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I HC THI NGUYN
TRNG I HC KHOA HC
Bi c Dng
V MT PHNG PHP GII TON S CP
Chuyn ngnh:Phng Php Ton S Cp
M s: 60 46 0113
LUN VN THC S TON HC
Ngi hng dn khoa hc
GS.TSKH. H Huy Khoi
Thi Nguyn - 2012
1S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
1
Li cm n
Lun vn c thc hin v hon thnh ti trng i hc Khoa hc
- i hc Thi Nguyn di s hng dn khoa hc ca GS. TSKH. H
Huy Khoi. Qua y, tc gi xin c gi li cm n su sc n thy gio,
ngi hng dn khoa hc ca mnh, GS.TSKH. H Huy Khoi, ngi
a ra ti v tn tnh hng dn trong sut qu trnh nghin cu ca
tc gi. ng thi tc gi cng chn thnh cm n cc thy c trong khoa
Ton - Tin hc trng i hc Khoa hc, i hc Thi Nguyn, to
mi iu kin cho tc gi v ti liu v th tc hnh chnh tc gi hon
thnh bn lun vn ny. Tc gi cng gi li cm n n gia nh, BGH
trng THPT Yn Thy B-Yn Thy-Ha Bnh v cc bn trong lp Cao
hc K4, ng vin gip tc gi trong qu trnh hc tp v lm lun
vn.
1
Mc lc
M u 3
1 nh ngha v tnh cht ca s phc 5
1.1 nh ngha . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 Tnh cht s phc . . . . . . . . . . . . . . . . . . . . . . . 6
1.2.1 Cc tnh cht lin quan n php cng . . . . . . . 6
1.2.2 Cc tnh cht lin quan n php nhn . . . . . . . 6
1.3 Dng i s ca s phc . . . . . . . . . . . . . . . . . . . 7
1.3.1 nh ngha v tnh cht . . . . . . . . . . . . . . . . 7
1.3.2 Gii phng trnh bc hai . . . . . . . . . . . . . . . 10
1.3.3 ngha hnh hc ca cc s phc v modun . . . . 12
1.3.4 ngha hnh hc ca cc php ton i s . . . . . 13
1.4 Dng lng gic ca s phc . . . . . . . . . . . . . . . . . 15
1.4.1 Ta cc trong mt phng . . . . . . . . . . . . . 15
1.4.2 Ta cc ca s phc . . . . . . . . . . . . . . . . 16
1.4.3 Cc php ton s phc trong ta cc . . . . . . 16
1.4.4 ngha hnh hc ca php nhn . . . . . . . . . . 17
1.4.5 Cn bc n ca n v . . . . . . . . . . . . . . . . . 17
1.5 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2 S dng s phc trong gii ton s cp 25
2.1 S phc v cc bi ton hnh hc . . . . . . . . . . . . . . . 25
2.1.1 Mt vi khi nim v tnh cht . . . . . . . . . . . . 25
2.1.2 iu kin thng hng , vung gc v cng thuc
mt ng trn . . . . . . . . . . . . . . . . . . . . 30
2.1.3 Tam gic ng dng . . . . . . . . . . . . . . . . . 31
2.1.4 Tam gic u . . . . . . . . . . . . . . . . . . . . . 33
2
2.1.5 Hnh hc gii tch vi s phc . . . . . . . . . . . . 35
2.1.6 Tch thc ca hai s phc . . . . . . . . . . . . . . 39
2.1.7 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . 43
2.2 S phc v cc bi ton i s , lng gic . . . . . . . . . 45
2.2.1 Cc bi ton lng gic . . . . . . . . . . . . . . . 45
2.2.2 Cc bi ton i s . . . . . . . . . . . . . . . . . . 52
2.2.3 Bi tp . . . . . . . . . . . . . . . . . . . . . . . . . 54
2.3 S phc v cc bi ton t hp . . . . . . . . . . . . . . . 55
Kt lun 62
Ti liu tham kho 63
3
M U
1. L do chn ti
Trong chng trnh ton hc cp THPT s phc c a vo ging
dy phn gii tch ton lp 12. Ton b phn s phc mi ch a ra
nh ngha s phc v mt vi tnh cht n gin ca n. ng dng s
phc trong gii ton mi ch dng li mt vi bi tp hnh hc n gin.
Nhm gip cc em hc sinh kh gii c ci nhn ton din hn v s phc,
c bit s dng s phc gii mt s bi ton s cp: hnh hc, i
s, t hp, lng gic nn ti chn ti lun vn: V mt phng
php gii ton s cp.
2. Mc ch nghin cu
H thng ha cc dng bi tp hnh hc, i s, t hp, lng gic c
gii bng phng php s phc ng thi nm c mt s k thut tnh
ton lin quan.
3. Nhim v ti
a ra nh ngha v tnh cht ca s phc. c bit s dng s phc
gii mt s dng ton: hnh hc, i s, t hp, lng gic.
4. i tng v phm vi nghin cu
Nghin cu cc bi ton hnh hc, i s, t hp, lng gic trn tp
hp s phc v cc ng dng lin quan.
Nghin cu cc ti liu bi dng hc sinh gii, k yu hi tho chuyn
ton, t sch chuyn ton...
5. ngha khoa hc v thc tin ca ti
To c mt ti ph hp cho vic ging dy, bi dng hc sinh
trung hc ph thng. ti ng gp thit thc cho vic hc v dy cc
chuyn ton trong trng THPT, em li nim am m sng to trong
vic dy v hc ton.
6. Cu trc lun vn
Lun vn gm 3 chng
Chng 1: nh ngha v tnh cht ca s phc
Chng 2: Cc dng biu din s phc
Chng 3: S dng s phc trong gii ton s cp
4
Do thi gian v khi lng kin thc ln, chc chn bn lun vn khng
th trnh khi nhng thiu st, tc gi rt mong nhn c s ch bo tn
tnh ca cc thy c v bn b ng nghip, tc gi xin chn thnh cm n!
Thi Nguyn, nm 2012
Tc gi
5
Chng 1
nh ngha v tnh cht ca s phc
1.1 nh ngha
Gi thit ta bit nh ngha v cc tnh cht c bn ca tp s thc R
Ta xt tp hp
R
2
= R R = {(x, y) | x, y R} .
Hai phn t (x
1
, y
1
) v (x
2
, y
2
) bng nhau khi v ch khi
_
x
1
= x
2
y
1
= y
2
Cc php ton cng v nhn c nh ngha trn R
2
nh sau :
z
1
+ z
2
= (x
1
, y
1
) + (x
2
, y
2
) = (x
1
+x
2
, y
1
+y
2
) R
2
.
v
z
1
.z
2
= (x
1
, y
1
) . (x
2
, y
2
) = (x
1
x
2
y
1
y
2
, x
1
y
2
+ x
2
y
1
) R
2
.
vi mi z
1
= (x
1
, y
1
) R
2
v z
2
= (x
2
, y
2
) R
2
. Phn t z
1
+ z
2
gi l
tng ca z
1
, z
2
, phn t z
1
.z
2
R
2
gi l tch ca z
1
, z
2
.
Nhn xt
1) Nu z
1
= (x
1
, 0) R
2
v z
2
= (x
2
, 0) R
2
th z
1
z
2
= (x
1
x
2
, 0).
2))Nu z
1
= (0, y
1
) R
2
v z
2
= (0, y
2
) R
2
th z
1
z
2
= (y
1
y
2
, 0).
nh ngha 1.1.1. Tp hp R
2
cng vi php cng v nhn gi l tp s
phc, k hiu C. Mi phn t z = (x, y) C c gi l mt s phc.
K hiu C
ch tp hp C\ {(0, 0)} .
6
1.2 Tnh cht s phc
1.2.1 Cc tnh cht lin quan n php cng
Php cng cc s phc tha mn cc iu kin sau y
Tnh giao hon : z
1
+z
2
= z
2
+z
1
vi mi z
1
, z
2
C.
Tnh kt hp :(z
1
+ z
2
) + z
3
= z
1
+ (z
2
+z
3
) vi mi z
1
, z
2
, z
3
C.
Phn t n v : C duy nht mt s phc 0 = (0, 0) C z +0 = 0 +z
vi mi z = (x, y) C.
Phn t i : Mi s phc z = (x, y) C c duy nht s phc z =
(x, y) C sao cho z + (z) = (z) + z = 0.
1.2.2 Cc tnh cht lin quan n php nhn
Php nhn cc s phc tha mn cc iu kin sau y
Tnh giao hon:z
1
z
2
= z
2
z
1
vi mi z
1
, z
2
C.
Tnh kt hp:(z
1
z
2
)z
3
= z
1
(z
2
z
3
) vi mi z
1
, z
2
, z
3
C.
Phn t n v: C duy nht s phc 1 = (1, 0) C tha mn z.1 =
1.z = z. S phc 1 = (1, 0) gi l phn t n v vi mi z C.
Phn t nghch o:Mi s phc z = (x, y) C,z = 0 c duy nht s
phc z
1
= (x
,
, y
,
) C sao cho z.z
1
= z
1
z = 1 s phc z
1
= (x
,
, y
,
)
gi l phn t nghch o ca s phc z = (x, y) C.
Ly tha vi s m nguyn ca s phc z C
c nh ngha nh
sau z
0
= 1 ; z
1
= z ; z
2
= z.z ,v z
n
= z.z...z
. .
nl n
vi mi s nguyn n > 0
v z
n
= (z
1
)
n
vi mi s nguyn n < 0.
Mi s phc z
1
, z
2
, z
3
C
v mi s nguyn m, n ta c cc tnh cht

sau
1) z
m
.z
n
= z
m+n
;
2)
z
m
z
n
= z
mn
;
3) (z
m
)
n
= z
mn
;
4) (z
1
z
2
)
n
= z
n
1
z
n
2
;
5)
_
z
1
z
2
_
n
=
z
n
1
z
n
2
;
Khi z = 0 ta nh ngha 0
n
= 0 vi mi s nguyn n > 0.
7
Tnh phn phi : z
1
(z
2
+z
3
) = z
1
z
2
+ z
1
z
3
vi mi z
1
, z
2
, z
3
C
.
Trn y l nhng tnh cht ca php cng v php nhn,thy rng tp
hp C cc s phc cng vi cc php ton trn lp thnh mt trng.
1.3 Dng i s ca s phc
1.3.1 nh ngha v tnh cht
Mi s phc c biu din nh mt cp s sp th t, nn khi thc
hin cc bin i i s thng khng c thun li. l l do tm
dng khc khi vit
Ta s a vo dng biu din i s mi. Xt tp hp R{0} cng vi
php ton cng v nhn c nh ngha trn R
2
.
Hm s
f : R R {0} , f (x) = (x, 0)
l mt song nh v ngoi ra (x, 0) + (y, 0) = (x + y, 0) v (x, 0).(y, 0) =
(xy, 0).
Ngi c s khng sai lm nu ch rng cc php ton i s trn
R{0} ng nht vi cc php ton trn R; v th chng ta c th ng
nht cp s (x, 0) vi s x, vi mi x R. Ta s dng song nh trn v k
hiu (x, 0) = x.
Xt i = (0, 1) ta c
z = (x, y) = (x, 0) + (0, y) = (x, 0) + (y, 0).(0, 1)
= x + yi = (x, 0) + (0, 1).(y, 0)
T trn ta c mnh
Mnh 1.3.1. Mi s phc z = (x, y) c th biu din duy nht di
dng
z = x + yi
Vi x, y R.
H thc i
2
= 1 c suy ra t nh ngha php nhn i
2
= i.i =
(0, 1).(0, 1) = (1, 0) = 1.
8
Biu thc x + yi c gi l biu din i s (dng) ca s phc z =
(x, y). V th ta c th vit C =
_
x + yi |x R, y R, i
2
= 1
_
. T
gi ta k hiu z = (x, y) bi z = x + yi. S thc x = Re(z) c gi l
phn thc ca s phc z, y = Im(z) c gi l phn o ca z. S phc
c dng yi , y R
gi l s thun o, s phc i gi l s n v o.
T cc h thc trn ta d dng c cc kt qu sau:
a) z
1
= z
2
khi v ch khi Re(z
1
) = Re(z
2
) v Im(z
1
) = Im(z
2
).
b) z R khi v ch khi Im(z) = 0.
c) z C\R khi v ch khi Im(z) = 0.
S dng dng i s, cc php ton v s phc c thc hin nh sau:
Php cng
z
1
+z
2
= (x
1
+y
1
i) + (x
2
+ y
2
i) = (x
1
+ x
2
) + (y
1
+y
2
)i C.
D thy tng hai s phc l mt s phc c phn thc l tng cc phn
thc, c phn o l tng cc phn o:
Re(z
1
+z
2
) = Re(z
1
) + Re(z
2
);
Im(z
1
+ z
2
) = Im(z
1
) + Im(z
2
).
Php tr
z
1
z
2
= (x
1
+y
1
i) (x
2
+y
2
i) = (x
1
x
2
) + (y
1
y
2
)i C.
Ta c
Re(z
1
z
2
) = Re(z
1
) Re(z
2
);
Im(z
1
z
2
) = Im(z
1
) Im(z
2
).
Php nhn
z
1
.z
2
= (x
1
+y
1
i).(x
2
+y
2
i) = (x
1
x
2
y
1
y
2
) + (x
1
y
2
+ x
2
y
1
) i C.
Ta c
Re(z
1
z
2
) = Re(z
1
) Re(z
2
) Im(z
1
) Im(z
2
);
Im(z
1
z
2
) = Im(z
1
) Re(z
2
) + Im(z
2
) Re(z
1
).
Mi s thc , s phc z = x +yi, z = (x +yi) = x +yi C l
tch ca mt s thc vi mt s phc. Ta c cc tnh cht sau
9
1) (z
1
+ z
2
) = z
1
+ z
2
;
2)
1
(
2
z) = (
1
2
)z;
3)(
1
+
2
)z =
1
z +
2
z.
Ly tha ca s i
Cc cng thc cho s phc vi ly tha l s nguyn c bo ton i
vi dng i s z = x + yi. Xt z = i, ta thu c
i
0
= 1 ; i
1
= i ; i
2
= 1 ; i
3
= i
2
.i = i
i
4
= i
3
.i = 1; i
5
= i
4
.i = i ; i
6
= i
5
.i = 1; i
7
= i
6
.i = i
Ta c th tng qut cc cng thc trn i vi s m nguyn dng n
i
4n
= 1 ; i
4n+1
= i ; i
4n+2
= 1 ; i
4n+3
= i
V th i
n
{1 , 1 , i , i} vi mi s nguyn n 0. Nu n l s
nguyn m ta c:
i
n
=
_
i
1
_
n
=
_
1
i
_
n
= (i)
n
.
S phc lin hp
Mi s phc z = x+yi u c s phc z = xyi, s phc c gi
l s phc lin hp hoc s phc lin hp ca s phc z.
Mnh 1.3.2. 1) H thc z = z ng khi v ch khi z R;
2)Mi s phc z ta lun c ng thc z = z;
3)Mi s phc z ta lun c z.z l mt s thc khng m ;
4)z
1
+ z
2
= z
1
+z
2
(s phc lin hp ca mt tng bng tng cc s phc
lin hp);
5)z
1
.z
2
= z
1
.z
2
(s phc lin hp ca mt tch bng tch cc s phc lin
hp);
6)Mi s phc z khc 0 ng thc sau lun ng z
1
= z
1
;
10
7)
_
z
1
z
2
_
=
z
1
z
2
, z
2
= 0 (lin hp ca mt thng bng thng cc lin
hp);
8)Cng thc Re(z) =
z + z
2
v Im(z) =
z z
2i
, ng vi mi s phc
z C.
Ghi ch
a) phn t nghch o ca s phc z C
c th c tnh nh sau
1
z
=
z
z.z
=
x yi
x
2
+ y
2
=
x
x
2
+y
2

y
x
2
+ y
2
i.
b) S phc lin hp c s dng trong vic tm thng ca hai s phc
nh sau:
z
1
z
2
=
z
1
.z
2
z
2
z
2
=
(x
1
+y
1
i) (x
2
y
2
i)
x
2
2
+ y
2
2
=
x
1
x
2
+y
1
y
2
x
2
2
+y
2
2
+
x
1
y
2
+ x
2
y
1
x
2
2
+ y
2
2
i.
Modun ca s phc
S |z| =
_
x
2
+ y
2
c gi l modun ca s phc z = x +yi.
Mnh 1.3.3. 1) |z| Re(z) |z| v |z| Im(z) |z|;
2) |z| 0 , z C,ngoi ra |z| = 0 khi v ch khi z = 0;
3) |z| = |z| = |z|;
4) z.z = |z|
2
;
5)|z
1
z
2
| = |z
1
| . |z
2
| (m un ca mt tch bng tch cc m un);
6) |z
1
| |z
2
| |z
1
+ z
2
| |z
1
| + |z
2
|;
7)
z
1
= |z|
1
, z = 0;
8)
z
1
z
2
=
|z
1
|
|z
2
|
, z
2
= 0 (m un ca mt tch bng tch cc m un);
9)|z
1
| |z
2
| |z
1
z
2
| |z
1
| + |z
2
| .
1.3.2 Gii phng trnh bc hai
By gi chng ta c th gii phng trnh bc hai vi h s thc:
ax
2
+ bx +c = 0 , a = 0
11
trong trng hp bit thc = b
2
4ac nhn gi tr m.
Bng cch bin i, d dng a phng trnh v dng tng ng
sau
a
_
_
x +
b
2a
_
2
+

4a
2
_
= 0.
Do
_
x +
b
2a
_
2
i
2
_
2a
_
2
= 0.
V th
x
1
=
b + i
2a
, x
2
=
b i
2a
.
Cc nghim trn l cc s phc lin hp ca nhau v ta c th phn tch
thnh tha s nh sau
ax
2
+bx + c = a (x x
1
) (x x
2
) .
By gi chng ta xt phng trnh bc hai tng qut vi h s phc
az
2
+ bz + c = 0 , a = 0
S dng cc bin i i s nh trng hp phng trnh bc hai vi h
s thc ta c:
a
_
_
z +
b
2a
_
2
+

4a
2
_
= 0.
ng thc trn tng ng vi
_
z +
b
2a
_
2
=

4a
2
hoc (2az + b)
2
= .
Vi = b
2
4ac cng c gi l bit thc ca phng trnh bc hai.
t y = 2az +b phng trnh trn c rt gn v dng
y
2
= = u +vi
12
vi u,v l cc s thc
Phng trnh trn c li gii
y
1,2
=
_
_
r +u
2
+ (sgn v)
_
r u
2
i
_
,
Vi r = || ,v sgnv l du ca s thc v
Nghim ban u ca phng trnh l:
z
1,2
=
1
2a
(b + y
1,2
) .
Ta c mi lin h gia cc nghim v h s:
z
1
+ z
2
=
b
a
, z
1
.z
2
=
c
a
.
Khi phn tch ra tha s
az
2
+bz + c = a (z z
1
) (z z
2
).
Nh vy cc tnh cht trn c bo ton khi cc h s ca phng trnh
thuc trng s phc C.
1.3.3 ngha hnh hc ca cc s phc v modun
ngha hnh hc ca s phc
Chng ta nh ngha s phc z = (x, y) = x + yi l mt cp s thc
sp th t (x, y) RR, v th hon ton t nhin khi xem mi s phc
z = x + yi l mt im M(x, y) trong khng gian R R.
Xt P l tp hp cc im ca khng gian
vi h trc ta xOy
v song nh : C P , (z) = M (x, y) .
im M(x; y)c gi l dng hnh hc ca s phc z = x + yi. S
phc z = x + yi c gi l ta phc ca im M(x; y). Chng ta k
hiu M(z) ch ta phc ca imM l s phc z.
Dng hnh hc ca s phc lin hp z ca s phc z = x + yi l im
M
(x, y) i xng vi M(x, y) qua truc ta Ox.

13
Dng hnh hc ca s i -z ca s phc z = x+yi l im M(x, y)
i xng vi M(x, y) qua gc ta .
Song nh t tp R ln trc Ox ta gi l trc thc, ln trc Oy ta gi
l trc o.
Khng gian
cng vi cc im c ng nht vi s phc gi l

khng gian phc.
Ta cng c th ng nht cc s phc z = x +yi vi vc t

v =

OM
, vi M(x, y) l dng hnh hc ca s phc z.
Gi V
0
l tp hp cc vc t c im gc l gc ta O. Ta c th nh
ngha song nh
: C V
0
,
(z) =

OM = x

i +y

j , vi

i ,

j l cc
vc t n v trn trc ta Ox, Oy.
ngha hnh hc ca modun
Xt s phc z = x + yi biu din hnh hc trong mt phng lM(x, y).
Khong cch clit OM cho bi cng thc
OM =
_
(x
M
x
O
)
2
+ (y
M
y
O
)
2
.
V th OM =
_
x
2
+y
2
= |z| = |

v | m un |z| ca s phc z = x + yi
l di ca on thng OM hoc l ln ca vc t

v = x

i + y

j .
Ch
a) Mi s thc dng r, tp hp cc s phc c m un r tng ng
vi ng trnC (O; r) tm O bn knh r trong mt phng.
b) Cc s phc z vi |z| < r l cc im nm bn trong ng trn
C(O; r). Cc s phc z vi|z| > r l cc im nm bn ngoi ng trn
C(O; r).
1.3.4 ngha hnh hc ca cc php ton i s
a) Php cng v php tr
Xt hai s phc z
1
= x
1
+y
1
i v z
2
= x
2
+y
2
i tng ng vi hai vc
t

v
1
= x
1

i + y
2

j v

v
2
= x
2

i + y
2

j .
14
Tng ca hai s phc l
z
1
+ z
2
= (x
1
+x
2
) + (y
1
+ y
2
) i.
Tng hai vc t

v
1
+

v
2
= (x
1
+ x
2
)

i + (y
1
+ y
2
)

j .
V th z
1
+z
2
tng ng vi

v
1
+

v
2
.
Hon ton tng t i vi php tr
Hiu ca hai s phc l
z
1
z
2
= (x
1
x
2
) + (y
1
y
2
) i.
Hiu hai vc t

v
1

v
2
= (x
1
x
2
)

i + (y
1
y
2
)

j .
V th z
1
z
2
tng ng vi

v
1

v
2
.
Ch
Khong cch gia M
1
(x
1
, y
1
) v M
2
(x
2
, y
2
) bng m un ca s phc
z
1
z
2
hoc di ca vc t

v
1

v
2
. Vy :
M
1
M
2
= |z
1
z
2
| = |

v
1

v
2
| =
_
(x
2
x
1
)
2
+ (y
2
y
1
)
2
.
b) Tch ca s thc v s phc
Xt s phc z = x + yi tng ng vi vc t

v = x

i + y

j . Nu
l s thc , th tch s thc z = x + yi tng ng vi vc t
v = x

i + y

j .
Ch : Nu > 0 th vc t

v v

v cng hng v |

v | = |

v |,
nu < 0 th vc t

v v

v ngc hng v |

v | = |

v |. Tt nhin
= 0 th

v =

0 .
15
1.4 Dng lng gic ca s phc
1.4.1 Ta cc trong mt phng
Xt mt phng ta vi M(x, y) khng trng gc ta . S thc
r =
_
x
2
+ y
2
gi l bn knh cc ca im M. Gc nh hng t
[0, 2)
gia vc t
OM vi chiu dng ca trc ta Ox gi l argumen cc ca

im M. Cp s (r, t
) gi l ta cc ca im M. Ta s vit M (r, t
).
Ch hm s
h : R R\ {(0, 0)} (0, ) x [0, 2) , h((x, y)) = (r, t
)
l song nh.
Gc ta O l im duy nht sao cho r = 0 , argumen t
ca gc
khng c nh ngha.
Mi im M trong mt phng , c duy nht giao im P ca tia vi
ng trn n v gc O. im P ging nh argument cc t
. S dng
nh ngha hm sin v cos ta c
x = r cos t
, y = r sin t
.
V th ta d dng c ta Cc ca mt im t ta cc
Ngc li, xt im M(x, y). Bn knh cc l r =
_
x
2
+ y
2
. Ta xc
nh argument cc trong cc trng hp sau
a)Nu x = 0 , t tan t
=
y
x
ta suy ra
t
= arctan
y
x
+k
Vi
k =
_
_
0 khi x > 0 , y 0
1 khi x < 0 , y R
2 khi x > 0 , y < 0
b)Nu x = 0 v y = 0 th
t
=
_
2
khi y > 0
3
2
khi y < 0
16
1.4.2 Ta cc ca s phc
Mi s phc z = x + yi ta c th vit di dng cc
z = r (cos t
+ i sin t
) ,
vi r [0, ) v t
[0, 2) l ta cc dng hnh hc ca s phc

z.
Argument cc ca dng hnh hc ca s phc z c gi l argument
ca z, k hiu l arg z. Bn knh cc ca dng hnh hc ca s phc z
bng m un cua z. Khi z = 0 m un v argument ca z c xc nh
mt cch duy nht.
Xt z = r (cos t
+ i sin t
) v t = t
+ 2k vi k l s nguyn th
z = r (cos (t 2k) + i sin (t 2k)) = r (cos t + i sin t) .
Mi s phc z c th biu din nh z = r (cos t + i sin t) vi r 0 v
t R. Tp hp Arg z = {t = t
+ 2k , k Z} c gi l arguent m
rng ca s phc z.
V th, hai s phc z
1
, z
2
= 0 c dng
z
1
= r
1
(cos t
1
+ i sin t
1
) v z
2
= r
2
(cos t
2
+ i sin t
2
)
bng nhau khi v ch khi r
1
= r
2
v t
1
t
2
= 2k, vi k l s nguyn.
Ch Cc dng sau nn nh
1 = cos0 + i sin 0 , i = cos
2
+ i sin

2
1 = cos + i sin , i = cos
3
2
+ i sin
3
2
.
1.4.3 Cc php ton s phc trong ta cc
Php nhn Gi s rng
z
1
= r
1
(cos t
1
+ i sin t
1
) v z
2
= r
2
(cos t
2
+ i sin t
2
)
th
z
1
z
2
= r
1
r
2
(cos (t
1
+t
2
) + i sin (t
1
+t
2
)) .
17
Ly tha ca mt s phc (De moirve) Cho z = r (cos t + i sin t) ,
n N, ta c
z
n
= r
n
(cos nt + i sin nt) .
Php chia Gi s rng
z
1
= r
1
(cos t
1
+ i sin t
1
) v z
2
= r
2
(cos t
2
+ i sin t
2
)
th
z
1
z
2
=
r
1
r
2
(cos (t
1
t
2
) + i sin (t
1
t
2
)) .
1.4.4 ngha hnh hc ca php nhn
Xt
z
1
= r
1
(cos t
1
+ i sin t
1
)
v
z
2
= r
2
(cos t
2
+ i sin t
2
) .
Biu din hnh hc ca chng l M
1
(r
1
, t
1
) , M
2
(r
2
, t
2
). Gi P
1
, P
2
ln
lt l giao im ca C(O, 1) vi cc tia (OM
1
v (OM
2
. Ly P
3
C(O, 1)
vi argument cc l t
1
+t
2
v chn M
3
(OP
3
sao cho OM
3
= OM
1
.OM
2
.
Ly z
3
c ta M
3
. im M
3
(r
1
r
2
, t
1
+t
2
) l dng hnh hc z
1
.z
2
Ly A l dng hnh hc ca s phc 1 . V
OM
3
OM
1
=
OM
2
1

OM
3
OM
2
=
OM
2
OA
v

M
2
OM
3
=

AOM
1
nn hai tam gic M
2
OM
3
v AOM
1
ng dng.
Khi biu din dng hnh hc ca mt thng ch rng dng hnh hc
ca
z
3
z
2
l im M
1
.
1.4.5 Cn bc n ca n v
Cho s nguyn dng n 2 v s phc z
0
= 0, ging nh trn trng
s thc, phng trnh
Z
n
z
0
= 0
18
c s dng nh ngha cn bc n ca s z
0
. V vy mi mt gi tr Z
tha mn phng trnh trn l mt cn bc n ca z
0
.
nh l 1.4.1. Cho z
0
= r (cos t
+ i sin t
) l s phc vi r > 0 v
t [0, 2) S phc z
0
c n cn bc n phn bit cho bi cng thc
Z
k
=
n
r
_
cos
t
+ 2k
n
+ i sin
t
+ 2k
n
_
vi k = 0, n 1.
Chng minh:S dng dng cc ca s phc vi argument xc nh
Z = (cos + i sin ) . Theo nh ngha Z
n
= z
0
hay
n
(cosn + i sin n) = r (cos t
+ i sin t
) .
Ta c
n
= r v n = t
+ 2k vi k Z . V th =
n
r v
k
=
t
n
+ k.
2
n
vi k Z. Do nghim ca (1) l
Z
k
=
n
r
_
cos
t
+ 2k
n
+ i sin
t
+ 2k
n
_
vi k Z.
Nhn thy rng 0
0
<
1
... <
n1
, v th cc s
k
, k
{0, 1...., n 1} chnh l cc argument v
k
=
k
. Ta c n gi tr cn phn
bit ca z
0
:Z
0
, Z
1
, ...., Z
n1
. Cho k l s nguyn v r {0, 1, ..., n 1},
th r ng d vi k theo modn. Khi k = nq + r Z v
k
=
t
n
+ (nq + r)
2
n
=
t
n
+r
2
n
+ 2q =
r
+ 2q.
Nhn thy Z
k
= Z
r
do
{Z
k
: k Z} = {Z
0
, Z
1
, ..., Z
n1
} .
Vy c chnh xc n gi tr phn bit ca cn bc n.
Biu din hnh hc cc gi tr ca cn bc n l cc nh ca mt n gic
u ni tip trong ng trn c tm l gc ta , bn knh l
n
r.
19
Ta chng minh iu trn nh sau, k hiu M
0
, M
1
, ..., M
n1
l cc
im c ta phc Z
0
, Z
1
, ..., Z
n1
. V OM
k
= |Z
k
| =
n
r vi k
{0, 1, ..., n 1} nn cc im M
k
nm trn ng trn C (O,
r
n). Bn
cnh , s o ca cung M
k
M
k+1
bng
arg Z
k+1
arg Z
k
=
t
+ 2 (k + 1) (t
+ 2k)
n
=
2
n
,
vi k {0, 1, ...., n 2} v s o cung M
n1
M
0
l
2
n
= 2 (n 1)
2
n
.
V tt c cc cung M
1
M
2
, ..., M
n1
M
0
u bng nhau nn a gic
M
0
M
1
...M
n1
l a gic u.
Cn bc n ca n v
Cc nghim phng trnh Z
n
1 = 0 c gi l cc cn bc n ca
n v.V 1 = cos0 + i sin 0 nn t cng thc cn bc n ca s phc ta c
cn bc n ca n v
k
= cos
2k
n
+ i sin
2k
n
, k {0, 1, ..., n 1} .
C th ta c
0
= cos 0 + i sin 0 = 1;
1
= cos
2
n
+ i sin
2
n
= ;
2
= cos
4
n
+ i sin
4
n
=
2
;
. . .
n1
= cos
2 (n 1)
n
+ i sin
2 (n 1)
n
=
n1
.
Tp hp
_
1, ,
2
, ...,
n1
_
k hiu U
n
. Ta c tp hp U
n
c sinh bi
, mi phn t ca U
n
l mt ly tha ca .
Ging nh trc, biu din hnh hc cc cn bc n ca mt s phc l
cc nh ca mt a gic u n cnh, ni tip trong ng trn n v m
c mt nh l 1. Ta xt mt vi gi tr ca n
20
i) vi n = 2, phng trnh Z
2
1 = 0 c cc nghim 1 v 1 y l
cc cn bc hai ca n v
ii) vi n = 3, phng trnh Z
3
1 = 0 c cc nghim cho bi cng thc
k
= cos
2k
3
+ i sin
2k
3
vi k {0, 1, 2} .
V th
0
= 1 ,
1
= cos
2
3
+ i sin
2
3
=
1
2
+i
3
2
,
v
2
= cos
4
3
+ i sin
2
3
=
1
2
i
3
2
.
y l cc nh ca tam gic u ni tip ng trn C (O, 1) .
iii) vi n = 4 ,cc cn bc 4 l
k
= cos
2k
4
+ i sin
2k
4
vi k {0, 1, 2, 3} .
C th nh sau
0
= 1 ,
1
= cos
2
+ i sin

2
= i
2
= cos + i sin = 1 ,
3
= cos
3
2
+ i sin
3
2
= i.
Ta c
U
4
=
_
1, i, i
2
, i
3
_
= {1, i, 1, i} .
Biu din hnh hc ca cc cn bc bn l cc nh ca hnh vung ni
tip ng trn C (O, 1)c mt nh l 1.
Cn
k
U
n
c gi l cn nguyn thy nu mi s nguyn dng
m < n ta c
m
k
= 1.
Mnh 1.4.2. 1) Nu n|q , mi nghim ca phng trnh Z
n
1 = 0
l nghim ca phng trnh Z
q
1 = 0;
21
2) Nghim chung ca phng trnh Z
m
1 = 0 v Z
n
1 = 0 l cc
nghim ca phng trnh Z
d
1 = 0; vi d = gcd(m, n) (d:c chung ln
nht), U
m
U
n
= U
d
;
3)Cc nghim nguyn thy ca phng trnh Z
m
1 = 0 l
k
= cos
2k
m
+ i sin
2k
m
;
vi 0 k m v gcd (k, m) = 1.
Mnh 1.4.3. Nu U
n
l mt cn nguyn thy ca n v th tt
c cc nghim ca phng trnh Z
n
1 = 0 l
r
,
r+1
, ...,
r+n1
vi r
l s nguyn dng ty .
Mnh 1.4.4. Cho
0
,
1
, ....,
n1
l cc cn bc n ca n v . Vi
mi s nguyn dng n ta lun c h thc
n1
j=0
k
j
=
_
n, n|k ;
0, n |k.
Mnh 1.4.5. Cho p l s nguyn t v = cos
2
p
+ i sin
2
p
. Nu
a
0
, a
1
, ..., a
p1
l cc s nguyn khc khng ,h thc
a
0
+ a
1
+ ... +a
p1
p1
= 0
ng khi v ch khi a
0
= a
1
= ... = a
p1
.
1.5 Bi tp
Bi 1 Cho cc s phc z
1
= (1, 2) , z
2
= (2, 3) , z
3
= (1 1) hy
tnh cc tng sau:
a) z
1
+ z
2
+z
3
; b) z
1
z
2
+ z
2
z
3
+z
3
z
1
; c) z
1
z
2
z
3
;
d) z
2
1
+z
2
2
+z
2
3
; e)
z
2
1
+ z
2
2
z
2
2
+ z
2
3
; f)
z
1
z
2
+
z
2
z
3
+
z
3
z
1
.
Bi 2 Gii cc phng trnh sau :
a) z + (5, 7) = (2, 1) ; b) (2, 3) + z = (5, 1) ;
c) z. (2, 3) = (4, 5) ; d)
z
(1, 3)
= (3, 2)
22
Bi 3 Gii cc phng trnh sau trn tp C
a) z
2
+ z + 1 = 0 ; b) z
3
+ 1 = 0 .
Bi 4 Cho z
0
= (a, b) C . Tm s phc z tha mn z
2
= z
0
.
Bi 5 Tm cc s thc x, y trong cc trng hp sau:
a) (1 2i) x + (1 + 2i) y = 1 +i ; b)
x 3
3 + i
+
y 3
3 i
= i ;
c) (4 3i) x
2
+ (1 + 2i) xy = 4y
2
1
2
x
2
+
_
3xy 2y
2
_
i .
Bi 7 Tnh :
a) (2 i) (3 + 2i) (5 4i) ; b) (2 4i) (5 + 2i) + (3 + 4i) (6 i) ;
c)
_
1 + i
1 i
_
16
+
_
1 i
1 + i
_
16
; d)
_
1 + i
3
2
_
6
+
_
1 i
7
2
_
6
;
e)
3 + 7i
2 + 3i
+
5 8i
2 3i
.
Bi 8 Tnh:
a) i
2000
+i
1999
+ i
201
+i
82
+ i
47
; b) E
n
= 1 +i + i
2
+ ... +i
n
, 1 n N;
c) i
1
.i
2
.i
3
...i
2000
; d) i
5
+ (i)
7
+ (i)
13
+ i
100
+ (i)
94
.
Bi 9 Tm tt c cc s phc z = 0 tha mn z +
1
z
R.
Bi 10Chng minh rng:
a) E
1
=
_
2 + i
5
_
7
+
_
2 i
5
_
7
R ;
b) E
2
=
_
19 + 7i
9 i
_
n
+
_
20 + 5i
7 + 6i
_
n
R .
Bi 11 Cho z C
tha mn
z
3
+
1
z
3
2 .Chng minh rng
z +
1
z
2 .
Bi 12 Tm cc s phc z tha mn |z| = 1 v
z
2
+ z
2
= 1.
Bi 13Tm cc s phc z tha mn 4z
2
+ 8 |z|
2
= 8.
Bi 14Tm cc s phc z tha mn z
3
= z.
Bi 15 Cho z C vi Re (z) > 1 .Chng minh rng
1
z

1
2
<
1
2
.
Bi 16 Cho a, b, c l cc s thc v =
1
2
+ i
3
2
.Tnh tng
_
a + b + c
2
_ _
a + b
2
+ c
_
.
Bi 17 Chng minh cc ng thc sau :
23
a) |z
1
+ z
2
|
2
+|z
2
+z
3
|
2
+|z
3
+ z
1
|
2
= |z
1
|
2
+|z
2
|
2
+|z
3
|
2
+|z
1
+ z
2
+z
3
|
2
;
b) |1 + z
1
z
2
|
2
+|z
1
z
2
|
2
=
_
1 +|z
1
|
2
__
1 +|z
2
|
2
_
;
c) |z
1
+z
2
+ z
3
| +|z
1
+z
2
+ z
3
| +|z
1
z
2
+ z
3
| +|z
1
+ z
2
z
3
| =
= 4
_
z
2
1
z
2
2
z
2
3
_
.
Bi 18 Tm tt c cc s nguyn dng n sao cho :
_
1 + i
3
2
_
n
+
_
1 i
3
2
_
n
= 2.
Bi 19 Cho z
1
, z
2
, z
3
l cc s phc tha mn |z
1
| = |z
2
| = |z
3
| = R > 0.
Chng minh rng :
|z
1
z
2
| . |z
2
z
3
| +|z
3
z
1
| . |z
1
z
2
| +|z
2
z
3
| . |z
3
z
1
| 9R
2
.
Bi 20 Cho z
1
, z
2
, ..., z
n
l cc s phc tha mn
|z
1
| = |z
2
| = ... = |z
3
| = r > 0.chng minh rng :
E =
(z
1
+ z
2
) (z
2
+ z
3
) ... (z
n1
+z
n
) (z
n
+z
1
)
z
1
z
2
...z
n
.
Bi 21 (Bt ng thc Hlawas) Cho z
1
, z
2
, z
3
l cc s phc. Chng
minh rng :
|z
1
+ z
2
| +|z
2
+ z
3
| +|z
3
+ z
1
| |z
1
| +|z
2
| +|z
3
| +|z
1
+ z
2
+ z
3
| .
Bi 22 Cho x
1
, x
2
l nghim phng trnh x
2
x + 1 = 0. Hy tnh :
a) x
2000
1
+ x
2000
2
; b) x
1999
1
+ x
1999
2
; c) x
n
1
+ x
n
2
, n N.
Bi 23 Tm dng ta cc ca cc s phc sau:
a) z
1
= 6 + 6i
3 ; b) z
2
=
1
4
+i
3
4
; c) z
3
=
1
2
i
3
2
;
d) z
4
= 9 9i
3 ; e) z
5
= 3 2i ; f) z
6
= 4i .
Bi 24 Tm dng ta cc ca cc s phc sau:
a) z
1
= cos a i sin a , a [0, 2) ;
b) z
2
= sin a + i (1 + cos a) , a [0, 2) ;
c) z
3
= cos a + sin a + i (sin a cos a) , a [0, 2) ;
d) z
4
= 1 cos a + i sin a , a [0, 2) .
Bi 25 S dng dng cc ca s phc,hy tnh cc tng sau:
24
a)
_
1
2
i
3
2
_
(3 + 3i)
_
2
3 + 2i
_
; b) (1 +i) (2 2i) ;
c) 2i
_
4 + 4
3i
_
(3 + 3i) ; d) 3 (1 i) (5 + 5i) .
Bi 26 Hy tm modun v argument ca cc s phc z:
a) z =
_
2
3 + 2i
_
8
(1 i)
6
+
(1 + i)
6
_
2
3 2i
_
8
;
b) z =
(1 + i)
4
_
3 i
_
10
+
1
_
2
3 + 2i
_
4
;
c) z =
_
1 + i
3
_
n
+
_
1 i
3
_
n
.
Bi 27 Tm cn bc hai ca cc s phc sau:
a) z = 1 +i ; b) z = i ; c) z =
1
2
+
i
2
;
d) z = 2
_
1 + i
3
_
; e) z = 7 + 24i ;
Bi 28 Tm cn bc ba ca cc s phc sau:
a) z = i ; b) z = 27 ; c) z = 2 + 2i ;
d) z =
1
2
i
3
2
; e) z = 18 + 26i .
Bi 29 Tm cn bc bn ca cc s phc sau:
a) z = 2 i
12 ; b) z =
3 + i ; c) z = i ;
d) z = 2i ; e) z = 7 + 24i .
Bi 30 Gii cc phng trnh sau:
a) z
3
125 = 0 ; b) z
4
+ 16 = 0 ;
c) z
3
+ 64i = 0 ; d) z
3
27i = 0 ;
Bi 31 Gii cc phng trnh sau:
a) z
7
2iz
4
iz
3
2 = 0 ; b) z
6
+ iz
3
+ i 1 = 0 ;
c) (2 3i) z
6
+ 1 + 5i = 0 ; d) z
10
+ (2 + i) z
5
2i = 0 .
25
Chng 2
S dng s phc trong gii ton s
cp
2.1 S phc v cc bi ton hnh hc
2.1.1 Mt vi khi nim v tnh cht
Khong cch gia hai im
Gi s cc s phc z
1
v z
2
c biu din hnh hc l cc im M
1
v M
2
khi khong cch gia hai im M
1
v M
2
c cho bi cng thc
M
1
M
2
= |z
1
z
2
|
on thng, tia, ng thng
ChoA v B l hai im phn bit, trong mt phng phc c ta l a
v b. Ta ni im M c ta z nm gia A v B nu z = a , z = b v
h thc sau tha mn
|a z| +|z b| = |a b| .
Ta s dng k hiuA M B.
Tp hp (AB) = {M : A M B} c gi l on thng m xc nh
bi imA v B.
Tp hp [AB] = (AB) {A, B} c gi l on thng ng xc nh
bi imA v B.
nh l 2.1.1. Gi s A(a), B(b) l hai im phn bit. Khi cc
trnh by di y l tng ng
1)M (AB) ;
2)C s thc dng k sao cho z a = b (k z) ;
26
3)C s thc t (0, 1) sao cho z = (1 t) a + tb; vi z l ta phc
ca M.
1) M (AB;
2)C s thc dng t sao cho z = (1 t) a + tb, vi z l ta phc ca
M;
3) arg (z a) = arg (z b) ;
4)
z a
b a
R
+
.
1) M nm trn ng thng AB;
2)
z a
b a
R;
3)C s thc t sao cho z = (1 t) a + tb;
4)
z a z a
b a b a
= 0;
5)
z z 1
a a 1
b b 1
= 0.
Chia on thng theo mt t s
Cho hai im A(a), B(b) phn bit. Mt im M(z) nm trn ng
thng AB chia onAB theo t s k R\ {1} khi h thc vc t sau
tha mn:
MA = k.
MB.
S dng ta h thc trn c th vit a z = k (b z) hoc
(1 k) .z = a k.b. V th ta c
z =
a kb
1 k
.
Khi k < 0 im M nm trn on thng ni A v B. Nu k (0, 1), th
M (BA\ [AB] .
Trng hp cn li k > 1 th M (AB\ [AB]
27
Gc nh hng
Nh li rng, mt tam gic c nh hng nu nh cc nh ca n
c ch r th t. Tam gic c hng dng nu hng cc nh ngc
chiu kim ng h, hng ngc li l hng m. Ly M
1
(z
1
) v
M
2
(z
2
) l hai im phn bit khc gc ta trong mt phng phc.
Gc

M
1
OM
2
c gi l nh hng nu cc im M
1
v M
2
c th t
thun chiu kim ng h.
Mnh 2.1.4. S o gc nh hng

M
1
OM
2
bng
arg
z
2
z
1
.
nh l 2.1.5. Cho ba im phn bit M
1
(z
1
) , M
2
(z
2
) , M
3
(z
3
). S
o gc nh hng

M
2
M
1
M
3
l
arg
z
3
z
1
z
2
z
1
.
Gc gia hai ng thng
Cho bn im M
i
(z
1
) , i {1, 2, 3, 4}. S o gc xc nh bi ng
thng M
1
M
3
v M
2
M
4
bng
arg
z
3
z
1
z
4
z
2
hoc
arg
z
4
z
2
z
3
z
1
.
Php quay mt im Xt gc v s phc cho bi = cos + i sin
Ly z = r (cos t + i sin t) l s phc v M l biu din hnh hc .
Dng tch z = r (cos (t +) + i sin (t + )) , ta c |r| = r v
arg (z) = arg z +.
Gi M
l biu din hnh hc ca z, ta thy rng im M
l nh ca M
qua php quay tm O ( gc ta ) gc quay l .
Mnh 2.1.6. Gi s im C l nh ca B qua php quay tm A gc
quay . Nu a, b, c l cc ta ca A, B, C phn bit th:
c = a + (b a) vi = cos + i sin .
28
Bi ton 1 Cho ABCD v BNMK l hai hnh vung khng trng
nhau . E l trung im ca AN, F l hnh chiu vung gc ca B ln
ng thng CK. Chng minh rng E, F, B thng hng.
Gii
Xt khng gian phc gc F cc trc ta CK v FB vi FB l trc
o. Lyc, k, bi l ta cc im C, B, K vi c, k, b R. Php quay tm
B gc quay =

2
bin im C thnh im A v th A c ta l
a = b (1 i) + ci . Tng t N l nh ca im B gc quay =
2
v
c ta phc l
n = b (1 + i) ki.
Trung im E ca on thng AN c ta phc l
e =
a + n
2
= b +
c k
2
i.
Vy E nm trn ng thng FB, ta c pcm.
Bi ton 2 Cho t gic ABCD , trn cc cnh AB, BC, CD, DA ta ln
lt dng v pha ngoi ca t gic cc hnh vung c tm O
1
, O
2
, O
3
, O
4
phn bit. Chng minh rng O
1
O
3
O
2
O
4
v O
1
O
3
= O
2
O
4
.
Gii LyABMM
, BCNN
, CDPP
, DAQQ
ln lt l cc hnh vung
c tm O
1
, O
2
, O
3
, O
4
.
im M l nh ca A qua php quay tm B gc quay =

2
; v th
m = b + (a b) i.
Tng t
n = c + (b c) i , p = (c d) i , q = a + (d a) i.
T ta c
o
1
=
a + m
2
=
a + b + (a b) i
2
, o
2
=
b +c + (b c) i
2
o
3
=
c +d + (c d) i
2
, o
4
=
d +a + (d a) i
2
.
29
V
o
3
o
1
o
4
o
2
=
c + d a b +i (c d a + b)
a +d b c +i (d a b + c)
= i iR
nn O
1
O
3
O
2
O
4
.
Ngoi ra
o
3
o
1
o
4
o
2
= |i| = 1
nn O
1
O
3
= O
2
O
4
.
Bi ton 3 V pha ngoi tam gic ABC ta dng cc tam
gicABR, BCP, CAQ sao cho
PBC =

CAQ = 45
o
BCP =

QCA = 30
o
ABR =

RAB = 15
o
Chng minh rng

QRP = 90
o
v RQ = RP.
Gii
Xt mt phng phc vi gc ta l R, gi M l hnh chiu vung gc
ca P ln BC.
T MP = MB v
MC
MP
=
3 ta c
p m
b m
= i v
c m
p m
= i
3
V th
p =
c +
3b
1 +
3
+
b c
1 +
3
i.
Tng t
q =
c +
3a
1 +
3
+
a c
1 +
3
i.
im B c c t im A bng cch quay quanh R mt gc = 150
o
v
th
b = a
_
3
2
+
1
2
i
_
.
30
S dng bin i i s ta c
p
q
= i iR
, v th QRPR. Ngoi ra
|p| = |iq| = |q| nn QR = PR. pcm
2.1.2 iu kin thng hng , vung gc v cng thuc mt
ng trn
Xt bn im phn bit M
i
(z
i
) , i {1, 2, 3, 4}
Mnh 2.1.7. Cc im M
1
, M
2
, M
3
thng hng khi v ch khi
z
3
z
1
z
2
z
1
R
.
Mnh 2.1.8. ng thng M
1
M
2
v M
3
M
4
vung gc khi v ch khi
z
1
z
2
z
3
z
4
iR
.
Mnh 2.1.9. Bn im phn bit M
1
(z
1
) , M
2
(z
2
) , M
3
(z
3
) , M
4
(z
4
)
thng hng hoc cng nm trn mt ng trn khi v ch khi
k =
z
3
z
2
z
1
z
2
:
z
3
z
4
z
1
z
4
R
S k c gi l t s kp ca bn im
M
1
(z
1
) , M
2
(z
2
) , M
3
(z
3
) , M
4
(z
4
) .
Ch 1) cc im M
1
(z
1
) , M
2
(z
2
) , M
3
(z
3
) , M
4
(z
4
) thng hng khi v
ch khi
z
3
z
2
z
1
z
2
R
v
z
3
z
4
z
1
z
4
R
.
2) Cc im M
1
(z
1
) , M
2
(z
2
) , M
3
(z
3
) , M
4
(z
4
) nm trn mt ng trn
khi v ch khi
k =
z
3
z
2
z
1
z
2
:
z
3
z
4
z
1
z
4
R
vi
z
3
z
2
z
1
z
2
/ R v
z
3
z
4
z
1
z
4
R.
31
2.1.3 Tam gic ng dng
Xt su im A
1
(a
1
) , A
2
(a
2
) , A
3
(a
3
) , B
1
(b
1
) , B
2
(b
2
) , B
3
(b
3
) trong
mt phng phc. Ta ni rng tam gic A
1
A
2
A
3
v B
1
B
2
B
3
ng dng
vi nhau nu gc A
k
bng gc B
k
, k {1, 2, 3} .
Mnh 2.1.10. Tam gic A
1
A
2
A
3
v B
1
B
2
B
3
ng dng v c cng
hng khi v ch khi:
a
2
a
1
a
3
a
1
=
b
2
b
1
b
3
b
1
.
Mnh 2.1.11. Tam gic A
1
A
2
A
3
v B
1
B
2
B
3
ng dng v c ngc
hng khi v ch khi:
a
2
a
1
a
3
a
1
=
b
2
b
1
b
3
b
1
.
Bi ton 1 Trn cc cnh AB, BC, CA ca tam gic ABC ta dng cc
tam gic ng dng v c cng hngADB, BEC, CFA. Chng minh
rng tam gic ABC v tam giac DEF c cng trng tm.
Gii V cc tam gic ADB, BEC, CFA ng dng v c cng hng nn
d a
b a
=
e b
c b
=
f c
a c
= z
T y ta c
d = a + (b a)z , e = b + (c b) z , f = c + (a c) z.
Suy ra
d +e +f
3
=
a +b +c
3
.
Vy tam gic ABC v tam giac DEF c cng trng tm.
Bi ton 2 Cho tam gic ABC trung im cc cnh AB, BC, CA ln
lt l M, N, P. Trn ng trung trc ca cc on thng
[AB] , [BC] , [CA] cc im A,B,C c chn pha trong tam gic sao
cho
32
MC
AB
=
NA
BC
=
PB
CA
.
Chng minh rng tam gic ABCvA
c cng trng tm.

Gii T
MC
AB
=
NA
BC
=
PB
CA
.
Ta c tan(
AB) = tan(
BC) = tan(
CB
A). V th tam gic

AC
B, BA
C, CB
A ng dng vi nhau . p dng kt qu bi trn ta

c li gii .
Bi ton 3 Cho tam gic ABO u vi tm S, tam gic u khc
A
O c cng hng vi tam giac ABO v S = A
, S = B
. Gi M v
N ln lt l trung im A
B v AB
. Chng minh rng tam gic SB
M
v SA
N ng dng.
( 30
th
IMO-Shortlist)
Gii
Gi R l bn knh ng trn ngoi tip tam gic ABO v
= cos
2
3
+ i sin
2
3
.
Xt mt phng phc vi gc S sao cho O nm trn chiu dng ca trc
thc. Ta ca cc im O, A, B l R, R, R
2
.
Xt R +z l ta ca B
khi ta A
l R z. Ta trung
im M, N l
z
M
=
z
B
+z
A
2
=
R
2
+R z
2
=
R(
2
+) z
2
=
R z
2
=
(R +z)
2
.
v
z
N
=
z
A
+z
B
2
=
R +R +z
2
=
R( + 1) + z
2
=
R
2
+ z
2
=
z
R
2
=
R z
2
.
Khi ta c
z
B
z
S
z
M
z
S
=
z
A
z
S
z
N
z
S
R + z
(R+z)
2
=
R z
Rz
2
. = 1 ||
2
= 1.
Vy gic SB
M v SA
N ng dng v ngc hng.

33
2.1.4 Tam gic u
Mnh 2.1.12. Gi s z
1
, z
2
, z
3
l ta cc nh ca tam gic
A
1
A
2
A
3
. Khi cc khng nh sau tng ng:
1) A
1
A
2
A
3
l tam gic u;
2) |z
1
z
2
| = |z
2
z
3
| = |z
3
z
1
| ;
3) z
2
1
+ z
2
2
+z
2
3
= z
1
z
2
+ z
2
z
3
+ z
3
z
1
;
4)
z
2
z
1
z
3
z
2
=
z
3
z
2
z
1
z
2
;
5)
1
z z
1
+
1
z z
2
+
1
z z
3
= 0 vi z =
z
1
+z
2
+ z
3
3
;
6)
_
z
1
+z
2
+
2
z
3
_ _
z
1
+
2
z
2
+ z
3
_
= 0 vi = cos
2
3
+ i sin
2
3
;
7)
1 1 1
z
1
z
2
z
3
z
2
z
3
z
1
= 0.
Mnh 2.1.13. Gi s z
1
, z
2
, z
3
l ta cc nh ca tam gic c
hng dng A
1
A
2
A
3
. Khi cc khng nh sau tng ng:
1) A
1
A
2
A
3
l tam gic u;
2) z
3
z
1
= (z
2
z
1
) vi = cos
3
+ i sin

3
;
3) z
2
z
1
= (z
3
z
1
) vi = cos
5
3
+ i sin
5
3
;
4) z
1
+z
2
+
2
z
3
= 0 vi = cos
2
3
+ i sin
2
3
.
Mnh 2.1.14. Gi s z
1
, z
2
, z
3
l ta cc nh ca tam gic c
hng m A
1
A
2
A
3
. Khi cc khng nh sau tng ng :
1) A
1
A
2
A
3
l tam gic u;
2) z
3
z
1
= (z
2
z
1
) vi = cos
5
3
+ i sin
5
3
;
3) z
2
z
1
= (z
3
z
1
) vi = cos
3
+ i sin

3
;
4) z
1
+
2
z
2
+ z
3
= 0 vi = cos
2
3
+ i sin
2
3
.
Mnh 2.1.15. Gi s z
1
, z
2
, z
3
l ta cc nh ca tam gic u
A
1
A
2
A
3
. Khi cc khng nh sau tng ng :
1) A
1
A
2
A
3
l tam gic u;
2) z
1
z
2
= z
2
z
3
= z
3
z
1
;
34
3) z
2
1
= z
2
z
3
v z
2
2
= z
1
z
3
.
Bi ton 1 V pha ngoi tam gic ABC dng ba tam gic u c
hng dng AC
B, BA
C, CB
A. Chng minh rng cc trng tm ca

ba tam gic l cc nh ca mt tam gic u.
(Napoleons problem)
Gii Gi a, b, c l ta ba nh A, B, C. S dng mnh trn ta c
a +c
+b
2
= 0 , b + a
+ c
2
= 0 , c + b
+a
2
= 0 (1)
Vi a
, b
, c
,l ta cc im A
, B
, C
. Trng tm cc tam gic

AC
B, BA
C, CB
A c ta l
a
=
1
3
(a
+b +c) , b
=
1
3
(a + b
+ c) , c
=
1
3
(a + b + c
).
Ta s kim tra c
+a
+ b
2
= 0. Tht vy
_
c
+ a
+ b
2
_
= (a +b +c
) + (a
+b +c) + (a +b
+c)
2
=
_
b + a
+ c
2
_
+
_
a +c
+ b
2
_
+
_
c + b
+ a
2
_
= 0.
.
Bi ton 2 Trn cc cnh ca tam gic, v pha ngoi ta dng ba a
gic u n cnh. Tm tt c cc gi tr ca n sao cho tm ca cc hnh n
gic l cc nh ca mt tam gic u.
(Balkan Mathematical Olympiad-Shortlist)
Gii Xt A
0
, B
0
, C
0
l trng tm ca cc n gic trn cc cnh
BC, CA, AB S o cc gc

AC
0
B ,

BA
0
C ,

AB
0
C l
2
n
K hiu a , b , c , a
0
, b
0
, c
0
l ta cc im A, B, C, A
0
, B
0
, C
0
. S
dng cc cng thc php quay ta c
a = c
0
+ (b c
0
) ;
b = a
0
+ (c a
0
) ;
c = b
0
+ (a b
0
) ;
35
Suy ra
a
0
=
b c
1
, b
0
=
c a
1
, c
0
=
a b
1
.
Tam gic A
0
B
0
C
0
u khi v ch khi
a
2
0
+ b
2
0
+ c
2
0
= a
0
b
0
+b
0
c
0
+ c
0
a
0
Thay cc gi tr a
0
, b
0
, c
0
ta c
(b c)
2
+ (c a)
2
+ (a b)
2
=
(b c) (c a) + (c a) (a b) + (a b) (b c) .
ng thc trn tng ng vi
_
1 + +
2
_
_
(a b)
2
+ (b c)
2
+ (c a)
2
_
= 0.
Suy ra 1 + +
2
= 0
2
n
=
2
3
ta c n = 3. Vy n = 3 tha mn yu
cu bi.
2.1.5 Hnh hc gii tch vi s phc
Phng trnh ng thng phng trnh ca ng thng trong mt
phng phc l
.z +z + = 0
vi C
, Rv z = x +yi C.
Mnh 2.1.16. Cho hai ng thng d
1
v d
2
c phng trnh
1
.z +
1
.z +
1
= 0 v
2
.z +
2
.z +
2
= 0
ng thng d
1
v d
2
l:
1) song song khi v ch khi

1
1
=

2
2
;
2) vung gc khi v ch khi

1
1
+

2
2
= 0;
3) Ct nhau khi v ch khi

1
1
=

2
2
;
36
Phng trnh ng thng xc nh bi hai im: Phng trnh
mt ng thng c xc nh bi hai im
P
1
(z
1
) , P
2
(z
2
) l
z
1
z
1
1
z
2
z
2
1
z z 1
= 0.
Din tch tam gic
nh l 2.1.17. Din tch tam gic A
1
A
2
A
3
vi ta cc nh z
1
, z
2
, z
3
bng modun ca s
i
4
z
1
z
1
1
z
2
z
2
1
z
3
z
3
1
.
H qu Din tch tam gic nh hng A
1
A
2
A
3
vi ta cc nh
z
1
, z
2
, z
3
l
area [A
1
A
2
A
3
] =
1
2
Im(z
1
z
2
+ z
2
z
3
+z
3
z
1
) .
Bi ton 1 Cho tam gic A
1
A
2
A
3
v cc im M
1
, M
2
, M
3
nm trn
ng thng A
2
A
3
, A
3
A
1
, A
1
A
2
. Gi s rng M
1
, M
2
, M
3
chia on thng
[A
2
A
3
] , [A
3
A
1
] , [A
1
A
2
] theo t s
1
,
2
,
3
. Chng minh rng :
area [M
1
M
2
M
3
]
area [A
1
A
2
A
3
]
=
1
1
3
(1
1
) (1
2
) (1
3
)
.
Gii Ta cc im M
1
, M
2
, M
3
l:
m
1
=
a
2
1
a
1
1
1
, m
2
=
a
3
2
a
1
1
2
, m
3
=
a
1
3
a
2
1
3
p dng cng thc (2) ta c
37
area [M
1
M
2
M
3
] =
1
2
(m
1
m
2
+ m
2
m
3
+ m
3
m
1
)
=
1
2
_
(a
2
1
a
3
) (a
3
2
a
1
)
(1
1
) (1
2
)
+
(a
3
2
a
1
) (a
1
3
a
2
)
(1
2
) (1
3
)
+
(a
1
3
a
2
) (a
2
1
a
3
)
(1
3
) (1
1
)
_
=
1
1
3
(1
1
) (1
2
) (1
3
)
area [A
1
A
2
A
3
]
Ch Cng thc dng (3) ta suy ra nh l Menelaus :Cc im
M
1
, M
2
, M
3
thng hng khi v ch khi
1
3
= 1 tng ng vi:
M
1
A
2
M
1
A
3
.
M
2
A
3
M
2
A
1
.
M
3
A
1
M
3
A
2
= 1.
Bi ton 2 Cho a, b, c l ta cc nh A, B, C ca mt tam gic. Bit
rng |a| = |b| = |c| = 1 v tn ti gc
_
0,

2
_
sao cho
a +b cos + c sin = 0. Chng minh rng
1 < area [ABC]
1 +
2
2
.
Gii Ta c
1 = |a|
2
= |b cos +c sin |
2
= (b cos +c sin )
_
b cos + c sin
_
= |b|
2
cos
2
+|c|
2
sin
2
+
_
bc + bc
_
sin cos
= 1 +
b
2
+ c
2
bc
sin cos.
T trn ta c b
2
+ c
2
= 0 nn b = ic. p dng h thc (2) ta c
38
area [ABC] =
1
2
Im
_
ab +bc +ca
_
=
1
2
Im
__
b cos c sin
_
b +bc c (b cos +c sin )
=
1
2
Im
_
cos sin bc sin bc cos + bc
_
=
1
2
Im
_
bc (sin + cos) bc
=
1
2
Im
_
(1 + sin +cos) bc
=
1
2
(1 + sin + cos) Im
_
bc
_
=
1
2
(1 + sin + cos) Im(icc)
=
1
2
(1 + sin + cos) Im(i) =
1
2
(1 + sin + cos)
=
1
2
_
1 +
2
_
2
2
sin +
2
2
cos
__
=
1
2
_
1 +
2 sin
_
+

4
__
Ta thy rng

4
< +

4
<
3
4
. V th
2
2
< sin
_
+

4
_
1. Nh vy
1 < area [ABC]
1 +
2
2
.
Phng trnh ng thng c xc nh bi im i qua v
phng
Mnh 2.1.18. Cho ng thng d : z +z + = 0 v im
P
0
(z
0
). Phng trnh ng thng di qua P
0
(z
0
) v song song vi d l :
z z
0
=
(z z
0
) .
Mnh 2.1.19. Cho ng thng d : z +z + = 0 v im
P
0
(z
0
). Phng trnh ng thng di qua P
0
(z
0
) v vung gc vi d l :
z z
0
=

(z z
0
) .
Hnh chiu vung gc ca mt im ln mt ng thng
Mnh 2.1.20. Cho im P
0
(z
0
) ng thng d:P
0
(z
0
) . Ta hnh
chiu ca im P
0
(z
0
) ln ng thng d l:
z =
z
0
z
0
2
.
39
Khong cch t mt im n mt ng thng
Mnh 2.1.21. Khong cch t im P
0
(z
0
) ti ng thng d:
z + z + = 0 , C
bng
D =
|z
0
+z
0
+ |
2
.
.
ng trn
Mnh 2.1.22. Phng trnh ca mt ng trn trong mt phng l
z.z + .z + .z + = 0,
vi C v R.
Phng tch ca mt im i vi mt ng trn
Mnh 2.1.23. Cho mt im P
0
(z
0
) v mt ng trn c phng
trnh z.
z +.z +.z + = 0
Vi C v R
Phng tch ca im P
0
vi ng trn trn l
(z
0
) = z
0
.z
0
+.z
0
+ .z
0
+.
2.1.6 Tch thc ca hai s phc
Khi nim tch v hng ca hai vc t tr thnh kinh in. Ta s
gii thiu khi nim i vi cc s phc. Trong nhiu trng hp s
dng tch ny lm cho li gii bi ton n gin i ng k. Xt a v b l
cc s phc
nh ngha 2.1.24. Ta gi tch thc ca hai s phc l s cho bi cng
thc
a.b =
1
2
_
ab + ab
_
.
40
D dng thy rng
a.b =
1
2
_
ab +ab
_
= a.b;
V th a.b l s thc, n th c th hin trong tn gi.
Nhng tnh cht di y ta d dng kim tra.
Mnh 2.1.25. Mi s phc a, b, c, z ta c cc h thc sau
1) a.a = |a|
2
;
2) a.b = b.a (tch thc c tnh cht giao hon);
3) a. (b + c) = a.b + a.c (tch thc phn phi i vi php cng);
4) (a) .b = (a.b) = a. (b) , R;
5) a.b = 0 khi v ch khi OAOB,vi A c ta a, B c ta b.
6) (az) . (bz) = |z|
2
(a.b) .
Mnh 2.1.26. Gi s A(a), B(b), C(c), D(d) l bn im phn bit.
Khi cc khng nh di y l tng ng:
1) ABCD ;
2) (b a) (c d) = 0 ;
3)
b a
d c
iR
(hoc l Re
_
b a
d c
_
= 0 ).
Mnh Cho tam gic ABC c tm ng trn ngoi tip l gc ta
trong mt phng phc. Nu a, b, c l ta ca A, B, Cth trc tm H
c ta l h = a +b +c.
Bi ton 1 Cho ABCD l cc nh ca mt t gic. Chng minh rng
AB
2
+ CD
2
= AD
2
+BC
2
Khi v ch khi ACBD.
Gii S dng tnh cht tch thc ca s phc ta c
AB
2
+ CD
2
= AD
2
+BC
2
khi v ch khi
(b a) . (b a) + (d c) . (d c) = (c b) . (c b) + (a d) . (a d)
a.b +c.d = b.c +d.a
(c a) . (d b) = 0.
41
Tng ng vi ACBD. iu phi chng minh.
Bi ton 2 Cho M, N, P, Q, R, S l trung im ca cc cnh
AB, BC, CD, DE, EF, FA ca mt luc gic. Chng minh rng
RN
2
= MQ
2
+ PS
2
khi v ch khi MQPS.
Gii Gi a, b, c, d, e, f l ta cc nh ca lc gic. Cc im
M, N, P, Q, R, S c ta l
m =
a + b
2
, n =
b +c
2
, p =
c +d
2
,
q =
d + e
2
, r =
e + f
2
, s =
f + a
2
S dng tnh cht tch thc ca s phc ta c
(e +f b c) . (+f b c)
= (d + e a b) . (d +e a b) + (f + a c d) . (f + a c d)
hay
(d + e a b) . (f + a c d) = 0
V th MQPS, pcm.
Bi ton 3 Cho A
1
A
2
...A
n
l mt a gic u ni tip trong ng
trn tm O bn knh R. Chng minh rng mi im M trong mt phng
ta u c h thc
n
k=1
MA
2
k
= n
_
OM
2
+ R
2
_
.
Gii Xt mt phng phc vi O l gc v R
k
l ta nh A
k
, vi
k
l cc cn bc n ca n v. Ly m l ta im M
S dng cc tnh cht tch thc ca cc s phc ta c
42
n
k=1
MA
2
k
=
n
k=1
(mR
k
). (mR
k
)
=
n
k=1
_
m.m2R
k
.m+ R
2
k
.
k
_
= n|m|
2
2R
_
n
k=1
k
_
.m+ R
2
n
k=1
|
k
|
2
= n.OM
2
+nR
2
= n
_
OM
2
+R
2
_
, v
n
k=1
k
= 0.
Ch Nu M nm trn ng trn ngoi tip a gic th
n
k=1
MA
2
k
= 2nR
2
Bi ton 4 Cho O l tm ng trn ngoi tip tam gic ABC , gi D
l trung im ca on thng AB, v E l trng tm tam gic ACD.
Chng minh rng CD v OE vung gc khi v ch khi AB = AC.
Gii Ly O l gc ta ca mt phng phc v a, b, c, d, e l ta cc
im A, B, C, D, E. Khi
d =
a +b
2
, e =
a +c +d
3
=
3a +b + 2c
3
.
S dng tch thc ca cc s phc, nu R l bn knh ng trn ngoi
tip tam gic ABC th a.a = b.b = c.c = R
2
ng thng CD v OE vung gc khi v ch khi (d c) .e = 0 v th
(a +b 2c) . (3a + b + 2c) = 0.
3a.a +a.b + 2a.c + 3a.b +b.b + 2b.c 6a.c 2b.c 4c.c = 0
a.b = a.c (1)
Mt khc, AB = AC tng ng vi
|b a|
2
= |c a|
2
(b a) . (b a) = (c a) . (c a)
b.b a.b a.b + a.a = c.c a.c. a.c + a.a
a.b = a.c (2)
43
T (1) v (2) thy rng CD vung gc vi OE khi v ch khi AB = AC.
Bi 5 Cho E,F,G,H l trung im cc cnh AB,BC,CD,DA ca t gic
ABCD. Chng minh rng ng thng AB v CD vung gc khi v ch
khi
BC
2
+AD
2
= 2
_
EG
2
+FH
2
_
.
Gii K hiu ch thng l ta ca cc im ch hoa tng ng. Ta c
e =
a +b
2
, f =
b +c
2
, g =
c + d
2
, h =
d + a
2
.
S dng tch thc ca s phc, h thc
BC
2
+AD
2
= 2
_
EG
2
+FH
2
_
.
tr thnh
(c b) . (c b) + (d a) . (d a) =
=
1
2
(c + d a b) . (c +d a b) +
1
2
(a + d b c) . (a +d b c)
c.c + b.b +d.d +a.a 2b.c 2a.d = a.a + b.b +c.c + d.d 2a.c 2b.d
a.d + b.c = a.c +b.d
.
H thc trn tr thnh (a b) . (d c) = 0 khi v ch khi ng thng
AB v CD vung gc.
2.1.7 Bi tp
Bi 1 Cho ABCD l mt hnh vung c nh. Xt tt c cc hnh vung
PQRS sao cho P, Q nm trn hai cnh khc nhau v Q nm trn mt
ng cho ca hnh vung ABCD. Tm tt c cc v tr c th c
ca S.
Bi 2 Cho t gic ABCD. Gi E, F, G, H theo th t l tm cc hnh
vung vi cc cnh AB, BC, CD, DA dng ra pha ngoi t gic. Chng
minh rng:
a) Trung im cc ng cho ca hai t gic ABCD, EFGHl nh
ca mt hnh vung.
b) EF v GH vung gc v bng nhau.
44
Bi 3 Trn ng trn cho trc hai im A, B c nh v mt im
M di ng trn . Trn tia MA ly im P sao cho MP = MB. Tm
qu tch im P.
Bi 4 Trn cc cnh AB, BC, CA ca tam gic ABC, dng ra pha
ngoi ba tam gic ng dng ABC
1
, A
1
BC, AB
1
C. Chng minh rng
trng tm hai tam gic ABC, A
1
B
1
C
1
c cng trng tm. Hi kt lun
bi ton cn ng khng nu cc tam gic ABC
1
, A
1
BC, AB
1
C dng
vo pha trong tam gic ABC?
Bi 5 Xt im M trn ng trn ngoi tip tam gic ABC, khng
trng vi cc nh ca tam gic. Chng minh rng tm ng trn Euler
ca cc tam gic MBC, MCA, MAB l nh ca mt tam gic ng
dng vi tam gic ABC.
Bi 6 Cc cnh AB, BC, CA ca tam gic ABC c chia thnh ba
on bng nhau bi cc im M, N; P, Q v R, S. V pha ngoi tam
gic ABC dng cc tam gic u MND, PQE, RSF. Chng minh rng
DEF l tam gic u.
Bi 7 V pha ngoi tam gic ABC dng cc hnh vung ABEF v
ADGH ln lt c tm l O v Q, M l trung im ca on BD.
Chng minh rng OMQ l tam gic vung cn ti M.
Bi 8 V pha ngoi t gic li ABCD, ta dng cc tam gic u
ABM, CDP; v pha trong tam gic, ta dng cc tam gic u
BCN, ADQ. Chng minh rng MNPQ l hnh bnh hnh.
Bi 9 Cho ABC l ba nh lin tip ca mt n gic u, M l mt im
nm trn ng trn ngoi tip n-gic u sao cho B v M nm khc
pha i vi AC. Chng minh rng: MB +MC = 2MBcos

n
.
Bi 10 Cho P l mt im bt k nm trn ng trn ngoi tip hnh
vung ABCD. Tm tt c cc s nguyn dng n sao cho i lng
:PA
n
+ PB
n
+ PC
n
+PD
n
khng ph thuc vo v tr ca P.
Bi 11 Cho tam gic ABC c cnh BC = a, CA = b, AB = c.M l mt
im bt k trong mt phng tam gic . Chng minh rng:
a) a.MA
2
+ b.MB
2
+c.MC
2
abc ;
b) a.MB.MC + b.MC.MA + c.MA.MA abc .
Bi 12 Cho tam gic nhn ABC c cnh BC = a, CA = b, AB = c.P l
45
mt im bt k trong mt phng tam gic. Chng minh rng:
a.PB.PC + b.PC.PA + c.PA.PA = abc
khi v ch khi P trng vi trc tm ca tam gic ABC.
Bi 13 Cho tam gic ABC c trng tm G. Gi R, R
1
, R
2
, R
3
ln lt l
bn knh ng trn ngoi tip tam gic ABC, GBC, GCA, GAB.
Chng minh rng R
1
+ R
2
+ R
3
3R .
2.2 S phc v cc bi ton i s , lng gic
2.2.1 Cc bi ton lng gic
Bi ton 1 Chng minh rng
cos

11
+ cos
3
11
+ cos
5
11
+cos
7
11
+ cos
9
11
=
1
2
.
Gii: t z = cos

11
+ i sin

11
ta c
z + z
3
+ z
5
+z
7
+z
9
=
z
11
z
z
2
1
=
1 z
z
2
1
=
1
1 z
.
Ta chng minh kt qu sau.
Nu z = cos t + i sin t v z = 1 th Re
_
1
1 z
_
=
1
2
Tht vy
1
1 z
=
1
1 (cos t + i sin t )
=
1
(1 cos t) i sin t
=
1
2 sin
2 t
2
2i sin
t
2
cos
t
2
=
1
2 sin
t
2
(sin
t
2
i cos
t
2
)
=
sin
t
2
+ i cos
t
2
2 sin
t
2
=
1
2
+ i
cos
t
2
2 sin
t
2
.
Suy ra Re
_
1
1 z
_
=
1
2
.
46
Bi ton 2 Tnh tch
P = cos20
o
.cos40
o
.cos80
o
Gii:t z = cos20
o
+ i sin 20
o
ta c z
9
= 1v z = cos20
o
- i sin 20
o
v
cos20
o
=
z
2
+ 1
2z
, cos40
o
=
z
4
+ 1
2z
2
, cos80
o
=
z
8
+ 1
2z
4
.
nn
P =
_
z
2
+ 1
_ _
z
4
+ 1
_ _
z
8
+ 1
_
8z
7
=
_
z
2
1
_ _
z
2
+ 1
_ _
z
4
+ 1
_ _
z
8
+ 1
_
8z
7
(z
2
1)
=
z
16
1
8 (z
9
z
7
)
=
z
7
1
8 (1 z
7
)
=
1
8
.
Cch gii 2
sin 20
o
.P = sin 20
o
.cos20
o
.cos40
o
.cos80
o
=
1
2
sin 40
o
.cos40
o
.cos80
o
=
1
4
sin 80
o
.cos80
o
=
1
8
sin 160
o
=
1
8
sin 20
o
.
Vy P =
1
8
.
Bi ton 3 Cho x, y, z l cc s thc tha mn sin x + sin y + sin z = 0
v cosx +cosy + cosz = 0.
Chng minh rng : sin 2x+sin 2y+sin 2z = 0 v cos2x+cos2y+cos2z = 0
Gii: t z
1
= cosx + i sin x , z
2
= cosy + i sin y , z
3
= cosz + i sin z , ta
c z
1
+ z
2
+z
3
= 0 v |z
1
| = |z
2
| = |z
3
| = 1 . T y
z
2
1
+z
2
2
+ z
2
3
= (z
1
+ z
2
+z
3
)
2
2 (z
1
z
2
+z
2
z
3
+ z
3
z
1
)
= 2z
1
z
2
z
3
_
1
z
1
+
1
z
2
+
1
z
3
_
= 2z
1
z
2
z
3
(z
1
+ z
2
+z
3
)
= 2z
1
z
2
z
3
(z
1
+ z
2
+ z
3
) = 0
Suy ra ( cos2x +cos2y + cos2z) + i (sin 2x + sin 2y + sin 2z) = 0 .
T y ta c iu phi chng minh.
47
cos
2
10
o
+ cos
2
50
o
+cos
2
70
o
=
3
2
.
Gii t z = cos10
o
+ i sin 10
o
ta c z
9
= i
cos10
o
=
z
2
+ 1
2z
, cos50
o
=
z
10
+ 1
2z
5
, cos70
o
=
z
14
+ 1
2z
7
.
ng thc cn chng minh tng ng vi
_
z
2
+ 1
2z
_
2
+
_
z
10
+ 1
2z
5
_
2
+
_
z
14
+ 1
2z
7
_
2
=
3
2
.
hay
z
16
+ 2z
14
+ z
12
+ z
24
+ 2z
14
+ z
4
+z
28
+ 2z
14
+ 1 = 6z
14
z
28
+z
24
+z
16
+z
12
+ z
4
+ 1 = 0.
S dng z
8
= 1 ta thu c
z
16
+ z
12
z
10
z
6
+ z
4
+ 1 = 0
_
z
4
+ 1
_ _
z
12
z
6
+ 1
_
= 0
_
z
4
+ 1
_ _
z
18
+ 1
_
z
6
+ 1
= 0.
Khng nh c chng minh.
Bi ton 5 Gii phng trnh
cos x +cos2x cos3x = 1.
Gii t z = cos x + i sin x ta c
cosx =
z
2
+ 1
2z
, cos2x =
z
4
+ 1
2z
2
, cos3x =
z
6
+ 1
2z
3
.
phng trnh tr thnh
48
z
2
+ 1
2z
+
z
4
+ 1
2z
2

z
6
+ 1
2z
3
= 1
z
4
+ z
2
+ z
5
+ z z
6
1 2z
3
0
_
z
6
z
5
z
4
+z
3
_
+
_
z
3
z
2
z + 1
_
= 0
_
z
3
+ 1
_ _
z
3
z
2
z + 1
_
= 0
_
z
3
+ 1
_
(z 1)
2
(z + 1) = 0
.
v z = 1 z = 1 , z
3
= 1 nn
x {2k |k Z} , x { + 2k |k Z} , x
_
+ 2k
3
|k Z
_
.
hp cc nghim ta c
x {k |k Z} , x
_
2k + 1
3
|k Z
_
.
Bi ton 6 Tnh tng S =
n
k=1
q
k
. cos kx v T =
n
k=1
q
k
. sin kx .
Gii Ta c
1 + S + iT =
n
k=1
q
k
(cos kx +i sin kx ) =
n
k=1
q
k
(cos x + i sin x)
k
=
1 q
n+1
(cos x + i sin x)
n+1
1 q cos x iq sin x
=
1 q
n+1
[cos (n + 1) x + i sin (n + 1) x]
1 q cos x iq sin x
=
_
1 q
n+1
cos (n + 1) x iq
n+1
sin (n + 1) x
[1 q cos x +iq sin x]

q
2
2q cos x + 1
.
v th
1 + S =
q
n+2
cos nx q
n+1
cos (n + 1) x q cos x + 1
q
2
2q cos x + 1
T =
q
n+2
sin nx q
n+1
sin (n + 1) x +q sin x
q
2
2q cos x + 1
.
49
Ch Khi q = 1 ta c h thc quen thuc
n
k=1
cos kx =
sin
nx
2
cos
(n+1)x
2
sin
x
2
v
n
k=1
sin kx =
sin
nx
2
sin
(n+1)x
2
sin
x
2
.
Tht vy
n
k=1
cos kx =
cos nx cos (n + 1) x (1 cos x)
2 (1 cos x)
=
2 sin
x
2
sin
(2n+1)x
2
2 sin
2 x
2
4 sin
2 x
2
=
sin
(2n+1)x
2
sin
x
2
2 sin
x
2
=
sin
nx
2
cos
(n+1)x
2
sin
x
2
.
v
n
k=1
sin kx =
sin nx sin (n + 1) x + sin x
2 (1 cos x)
=
2 sin
x
2
cos
x
2
2 sin
x
2
cos
(2n+1)x
2
4 sin
2 x
2
=
cos
x
2
cos
(2n+1)x
2
2 sin
x
2
=
sin
nx
2
sin
(n+1)x
2
sin
x
2
.
Bi ton 7 Cho cc im A
1
, A
2
, ...., A
10
, c xp theo th t v cch
u nhau trn ng trn bn knh R. Chng minh rng
A
1
A
4
A
1
A
2
= R.
Gii t z = cos

10
+ i sin

10
khng mt tnh tng qut ta cho R = 1.
Ta cn ch ra 2 sin
3
10
- sin

10
= 1.
Trong trng hp tng qut z = cos a+i sin a th sin a =
z
2
1
2iz
. ng
thc cn chng minh trn tr thnh
2
z
6
1
2iz
3

z
2
1
2iz
= 1 z
6
z
4
+ z
2
1 = 2iz
3
. V z
5
= i nn z
8
z
6
+ z
4
z
2
+ 1 = 0. ng thc ny ng v
_
z
8
z
6
+z
4
z
2
+ 1
_ _
z
2
+ 1
_
= z
10
+ 1 = 0 v z
2
+ 1 = 0
50
cos

7
cos
2
7
+ cos
3
7
=
1
2
.
_
5
th
IMO
_
Gii t z = cos
7
+ i sin

7
ta c z
7
+ 1 = 0 Do z = 1 v z
7
+ 1 =
(z + 1)
_
z
6
z
5
+ z
4
z
3
+ z
2
z + 1
_
= 0 nn z
6
z
5
+ z
4
z
3
+ =
0 z
_
z
2
z + 1
_
=
1
1 z
3
.
Ta c cos

7
cos
2
7
+ cos
3
7
= Re
_
z
3
z
2
+ z
_
.
Ta chng minh kt qu sau.
Nu z = cos t + i sin t v z = 1 th Re
_
1
1 z
_
=
1
2
Tht vy
1
1 z
=
1
1 (cos t + i sin t )
=
1
(1 cos t) i sin t
=
1
2 sin
2 t
2
2i sin
t
2
cos
t
2
=
1
2 sin
t
2
(sin
t
2
i cos
t
2
)
=
sin
t
2
+ i cos
t
2
2 sin
t
2
=
1
2
+ i
cos
t
2
2 sin
t
2
.
Suy ra Re
_
1
1 z
_
=
1
2
.
Bi ton 9 Chng minh rng:trung bnh cng ca cc s k sin k
o
(k = 2, 4, . . . , 180) l cot 1
o
.
(1996 USA Mathematical Olympiad)
Gii: t z = cos t + i sin t ta c ng thc
z + 2z
2
+... + nz
n
= (z +... + z
n
) +
_
z
2
+... + z
n
_
+ ... + z
n
=
1
z 1
__
z
n+1
z
_
+
_
z
n+1
z
2
_
+... +
_
z
n+1
z
n
_
=
nz
n+1
z 1

z
n+1
z
(z 1)
2
,
51
t y ta thu c hai ng thc
n
k=1
k cos kt =
(n + 1) sin
(2n+1)t
2
2 sin
t
2
1 cos (n + 1) t
4 sin
2 t
2
. (1)
v
n
k=1
k sin kt =
sin (n + 1) t
4 sin
2 t
2
ncos
(2n+1)t
2
2 sin
t
2
. (2)
S dng h thc (2) ta thu c
2 sin 2
o
+ 4 sin 4
o
+.... + 178 sin 178
o
= 2 (sin 2
o
+ 2 sin 2.2
o
+... + 89 sin 89.2
o
)
= 2
_
sin 90.2
o
4 sin
2
1
o

90 cos 179
o
2 sin 1
o
_
=
90 cos 179
o
sin 1
o
= 90 cot 1
o
.
Suy ra
1
90
(2 sin 2
o
+ 4 sin 4
o
+ .... + 178 sin 178
o
+ 180 sin 180
o
) = cot 1
o
.
Khng nh c chng minh.
Bi ton 10 Cho n l s nguyn dng. Tm s thc a
o
v
a
kl
, k, l = 1, n, k > l tha mn
sin
2
nx
sin
2
x
= a
0
+
1lkn
a
kl
cos2 (k l) x;
vi mi s thc x = m , m Z .
Gii: S dng ng nht thc
S
1
=
n
j=1
cos2jx =
sin nx cos (n + 1) x
sin x
S
2
=
n
j=1
sin 2jx =
sin nx sin (n + 1) x
sin x
Ta thu c
S
2
1
+S
2
2
=
_
sin nx
sin
_
2
.
52
mt khc
S
2
1
+S
2
2
= (cos2x +cos4x + ... + cos2nx)
2
+
(sin 2x + sin 4x +... + sin 2nx)
2
= n +
1lkn
(cos2kxcos 2lx + sin 2kxsin 2lx)
= n +
1lkn
cos2 (k l) x;
.
nh vy a
0
= n , a
kl
= 2 , 1 l < k n.
Bi ton 11 Tnh cc tng sau vi = 30
o
i) 1 +
cos
cos
+
cos (2)
cos
2
+
cos (3)
cos
3
+... +
cos ((n 1) )
cos
n1
;
ii) coscos +cos
2
cos (2) + cos
3
cos (3) + ... + cos
n
cos (n).
1 + cos
2n
_
n
_
+ cos
2n
_
2
n
_
+ ... +cos
2n
_
(n1)
n
_
= n.4
n
(2 + C
n
2n
)
vi mi s nguyn dng n 2 .
Bi ton 13 Cho s nguyn p 0. Tm cc s thc a
0
, a
1
, ..., a
p
vi
a
p
= 0 sao cho
cos 2p = a
0
+a
1
sin
2
+... + a
p
.
_
sin
2
_
p
vi mi R.
2.2.2 Cc bi ton i s
Bi ton 1(Vit Nam 1996) Gii h phng trnh
_
3x
_
1 +
1
x + y
_
= 2
_
7y
_
1
1
x +y
_
= 4
2
Gii Nhn thy x > 0, y > 0. t
x = u,
y = v. H cho tr thnh
_
_
u
_
1 +
1
u
2
+v
2
_
=
2
3
v
_
1
1
u
2
+ v
2
_
=
4
_
u +
1
u
2
+v
2
=
2
3
(1)
v
1
u
2
+ v
2
=
4
7
(2)
53
ly phng trnh th (2) nhn vi i sau cng vi phng trnh (1) ta
c u + iv +
u iv
u
2
+ v
2
=
2
3
+i
4
7
()
t z = u + iv th u
2
+v
2
= |z|
2
. Khi phng trnh (*) tr thnh
z +
z
|z|
2
=
2
3
+i
4
7
z +
1
z
=
2
3
+ i
4
7
z
2
_
2
3
+ i
4
7
_
z + 1 = 0
z =
1
21
+i
_
2
2
_
.
Suy ra
(u, v) =
_
1
21
,
2
2
_
.
H cho c hai nghim
(x, y) =
_
_
_
1
21
_
2
,
_
2
2
_
2
_
_
=
_
11
21

4
3
7
,
22
7

8
7
_
.
Bi ton 2 Gii h phng trnh
_
x
3
3xy
2
= 1
3x
2
y y
3
=
3
Gii
_
x
3
3xy
2
= 1
3x
2
y y
3
=
_
x
3
3xy
2
= 1
3x
2
(iy) + i
3
y
3
= i
_
x
3
+ 3x (iy)
2
= 1
3x
2
(iy) + (iy)
3
= i
_
(x + iy)
3
= 1 i
3 (1)
3x
2
(iy) + (iy)
3
= i
3 (2) .
54
t z = x +iy t (1) ta c z
3
= 1 i
3 = 2
_
cos
_
3
_
+ i sin
_
3
__
Suy ra
z =
3
2
_
cos
3
+ 2k
3
+ i sin
3
+ 2k
3
_
=
3
2
_
cos
+ 6k
9
+ i sin
+ 6k
9
_
vi k = 0, 1, 2.
T y ta c cc nghim ca h phng trnh cho l
(x, y) =
_
3
2cos
+ 6k
9
,
3
2 sin
+ 6k
9
_
vi k = 0, 1, 2.
2.2.3 Bi tp
Bi 1 Chng minh rng :
a) cos
2
+ cos
2
2 +... + cos
2
n =
sin (n + 1) cos n
2 sin
+
n 1
2
;
b) sin
2
+ sin
2
2 +... + sin
2
n =
n + 1
2

sin (n + 1) cosn
2 sin
;
.
a) cos
2
2n + 1
+ cos
4
2n + 1
+cos
6
2n + 1
+ ... +cos
2n
2n + 1
=
1
2
;
b) cos

16
=
_
_
2 +
2 + 2
2
;
c) sin
2
5
=
_
10 + 2
5
2
.
.
Bi 3 Chng minh rng vi mi n chn (n = 2m),ta u c:
cos n =
cos
n
C
2
n
cos
n2
sin
2
+ C
4
n
cos
n4
sin
4
... + (1)
m
C
n
n
sin
n
.
55
a) sin

2n + 1
sin
2
2n + 1
... sin
n
2n + 1
=
2n + 1
2
n
;
b) cos

2n + 1
cos
2
2n + 1
...cos
n
2n + 1
=
1
2
n
;
c) sin

2n
sin
2
2n
... sin
(n 1)
2n
=
n
2
n1
;
d) cos

2n
cos
2
2n
cos
(n 1)
2n
=
n
2
n1
.
Bi 5 Gii phng trnh
a) x
5
+ x
4
+x
3
+ x
2
+x + 1 = 0 ;
b) x
5
+x
4
+
2
x
3
+
3
x
2
+
4
x +
5
= 0 , 0 = C .
Bi 6 Chng minh rng vi mi n l (n = 2m+ 1),ta u c:
sin n =
C
1
n
cos
n1
sin C
3
n
cos
n3
sin
3
+... + (1)
m1
C
n1
n
sin
n3
cos
3
.
Bi 7 Gii h phng trnh :
_
_
x +
3x y
x
2
+ y
2
= 3
y
x + 3y
x
2
+ y
2
= 0.
Bi 8 Gii h phng trnh :
_
x
_
1
12
3x +y
_
= 2
y
_
1 +
12
3x +y
_
= 6.
2.3 S phc v cc bi ton t hp
Bi ton 1 Tnh tng
3n1
k=0
(1)
k
C
2k+1
6n
3
k
.
56
Gii Ta c
3n1
k=0
(1)
k
C
2k+1
6n
3
k
=
3n1
k=0
C
2k+1
6n
(3)
k
=
3n1
k=0
C
2k+1
6n
_
i
3
_
2k
=
1
i
3
3n1
k=0
C
2k+1
6n
_
i
3
_
2k+1
=
1
i
3
Im
_
1 + i
3
_
6n
=
1
i
3
Im
_
2
_
cos
3
+ i sin

3
__
6n
=
1
i
3
_
2
6n
(cos2n + i sin 2n)
= 0.
Bi ton 2 Tnh tng
S
n
=
n
k=0
C
k
n
cos k
vi [0, ] .
Gii Ta c t z = cos + i sin v T
n
=
n
k=0
C
k
n
sin k. Ta c
S
n
+ iT
n
=
n
k=0
C
k
n
(cos k +i sin k) =
n
k=0
C
k
n
(cos +i sin )
k
=
n
k=0
C
k
n
z
k
= (1 +z)
k
. (1)
Dng ta cc ca s phc 1 + z l
1 + cos + i sin = 2 cos
2

2
+ 2i sin

2
cos
2
= 2cos
2
_
cos
2
+i sin

2
_
.
V [0, ] ng thc (1) tr thnh
S
n
+ iT
n
=
_
2cos
2
_
n
_
cos
n
2
+ i sin
n
2
_
.
T y ta thu c
S
n
=
_
2cos
2
_
n
cos
n
2
T
n
=
_
2cos
2
_
n
sin
n
2
.
57
Biton 3 Chng minh ng thc
_
C
0
n
C
2
n
+C
4
n
....
_
2
+
_
C
1
n
C
3
n
+ C
5
n
....
_
2
= 2
n
.
Gii t x
n
= C
0
n
C
2
n
+ C
4
n
.... v y
n
= C
1
n
C
3
n
+ C
5
n
....ta c
(1 + i)
n
= x
n
+iy
n
Ly modun hai v ta thu c |x
n
+ iy
n
| = |(1 + i)
n
| = |1 + i|
n
= 2
n
iu ny tng ng vi x
2
n
+y
2
n
= 2
n
. Khng nh c chng minh.
Ch Ta c th xy dng cng thc tnh x
n
, y
n
nh sau
(1 + i)
n
=
_
2
_
cos
4
+ i sin

4
__
n
= 2
n
2
_
cos
n
4
+ i sin
n
4
_
.
Suy ra x
n
= 2
n
2
cos
n
4
v y
n
= 2
n
2
sin
n
4
.
Biton 4 Chng minh ng thc
C
m
n
+C
m+p
n
+ C
m+2p
n
+ ... =
2
n
p
p1
k=0
_
cos
k
p
_
n
cos
(n 2m) k
p
.
Gii Gi
0
,
1
, ...,
p1
l cc cn bc p ca n v. Khi
p1
k=0
m
k
(1 +
k
)
n
=
n
k=0
C
k
n
_
km
0
+... +
km
p1
_
. (1)
Ta c kt qu quen thuc sau:
km
0
+ ... +
km
p1
=
_
p , p | (k m)
0 , p | (k m)
.(2)
Xt
m
k
(1 +
k
)
n
=
_
cos
2mk
p
i sin
2mk
p
__
2 cos
k
p
_
n
_
cos
2nk
p
+ i sin
2nk
p
_
= 2
n
_
cos
k
p
_
n
_
cos
(n 2m) k
p
+ i sin
(n 2m) k
p
_
58
S dng (1) v (2) d dng suy ra h thc cn chng minh.
Ch : T trn ta c h thc lng gic th v sau :
p1
k=0
_
cos
k
p
_
n
sin
(n 2m) k
p
= 0
Bi ton 5 Cho a
n
, b
n
, c
n
l cc s nguyn vi
a
n
= C
0
n
+C
3
n
+ C
6
n
+ ;
b
n
= C
1
n
+C
4
n
+ C
7
n
+ ;
c
n
= C
2
n
+C
5
n
+ C
8
n
+ ;
Chng minh rng
1) a
3
n
+ b
3
n
+c
3
n
3a
n
b
n
c
n
= 2
n
;
2)a
2
n
+ b
2
n
+c
2
n
a
n
b
n
b
n
c
n
c
n
a
n
= 1.
3) Hai trong ba s nguyn a
n
, b
n
, c
n
bng nhau cn s th ba bng 1
Gii 1) Gi l cn bc ba ca n v v khc 1. Ta c
(1 + 1)
n
= a
n
+b
n
+c
n
, (1 + )
n
= a
n
+b
n
+c
n
2
,
_
1 +
2
_
n
= a
n
+b
n
2
+c
n
v th
a
3
n
+b
3
n
+ c
3
n
3a
n
b
n
c
n
= (a
n
+ b
n
+c
n
)
_
a
n
+ b
n
+ c
n
2
_ _
a
n
+b
n
2
+ c
n
2
n
_
= 2
n
(1 + )
n
_
1 +
2
_
n
= 2
n
_
2
_
n
()
n
= 2
n
2) S dng ng thc quen thuc
x
3
+y
3
+z
3
3xyz = (x + y +z)
_
x
2
+y
2
+z
2
xy yz zx
_
.
t h thc trn ta c
a
2
n
+b
2
n
+ c
2
n
a
n
b
n
b
n
c
n
c
n
a
n
= 1
3)a
2
n
+ b
2
n
+c
2
n
a
n
b
n
b
n
c
n
c
n
a
n
= 1
(a
n
b
n
)
2
+ (b
n
c
n
)
2
+ (c
n
a
n
)
2
= 2
t ng thc ny suy ra:hai trong ba s a
n
, b
n
, c
n
bng nhau v s th
ba bng 1.
Ch T bi ton 5 ta thu c
a
n
=
1
3
_
2
n
+cos
n
3
+ (1)
n
cos
2n
3
_
=
1
3
_
2
n
+ 2cos
n
3
_
b
n
=
1
3
_
2
n
+ cos
(n 2)
3
+ (1)
n
cos
(2n 4)
3
_
=
1
3
_
2
n
+ 2cos
(n 2)
3
_
c
n
=
1
3
_
2
n
+ cos
(n 4)
3
+ (1)
n
cos
(2n 8)
3
_
=
1
3
_
2
n
+ 2cos
(n 4)
3
_
.
59
D thy rng
a
n
= b
n
khi v ch khi n 1 mod (3) ;
b
n
= c
n
khi v ch khi n 2 mod (3) ;
c
n
= a
n
khi v ch khi n 3 mod (3) .
Bi ton 6 C bao nhiu s c n ch s chn t tp hp {2, 3, 7, 9} v
chia ht cho 3?
(Romanian Mathematical Regional Contest Trian Lalescu)
Gii:Gi x
n
, y
n
, z
n
ln lt l s cc s nguyn c n ch s chn t tp
hp {2, 3, 7, 9} v ln lt ng d vi 0, 1, 2 theo modun3. Ta cn tm
cc gi tr x
n
.
t = cos
2
3
+ i sin
2
3
, ta thy rng x
n
+ y
n
+z
n
= 4
n
v
x
n
+ y
n
+
2
z
n
=
j
1
+j
2
+j
3
+j
4
=n
2j
1
+3j
2
+7j
3
+9j
4
=
_
2
+
3
+
7
+
9
_
n
= 1.
hay x
n
1 + y
n
+
2
z
n
= 0. Ta thu c x
n
1 = y
n
= z
n
= k. Khi
3k = x
n
+y
n
+z
n
1 = 4
n
1 dn n
k =
1
3
(4
n
1). Suy ra x
n
= k + 1 =
1
3
(4
n
+ 2) .
Bi ton 7 Cho s nguyn t n v cc s nguyn dng a
1
, a
2
, ..., a
m
.
Gi f (k) l s cc b m s (c
1
, c
2
, ..., c
m
) tha mn iu kin 0 c
i
a
i
v c
1
+ c
2
+ ... +c
m
k (modm). Chng minh rng
f (0) = f (1) = ... = f (n 1) khi v ch khi n|a
j
vi j no thuc tp
{1, 2, ..., m} .
Gii Xt = cos
2
n
+ i sin
2
n
. rng ng thc sau lun ng
n
k=1
_
X +X
2
+ ... + X
a
k
_
=
1c
k
a
k
X
c
1
+...+c
n
v
f (0)+f (1) +...+f (n 1)
n1
=
1c
k
a
k
X
c
1
+...+c
n
=
m
k=1
_
+
2
+... +
a
k
_
60
suy ra f (0) = f (1) = ... = f (n 1) khi v ch khi f (0)+f (1) +...+
f (n 1)
n1
= 0. iu ny tng ng vi
m
k=1
_
+
2
+... +
a
k
_
= 0
tc l +
2
+... +
a
k
= 0 vi j no thuc tp {1, 2, ..., m}
Bi ton 8 Cho s nguyn t n v cc s nguyn dng a
1
, a
2
, ..., a
m
.
Gi f (k) l s cc b m s (c
1
, c
2
, ..., c
m
) tha mn iu kin 0 c
i
a
i
v c
1
+c
2
+ ... +c
m
k (modm). Chng minh rng
f (0) = f (1) = ... = f (n 1) khi v ch khi n|a
j
vi j no thuc tp
{1, 2, ..., m} .
_
36
th
IMO
_
Gii Trng hp p = 2 l tm thng. Xt p 3 v = cos
2
3
+i sin
2
3
.
K hiu x
j
l s cc tp con vi tnh cht |B| = p v m(B)
j (modp).Khi
n1
j=0
x
j
j
=
BA,|B|=p
m(B)
=
1c
1
<...<c
p
2p
c
1
+c
2
+...+c
p
tng trn chnh l h s ca X
p
trong a thc (X +)
_
X +
2
_
...
_
X +
2p
_
T h thc quen thuc X
p
1 = (X 1) (X ) ....
_
X
p1
_
ta
thu c (X + )
_
X +
2
_
...
_
X +
2p
_
= (X
p
+ 1)
2
, v th h s ca
X
p
l 2.Suy ra
p1
j=0
x
j
j
= 2 x
0
2 + x
1
+ ... + x
p1
p1
= 0 ta c
x
0
2 = x
1
= ... = x
p1
= k. Dn n pk = x
0
+x
1
+...+x
p1
2 = C
p
2p
2
hay k =
1
p
_
C
p
2p
2
_
Kt qu bi ton cho l x
0
= 2 +k = 2 +
1
p
_
C
p
2p
2
_
Bi ton 9 Chng minh rng s
n
k=0
_
C
2k+1
2n+1
_
2
3k
khng chia ht cho 5 vi
mi s nguyn n 0.
Bi ton 10 Tnh tng
n
k=0
_
C
k
n
_
2
cos kt
61
vi t [0, ]
Bi ton 11 Chng minh ng thc sau
1)C
0
n
+ C
4
n
+ C
8
n
+... =
1
4
_
2
n
+ 2
n
2
+1
+ cos
n
4
_
.
(Romanian Mathematicanl Olympiad Second Round 1981)
2)C
0
n
+C
5
n
+C
10
n
+... =
1
5
_
_
2
n
+
_
5 + 1
_
n
2
n1
cos
n
5
+
_
5 1
_
n
2
n1
cos
2n
5
]
Bi 12 Cho cc s nguyn A
n
, B
n
, C
n
xc nh nh sau
A
n
= C
0
n
C
3
n
+ C
6
n
... ,
B
n
= C
1
n
+C
4
n
C
7
n
+ ... ,
C
n
= C
2
n
C
5
n
+ C
8
n
... .
Chng minh cc ng thc sau
1) A
2
n
+ B
2
n
+ C
2
n
A
n
B
n
B
n
C
n
C
n
A
n
= 3
n
;
2) A
2
n
+ A
n
B
n
+B
2
n
= 3
n1
.
62
Kt lun
Lun vn trnh by hng tip cn mi mt s dng ton s cp
bng cng c s phc.Cc kt qu chnh ca lun vn:
S dng s phc gii ton hnh hc phng.
S dng s phc gii mt s lp ton i s ,lng gic.
S dng s phc gii mt s bi ton t hp.
Hng nghin cu tip theo ca lun vn l tip tc p dng s phc
gii cc bi ton:t mu, s hc, a thc , phng trnh hm.
63
Ti liu tham kho
[1] Titu Andreescu, Dorin Andrica, Complexnumbes from A to...Z ,
Birkhauser, 2006.
[2] Bchanu, Internatoonal Mathematical Olympiads 1959 2000. Proplem.
Solution. Results, Acdemic Distri Center, Free, USA, 2001
[3] on Qunh , S phc vi Hnh hc phng, Nh xut bn gio dc,
1998.
[4] Nguyn Huy oan (Ch Bin), Gii tch 12, Nh xut bn gio dc,
2008.
[5] Nguyn Thy Thanh, C s l thuyt hm bin phc, Nh xut bn
i hc quc gia H ni,2007

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I HC THI NGUYN

v mi s nguyn m, n ta c cc tnh cht

(x, y) i xng vi M(x, y) qua truc ta Ox.

cng vi cc im c ng nht vi s phc gi l

OM vi chiu dng ca trc ta Ox gi l argumen cc ca

[0, 2) l ta cc dng hnh hc ca s phc

2 .Chng minh rng

l biu din hnh hc ca z, ta thy rng im M

c cng trng tm.

A). V th tam gic

A ng dng vi nhau . p dng kt qu bi trn ta

O c cng hng vi tam giac ABO v S = A

. Chng minh rng tam gic SB

N ng dng v ngc hng.

A. Chng minh rng cc trng tm ca

. Trng tm cc tam gic

[1 q cos x +iq sin x]

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