) Find value , , , . . .




ar air 2 flosk f'0 3.
firlos Cr




I
f(l) fl0) filo) +...
-



3) ...
+ ,
p = +
(in formula bos is



p Fi ar + 1 I t -

T
I 1 -

5 terms cancel 2 # -


j e .
g. 1n(2 3x) + = 1n(2(1 +
/)) since-1 < (21 for expansion of In (1 + <)
out so just
~ consider terms


I 5- that are not !
3 - -

To =
1n(2) + 1n)1 +
3") -
(2) =
-




5423/

3 5- 4 - -


iz =
In (2) +
[( =) -() -I ... I
· : · · =
1n(2) + Ex -

-H x + +
...




I I
I I -



n 2(n 2)
-

n -
2 2n -
5 In 3 -
-
2 -
2n



(6 , )
I I
n -
1 an'3-an-1 n -
1 2(n-1)
-




2(n + 1) .
e g. en
=
[ 1n(1 +
2() -

1n(1 3x) -




n an-1 -


an + 1
n in-cintal = ((211) 121' (2011) - - +
...
) -
(1-311) -

131' 1-3111 + .

= 4x +
4 +
3 + ...




= in an viras =
* # -cintil-aint
-
1 2x + / =
-


< :
!
B
=
3R2+ +
2
-
1 -
3/21 =
-


5- ? 5




Chapter 3 : Methods in Calculus

If(l)di) is improper if :
(i) One or both of the limits is infinite

(ii) f(() is undefined at 1 = a ,
( = b, or another point in the interval [a b) .




dummy variable is


·
To find ( f(l) dis ,
determine lim It /
e .
g .


[in di =
/" i di
.
e
g. ) ? Tall =molidi Differentiating inverse trigonometric functions :




=
fimo [en ! Carosinas =

Jim
-fino [ -+1
1(mg) + 1) im o In +
1
(arccos) I
- - -
-
-
= -




51 -
1
O
= I ( convergent so exists (
=>
(divergent does not exist /



as + > 0
-

,
-

= 0 as + > 0
-
,
Int > 0
-
carctanc =
its


·

Similarly ,
for values where flil is undefined : e .
G.

e .
g. Jo i d =
fimo JI d candefined at x =
0 (i) y
=
arcsin() /similar to arccoss ( (ii) y
=
arctans (iii) y
= arcoss)"
I

sin () + any x # =

1
* 2x

-lim [ 1nxf y
= =
1 +
(12)
2/
1
cosy secy 1
= = -
simplicit differentiation ( (chain rule (
(1 + x4
A I

sits = secy
↓ ↓ I
-(0(lB) -
1nt) all =cosy S1-siny
=

1 +
+any
2
=

1 +12



= - -
(divergent (



+
as > 0
,
1n + >
- -
>
Integrating with inverse
trigonometry functions :




.
e
g. Show (11kd = arctan + C show /Jai-d =
arcsin() + C



·

Integrate between -- and 8 : let 11 =
Tan O (e + u =
4 = x =
an




I fill di =
moff fill all + S =o
fill dis di = sec o do dx = a du


& +tano seco do =
/1 do Said (Jarau = a du


·
Mean value of fill over the interval [a b) ,
: = 0 + C a
=

situz du



= =
Ja fll) di =
arctan() + C =>
arcsin U + C

b -
A
=
arcsin('a) + C



if f(l) has mean value F over interval [a b]
,
:
snow /and =
arctan (a) + c



·
f(()) + k has mean value Ftk over interval [a b] , let (1 = a tanu Said =
atanza a+
a secu du


·
kf(II) has mean value lif over interval [a b) , dx = a secu du =
) asecinnus du


·
-fill) has mean value -
↑ over interval [a b) ,
= / du
·
does NOT work for f(-1) or f(k))
=
-U + C



= arctan( # ) + C