Summary AQA A Level Further Maths Year 2 Pure Topics Notes

1 z4 1z1n
ROOTS COMPLEX NUMBERS




·
radig
=
OF Izwl = IZlIW)
arga
arg(z) nargz =


arglzw) argz
= +


Zu W =
the solutions of zn = w
write modulus argument form
w in form De moivre's theorem :
includes fractions
a regular polygon
Use de moivres e zn un(cosno + isinno negative
with vertices on circle
isino)"
=
a
zn =
(r(coso + = r(cosno + isinno) powers
compare moduli & arguments centred at the origin
Write n different solutions :
stil
THINK ABOUT (4i) =
4
ARGAND FOR START VALUE
- arg( +
i) = arg(i) =
/
z 45 4i arg(45 + 4i) F
=
= +

Iti(cos + isin)
,
=


r (10530 + isin30) +
/6 ,
-
2π 30 =




(l i) 32(05 isin)
+ = +

828




*
145 + 4il = =
2
= 32(205 /2
+
+ isin/2)
= 32 ;
Roots of mity ze
Euler's formula :



nth roots of unity - 1 ,
ee ... cio-COSO + isino
reio-r(coso + isino
form a
regular n-gon on an argand diagram reio z= reio
ei - z




2
=
=


nth
root
S
=
Wi =
(e)" =
Wi = 0 1 2 ,3
. .
... n -
1
further factorising Z-zez we zei e2 + 3i = 22g
It Wi 0
each ROU is (2 +... + Wn e ((053 + isin3)
+ =




(2) (e)
a "
(z W((z-W*) [Re(W) + IWR
=

power of W
,
-
= z zws
wa
-

-
... + Wh
=

It Wit Wat ...
= I + 2 + + Z
= - 7 32
.
+ 1 . 04i

= 2e
Z" -81 z in Cartesian
= -w =

+ 3i

(e(n3)2
-




Sle iπr4 8)r 3
+3i
1- w
rugino 8) = =
32
e
= -
= =


W 4
: ezn3 el3insli
=

48 = , 3 5
0 =
:



Fifth ROU
, ,




4(c0s[ + isin)
isin (In 27)
* :
WY = W
9((0S(In27) +
13 3 i
=
* =
z +
wi (w2) =



e e, e
=
=
-8 89-1 38i
I 25 - 21
. .




Re(w") Re(w) = ,
, =
=



Re(w3) Re(w) =
product of two real factors



isinos
COS (z zi)(z zu) z2 352z + 9
Geometry
cosRe(W) = -
-
=
Re(W)
-



=
1 +w+ 22 + 23 + w4 0
multiplication by ros +
2002
=



Re(WY Re(w) + Re(w") O
: - 1
(z zz)(z -zs) z = 352z + 9
-

= +
It Re(w)
=
+ +
= rotation through
cos 21052 112
0 3
+ Re(WY + Relw") + Re()
1+ Relws
(z) 352 + 9)(z2 + 352 + 9)
= -



+ 1 =

z" + 81
-


=
& about origin AC Other point in equilateral
I IRe(w) + [Re(w 0 =




Re(w) + Re(w2) =
-


1/2 cos :


I (0) & enlargement
Sf r
mangle e rotation TY3
modulus 1
,

X other point by (CSTs + isinTys)

, coso-ceo Sino-co
"
·
z = eio COS40 in terms of losO Z =
COSO + isino Z" = (losO + isinG)




Sto sSo
Licos'Osino-GcosOSinO-LicososinsO
sin50-Esin30 sino
z +
En =
Icosno BT : z" = COSYO + + sin "O




↳
Show that sinO = +
Cos40 + Isin40
Demoivres
zn-1 zisinno COS"O-Glos' Osino + sin "O
COS40
=



isino zu equate real parts :


z = COSO + = 8105"0 81050 + 1 -




= E Ssinso
(z E) z5 52 + 102 + -




-, , ,
-



cost0=0
-
- =




HOPE
=

Sijsin50-Esinso Esino +
COS40 : 80S"0 810520 + / -




Ho bisinos to Cosso Ecosso +
82-8c" [0 i)
Rising Risingos-5cisingos
-
roots of =
1 ,

-



20isinO COSCO

EME
32isin50 Zisin50-10 ; sinso + 8(4 87 1 0 C
- = =
=

C COSO
COS COST
=

COS40 O
sin50-Esinso + sino
=

Sin5o =
J
,




TRIGONOMETRY-
&
- exact 8(4 - 82 1 0


i
- =


series
-

value of
Irig Isinlox-sinlx
Sin >
nix hence show sinx + sin2x +... + Sin10X =
positive solution
usin2 ( /2)
*
(ix + e2ix +... + e
e0ix)
Im (eix +... +

cos
-
=



(eix-1-e" 10 e'oix
(
..
+



(
: Im
Im
-


:
2- ICOSX

+ Sin10x
Cos30 + isin30 (Cos Otising) :


Snx-Sinix
cosso -3CS2 sin30 325-S3
(i) =
= :
=


Im

It /zeid /eliot+
elixelix)
Im) 2-eix-e-ix +
sinix tanzo :


3-392 -S
t = tano

4sin2 (X/2)
...

·




So :

Feio =eio eix + e-
:x
= 20SX
= 3t2 37 + 1 0
-
- = = ) = 1 = > +an3o =
1

Seio
Ceio+ 1)5 esio + 5 ea+ 100 + 10ezio +
+ 1

COSKO-Re (1 +/220+ yez0+ ) :
=

tanzo 10
Im (ei0 + 1)5
... :



Sin50 + 5 sin40 + 10 sin30 + 10sin20 + 5 sind =

0
4 2e-
iteio-ze-ia 1)
Re(eio) Re) 1)" (2010) e =tano-tan tan tan
-




*
ei0 + 1 (0s0+ 1) + isino :0 hence
(e
· =
= , ,
+ +
200518 + Lisinocos02
=

Im
↓
tanz + tan tan
:
+ 3 ptq = ba
(21030)" sin
=


20s02 (1050 + isin0/2)
Re( Im("")
=
=
=
=
210502 2 %2
:
tanz + tan 4 =




Sin50 + 5 sin40 + 10 sin30 + 10sin20 + 5 sinO
= 32(OS9 Jin