Về một phương pháp giải toán sơ cấp.pdf

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

1Li 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.


1Mc 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


22.1.5 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


3M 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


4Do 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


5Chng 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 RTa xt tp hp

R2 = R R = {(x, y) |x, y R } .Hai phn t (x1, y1) v (x2, y2) bng nhau khi v ch khi{

x1 = x2

y1 = y2

Cc php ton cng v nhn c nh ngha trn R2 nh sau :

z1 + z2 = (x1, y1) + (x2, y2) = (x1 + x2, y1 + y2) R2.v

z1.z2 = (x1, y1) . (x2, y2) = (x1x2 y1y2, x1y2 + x2y1) R2.vi mi z1 = (x1, y1) R2 v z2 = (x2, y2) R2. Phn t z1 + z2 gi ltng ca z1, z2 , phn t z1.z2 R2 gi l tch ca z1, z2.Nhn xt

1) Nu z1 = (x1, 0) R2 v z2 = (x2, 0) R2 th z1z2 = (x1x2, 0).2))Nu z1 = (0, y1) R2 v z2 = (0, y2) R2 th z1z2 = (y1y2, 0).nh ngha 1.1.1. Tp hp R2 cng vi php cng v nhn gi l tp sphc, 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)} .


61.2 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 : z1 + z2 = z2 + z1 vi mi z1, z2 C.Tnh kt hp :(z1 + z2) + z3 = z1 + (z2 + z3) vi mi z1, z2, z3 C.Phn t n v : C duy nht mt s phc 0 = (0, 0) C z+ 0 = 0 + zvi 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:z1z2 = z2z1 vi mi z1, z2 C.Tnh kt hp:(z1z2)z3 = z1(z2z3) vi mi z1, z2, z3 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 6= 0 c duy nht sphc z1 = (x,, y,) C sao cho z.z1 = z1z = 1 s phc z1 = (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 nhsau z0 = 1 ; z1 = z ; z2 = z.z ,v zn = z.z...z

n l n

vi mi s nguyn n > 0

v zn = (z1)n vi mi s nguyn n < 0.Mi s phc z1, z2, z3 C v mi s nguyn m,n ta c cc tnh cht

sau

1) zm.zn = zm+n;

2)zm

zn= zmn;

3) (zm)n = zmn;

4) (z1z2)n = zn1 z

n2;

5)

(z1z2

)n=zn1zn2

;

Khi z = 0 ta nh ngha 0n = 0 vi mi s nguyn n > 0.


7Tnh phn phi : z1 (z2 + z3) = z1z2 + z1z3 vi mi z1, z2, z3 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 viphp ton cng v nhn c nh ngha trn R2.

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 ngnht cp s (x, 0) vi s x, vi mi x R. Ta s dng song nh trn v khiu (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 i2 = 1 c suy ra t nh ngha php nhn i2 = i.i =

(0, 1).(0, 1) = (1, 0) = 1.


8Biu 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 , i2 = 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) z1 = z2 khi v ch khi Re(z1) = Re(z2) v Im(z1) = Im(z2).

b) z R khi v ch khi Im(z) = 0.c) z C\R khi v ch khi Im(z) 6= 0.S dng dng i s, cc php ton v s phc c thc hin nh sau:

Php cng

z1 + z2 = (x1 + y1i) + (x2 + y2i) = (x1 + x2) + (y1 + y2)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(z1 + z2) = Re(z1) + Re(z2);

Im(z1 + z2) = Im(z1) + Im(z2).

Php tr

z1 z2 = (x1 + y1i) (x2 + y2i) = (x1 x2) + (y1 y2)i C.Ta c

Re(z1 z2) = Re(z1) Re(z2);Im(z1 z2) = Im(z1) Im(z2).

Php nhn

z1.z2 = (x1 + y1i).(x2 + y2i) = (x1x2 y1y2) + (x1y2 + x2y1) i C.Ta c

Re(z1z2) = Re(z1) Re(z2) Im(z1) Im(z2);Im(z1z2) = Im(z1) Re(z2) + Im(z2) Re(z1).

Mi s thc , s phc z = x+ yi, z = (x+ yi) = x+ yi C ltch ca mt s thc vi mt s phc. Ta c cc tnh cht sau


91) (z1 + z2) = z1 + z2;

2) 1(2z) = (12)z;

3)(1 + 2)z = 1z + 2z.

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

i0 = 1 ; i1 = i ; i2 = 1 ; i3 = i2.i = i

i4 = i3.i = 1; i5 = i4.i = i ; i6 = i5.i = 1; i7 = i6.i = i

Ta c th tng qut cc cng thc trn i vi s m nguyn dng n

i4n = 1 ; i4n+1 = i ; i4n+2 = 1 ; i4n+3 = iV th in {1 , 1 ,i , i} vi mi s nguyn n > 0. Nu n l s

nguyn m ta c:

in =(i1)n

=

(1

i

)n= (i)n .

S phc lin hp

Mi s phc z = x+ yi u c s phc z = x yi, s phc c gil 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)z1 + z2 = z1 + z2 (s phc lin hp ca mt tng bng tng cc s phc

lin hp);

5)z1.z2 = z1.z2(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 z1 = z1;


10

7)

(z1z2

)=

z1z2, z2 6= 0 (lin hp ca mt thng bng thng cc lin

hp);

8)Cng thc Re(z) =z + z

2v Im(z) =

z z2i

, ng vi mi s phc

z C.

Ghi ch

a) phn t nghch o ca s phc z C c th c tnh nh sau1

z=

z

z.z=

x yix2 + y2

=x

x2 + y2 yx2 + y2

i.

b) S phc lin hp c s dng trong vic tm thng ca hai s phc

nh sau:

z1z2

=z1.z2z2z2

=(x1 + y1i) (x2 y2i)

x22 + y22

=x1x2 + y1y2x22 + y

22

+x1y2 + x2y1x22 + y

22

i.

Modun ca s phc

S |z| = x2 + y2 c gi l modun ca s phc z = x+ yi.Mnh 1.3.3. 1) |z| 6 Re(z) 6 |z| v |z| 6 Im(z) 6 |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)|z1z2| = |z1| . |z2| (m un ca mt tch bng tch cc m un);6) |z1| |z2| 6 |z1 + z2| 6 |z1|+ |z2|;7)z1 = |z|1 , z 6= 0;

8)

z1z2 = |z1||z2| , z2 6= 0 (m un ca mt tch bng tch cc m un);

9)|z1| |z2| 6 |z1 z2| 6 |z1|+ |z2| .

1.3.2 Gii phng trnh bc hai

By gi chng ta c th gii phng trnh bc hai vi h s thc:

ax2 + bx+ c = 0 , a 6= 0


11

trong trng hp bit thc = b2 4ac nhn gi tr m.Bng cch bin i, d dng a phng trnh v dng tng ng

sau

a

[(x+

b

2a

)2+4a2

]= 0.

Do (x+

b

2a

)2 i2

(2a

)2= 0.

V th

x1 =b+ i

2a, x2 =

b i2a

.

Cc nghim trn l cc s phc lin hp ca nhau v ta c th phn tch

thnh tha s nh sau

ax2 + bx+ c = a (x x1) (x x2) .

By gi chng ta xt phng trnh bc hai tng qut vi h s phc

az2 + bz + c = 0 , a 6= 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+4a2

]= 0.

ng thc trn tng ng vi(z +

b

2a

)2=

4a2

hoc (2az + b)2 = .

Vi = b2 4ac cng c gi l bit thc ca phng trnh bc hai.t y = 2az + b phng trnh trn c rt gn v dng

y2 = = u+ vi


12

vi u,v l cc s thc

Phng trnh trn c li gii

y1,2 = (

r + u

2+ (sgn v)

r u

2i

),

Vi r = || ,v sgnv l du ca s thc vNghim ban u ca phng trnh l:

z1,2 =1

2a(b+ y1,2) .

Ta c mi lin h gia cc nghim v h s:

z1 + z2 = ba, z1.z2 =

c

a.

Khi phn tch ra tha s

az2 + bz + c = a (z z1) (z z2).

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 phcz = 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 imM (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 gil 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 V0 l tp hp cc vc t c im gc l gc ta O. Ta c th nh

ngha song nh : C V0 , (z) = OM = xi + yj , vi i , j l ccvc 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 =

(xM xO)2 + (yM yO)2.

V th OM =x2 + y2 = |z| = |v | m un |z| ca s phc z = x + yi

l di ca on thng OM hoc l ln ca vc t v = xi + yj .

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 trnC(O; r). Cc s phc z vi|z| > r l cc im nm bn ngoi ng trnC(O; r).

1.3.4 ngha hnh hc ca cc php ton i s

a) Php cng v php tr

Xt hai s phc z1 = x1 + y1i v z2 = x2 + y2i tng ng vi hai vc

t v1 = x1i + y2j v v2 = x2i + y2j .


14

Tng ca hai s phc l

z1 + z2 = (x1 + x2) + (y1 + y2) i.

Tng hai vc t

v1 +v2 = (x1 + x2)i + (y1 + y2)j .

V th z1 + z2 tng ng viv1 +v2 .

Hon ton tng t i vi php tr

Hiu ca hai s phc l

z1 z2 = (x1 x2) + (y1 y2) i.

Hiu hai vc t

v1 v2 = (x1 x2)i + (y1 y2)j .

V th z1 z2 tng ng vi v1 v2 .

Ch

Khong cch gia M1 (x1, y1) v M2 (x2, y2) bng m un ca s phc

z1 z2 hoc di ca vc t v1 v2 . Vy :

M1M2 = |z1 z2| = |v1 v2 | =

(x2 x1)2 + (y2 y1)2.

b) Tch ca s thc v s phc

Xt s phc z = x + yi tng ng vi vc t v = xi + yj . Nu l s thc , th tch s thc z = x + yi tng ng vi vc tv = x

i + y

j .

Ch : Nu > 0 th vc tv v v cng hng v |v | = |v |,

nu < 0 th vc tv 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 =x2 + y2 gi l bn knh cc ca imM . Gc nh hng t [0, 2pi)

gia vc tOM 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, 2pi) , h ((x, y)) = (r, t)l song nh.

Gc ta O l im duy nht sao cho r = 0 , argumen t ca gckhng 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 dngnh 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 =x2 + y2. Ta xc

nh argument cc trong cc trng hp sau

a)Nu x 6= 0 , t tan t = yx

ta suy ra

t = arctany

x+ kpi

Vi

k =

0 khi x > 0 , y > 01 khi x < 0 , y R2 khi x > 0 , y < 0

b)Nu x = 0 v y 6= 0 th

t =

pi

2khi y > 0

3pi

2khi 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, 2pi) l ta cc dng hnh hc ca s phcz.

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 6= 0 m un v argument ca z c xc nhmt cch duy nht.

Xt z = r (cos t + i sin t) v t = t + 2kpi vi k l s nguyn th

z = r (cos (t 2kpi) + i sin (t 2kpi)) = 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 vt R. Tp hp Arg z = {t = t + 2kpi , k Z} c gi l arguent mrng ca s phc z.

V th, hai s phc z1, z2 6= 0 c dngz1 = r1 (cos t1 + i sin t1) v z2 = r2 (cos t2 + i sin t2)

bng nhau khi v ch khi r1 = r2 v t1 t2 = 2kpi, vi k l s nguyn.Ch Cc dng sau nn nh

1 = cos0 + i sin 0 , i = cospi

2+ i sin

pi

2

1 = cospi + i sin pi , i = cos3pi2

+ i sin3pi

2.

1.4.3 Cc php ton s phc trong ta cc

Php nhn Gi s rng

z1 = r1 (cos t1 + i sin t1) v z2 = r2 (cos t2 + i sin t2)

th

z1z2 = r1r2 (cos (t1 + t2) + i sin (t1 + t2)) .


17

Ly tha ca mt s phc (De moirve) Cho z = r (cos t+ i sin t) ,

n N, ta czn = rn (cosnt+ i sinnt) .

Php chia Gi s rng

z1 = r1 (cos t1 + i sin t1) v z2 = r2 (cos t2 + i sin t2)

thz1z2

=r1r2

(cos (t1 t2) + i sin (t1 t2)) .

1.4.4 ngha hnh hc ca php nhn

Xt

z1 = r1 (cos t1 + i sin t

1)

v

z2 = r2 (cos t2 + i sin t

2) .

Biu din hnh hc ca chng lM1(r1, t1) , M2(r2, t

2). Gi P1 , P2 ln

lt l giao im ca C(O, 1) vi cc tia (OM1 v (OM2 . Ly P3 C(O, 1)vi argument cc l t1+t

2v chnM3 (OP3 sao cho OM3 = OM1.OM2.

Ly z3 c ta M3. im M3 (r1r2, t1 + t

2) l dng hnh hc z1.z2

Ly A l dng hnh hc ca s phc 1 . V

OM3OM1

=OM2

1 OM3

OM2=OM2OA

v M2OM3 = AOM1 nn hai tam gic M2OM3 v AOM1 ng dng.

Khi biu din dng hnh hc ca mt thng ch rng dng hnh hc

caz3z2

l im M1.

1.4.5 Cn bc n ca n v

Cho s nguyn dng n > 2 v s phc z0 6= 0, ging nh trn trngs thc, phng trnh

Zn z0 = 0


18

c s dng nh ngha cn bc n ca s z0. V vy mi mt gi tr Z

tha mn phng trnh trn l mt cn bc n ca z0.

nh l 1.4.1. Cho z0 = r (cos t + i sin t) l s phc vi r > 0 v

t [0, 2pi) S phc z0 c n cn bc n phn bit cho bi cng thc

Zk =nr

(cos

t + 2kpin

+ i sint + 2kpi

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 Zn = z0 hay

n (cosn+ i sinn) = r (cos t + i sin t) .

Ta c n = r v n = t + 2kpi vi k Z . V th = nr vk =

t

n+ k.

2pi

nvi k Z. Do nghim ca (1) l

Zk =nr

(cos

t + 2kpin

+ i sint + 2kpi

n

)vi k Z.

Nhn thy rng 0 6 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 phnbit ca z0:Z0, Z1 , ...., Zn1 . 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 =tn

+ (nq + r)2pi

n=tn

+ r2pi

n+ 2qpi = r + 2qpi.

Nhn thy Zk = Zr do

{Zk : k Z} = {Z0 , Z1, ..., Zn1} .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 nr.


19

Ta chng minh iu trn nh sau, k hiu M0,M1, ...,Mn1 l ccim c ta phc Z0 , Z1, ..., Zn1. V OMk = |Zk| = n

r vi k

{0, 1, ..., n 1} nn cc im Mk nm trn ng trn C (O, rn). Bn

cnh , s o ca cung MkMk+1 bng

argZk+1 argZk = t + 2 (k + 1) pi (t + 2kpi)

n=

2pi

n,

vi k {0, 1, ...., n 2} v s o cung Mn1M0 l 2pin

= 2pi (n 1) 2pin.

V tt c cc cung M1M2, ...,Mn1M0 u bng nhau nn a gicM0M1...Mn1 l a gic u.

Cn bc n ca n v

Cc nghim phng trnh Zn 1 = 0 c gi l cc cn bc n can 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 = cos2kpi

n+ i sin

2kpi

n, k {0, 1, ..., n 1} .

C th ta c

0 = cos 0 + i sin 0 = 1;

1 = cos2pi

n+ i sin

2pi

n= ;

2 = cos4pi

n+ i sin

4pi

n= 2;

. . .

n1 = cos2 (n 1) pi

n+ i sin

2 (n 1)pin

= n1.

Tp hp{

1, , 2, ..., n1}k hiu Un. Ta c tp hp Un c sinh bi

, mi phn t ca Un 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 Z2 1 = 0 c cc nghim 1 v 1 y lcc cn bc hai ca n v

ii) vi n = 3, phng trnh Z31 = 0 c cc nghim cho bi cng thc

k = cos2kpi

3+ i sin

2kpi

3

vi k {0, 1, 2} .V th

0 = 1 , 1 = cos2pi

3+ i sin

2pi

3= 1

2+ i

3

2,

v

2 = cos4pi

3+ i sin

2pi

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 = cos2kpi

4+ i sin

2kpi

4

vi k {0, 1, 2, 3} .C th nh sau

0 = 1 , 1 = cospi

2+ i sin

pi

2= i

2 = cospi + i sin pi = 1 , 3 = cos3pi2

+ i sin3pi

2= i.

Ta c

U4 ={

1, i, i2, i3}

= {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 Un c gi l cn nguyn thy nu mi s nguyn dngm < n ta c mk 6= 1.Mnh 1.4.2. 1) Nu n|q , mi nghim ca phng trnh Zn 1 = 0l nghim ca phng trnh Zq 1 = 0;


21

2) Nghim chung ca phng trnh Zm 1 = 0 v Zn 1 = 0 l ccnghim ca phng trnh Zd 1 = 0; vi d = gcd(m,n) (d:c chung lnnht), Um Un = Ud;

3)Cc nghim nguyn thy ca phng trnh Zm 1 = 0 l

k = cos2kpi

m+ i sin

2kpi

m;

vi 0 6 k 6 m v gcd (k,m) = 1.

Mnh 1.4.3. Nu Un l mt cn nguyn thy ca n v th ttc cc nghim ca phng trnh Zn 1 = 0 l r, r+1 , ..., r+n1 vi rl s nguyn dng ty .

Mnh 1.4.4. Cho 0 , 1 , ...., n1 l cc cn bc n ca n v . Vimi s nguyn dng n ta lun c h thc

n1j=0

kj =

{n, n|k ;0, n 6 |k.

Mnh 1.4.5. Cho p l s nguyn t v = cos2pi

p+ i sin

2pi

p. Nu

a0 , a1 , ..., ap1 l cc s nguyn khc khng ,h thc

a0 + a1+ ...+ ap1p1 = 0

ng khi v ch khi a0 = a1 = ... = ap1.

1.5 Bi tp

Bi 1 Cho cc s phc z1 = (1, 2) , z2 = (2, 3) , z3 = (1 1) hytnh cc tng sau:a) z1 + z2 + z3 ; b) z1z2 + z2z3 + z3z1 ; c) z1z2z3 ;

d) z21 + z22 + z

23 ; e)

z21 + z22

z22 + z23

; f)z1z2

+z2z3

+z3z1.

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 Ca) z2 + z + 1 = 0 ; b) z3 + 1 = 0 .

Bi 4 Cho z0 = (a, b) C . Tm s phc z tha mn z2 = z0.Bi 5 Tm cc s thc x, y trong cc trng hp sau:

a) (1 2i)x+ (1 + 2i) y = 1 + i ; b) x 33 + i

+y 33 i = i ;

c) (4 3i)x2 + (1 + 2i)xy = 4y2 12x2 +

(3xy 2y2) 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 i1 + i

)16; d)

(1 + i3

2

)6+

(1 i7

2

)6;

e)3 + 7i

2 + 3i+

5 8i2 3i .

Bi 8 Tnh:a) i2000 + i1999 + i201 + i82 + i47 ; b)En = 1 + i+ i

2 + ...+ in , 1 6 n N;c) i1.i2.i3...i2000 ; d) i5 + (i)7 + (i)13 + i100 + (i)94 .Bi 9 Tm tt c cc s phc z 6= 0 tha mn z + 1z R.Bi 10Chng minh rng:

a) E1 =(

2 + i

5)7

+(

2 i

5)7 R ;

b) E2 =

(19 + 7i

9 i)n

+

(20 + 5i

7 + 6i

)n R .

Bi 11 Cho z C tha mn z3 + 1z3 6 2 .Chng minh rngz + 1z 6 2 .Bi 12 Tm cc s phc z tha mn |z| = 1 v z2 + z2 = 1.Bi 13Tm cc s phc z tha mn 4z2 + 8 |z|2 = 8.Bi 14Tm cc s phc z tha mn z3 = z.

Bi 15 Cho z C vi Re (z) > 1 .Chng minh rng 1z 12 < 12 .Bi 16 Cho a, b, c l cc s thc v = 12 + i

32 .Tnh tng(

a+ b + c2) (a+ b2 + c

).

Bi 17 Chng minh cc ng thc sau :


23

a) |z1 + z2|2 + |z2 + z3|2 + |z3 + z1|2 = |z1|2 + |z2|2 + |z3|2 + |z1 + z2 + z3|2 ;b) |1 + z1z2|2 + |z1 z2|2 =

(1 + |z1|2

)(1 + |z2|2

);

c) |z1 + z2 + z3|+ |z1 + z2 + z3|+ |z1 z2 + z3|+ |z1 + z2 z3| == 4

(z21+ z22+ z23) .Bi 18 Tm tt c cc s nguyn dng n sao cho :(

1 + i32

)n+

(1 i3

2

)n= 2.

Bi 19 Cho z1, z2, z3 l cc s phc tha mn |z1| = |z2| = |z3| = R > 0.Chng minh rng :

|z1 z2| . |z2 z3|+ |z3 z1| . |z1 z2|+ |z2 z3| . |z3 z1| 6 9R2 .

Bi 20 Cho z1, z2, ..., zn l cc s phc tha mn

|z1| = |z2| = ... = |z3| = r > 0.chng minh rng :

E =(z1 + z2) (z2 + z3) ... (zn1 + zn) (zn + z1)

z1z2...zn.

Bi 21 (Bt ng thc Hlawas) Cho z1, z2, z3 l cc s phc. Chng

minh rng :

|z1 + z2|+ |z2 + z3|+ |z3 + z1| 6 |z1|+ |z2|+ |z3|+ |z1 + z2 + z3| .

Bi 22 Cho x1, x2 l nghim phng trnh x2 x+ 1 = 0. Hy tnh :

a)x20001 + x20002 ; b) x

19991 + x

19992 ; c)x

n1 + x

n2 , n N.

Bi 23 Tm dng ta cc ca cc s phc sau:

a) z1 = 6 + 6i

3 ; b) z2 = 14

+ i

3

4; c) z3 = 1

2 i

3

2;

d) z4 = 9 9i

3 ; e) z5 = 3 2i ; f) z6 = 4i .Bi 24 Tm dng ta cc ca cc s phc sau:a) z1 = cos a i sin a , a [0, 2pi) ;b) z2 = sin a+ i (1 + cos a) , a [0, 2pi) ;c) z3 = cos a+ sin a+ i (sin a cos a) , a [0, 2pi) ;d) z4 = 1 cos a+ i sin a , a [0, 2pi) .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 =12

+i2

;

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) z3 125 = 0 ; b) z4 + 16 = 0 ;c) z3 + 64i = 0 ; d) z3 27i = 0 ;Bi 31 Gii cc phng trnh sau:a) z7 2iz4 iz3 2 = 0 ; b) z6 + iz3 + i 1 = 0 ;c) (2 3i) z6 + 1 + 5i = 0 ; d) z10 + (2 + i) z5 2i = 0 .


25

Chng 2

S dng s phc trong gii ton scp

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 z1 v z2 c biu din hnh hc l cc im M1 v M2khi khong cch gia hai im M1 v M2 c cho bi cng thc

M1M2 = |z1 z2|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 6= a , z 6= b vh thc sau tha mn

|a z|+ |z b| = |a b| .Ta s dng k hiuAM B.Tp hp (AB) = {M : AM B} c gi l on thng m xc nhbi imA v B.

Tp hp [AB] = (AB) {A,B} c gi l on thng ng xc nhbi 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 phcca M.



1) M (AB;2)C s thc dng t sao cho z = (1 t) a+ tb, vi z l ta phc caM ;

3) arg (z a) = arg (z b) ;4)

z ab a R

+.



1) M nm trn ng thng AB;

2)z ab a R;

3)C s thc t sao cho z = (1 t) a+ tb;4)

z a z ab a b a = 0;

5)

z z 1a a 1b 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 sautha 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 kb1 k .

Khi k < 0 im M nm trn on thng ni A v B. Nu k (0, 1), thM (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 M1 (z1) v

M2 (z2) l hai im phn bit khc gc ta trong mt phng phc.

Gc M1OM2 c gi l nh hng nu cc im M1 v M2 c th t

thun chiu kim ng h.

Mnh 2.1.4. S o gc nh hng M1OM2 bng

argz2z1.

nh l 2.1.5. Cho ba im phn bit M1 (z1) , M2 (z2) , M3 (z3). S

o gc nh hng M2M1M3 l

argz3 z1z2 z1 .

Gc gia hai ng thng

Cho bn im Mi (z1) , i {1, 2, 3, 4}. S o gc xc nh bi ngthng M1M3 v M2M4 bng

argz3 z1z4 z2

hoc

argz4 z2z3 z1 .

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 varg (z) = arg z + .

Gi M l biu din hnh hc ca z, ta thy rng im M l nh ca Mqua 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 tmB gc quay =

pi

2bin 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 = pi2v

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 k2

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 O1, O2, O3, O4phn bit. Chng minh rng O1O3O2O4 v O1O3 = O2O4.

Gii LyABMM , BCNN , CDPP , DAQQ ln lt l cc hnh vungc tm O1, O2 , O3 , O4.

im M l nh ca A qua php quay tm B gc quay = pi2 ; 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

o1 =a+m

2=a+ b+ (a b) i

2, o2 =

b+ c+ (b c) i2

o3 =c+ d+ (c d) i

2, o4 =

d+ a+ (d a) i2

.


29

Vo3 o1o4 o2 =

c+ d a b+ i (c d a+ b)a+ d b c+ i (d a b+ c) = i iR

nn O1O3O2O4 .Ngoi ra o3 o1o4 o2

= |i| = 1nn O1O3 = O2O4 .

Bi ton 3 V pha ngoi tam gic ABC ta dng cc tam

gicABR,BCP,CAQ sao cho

PBC = CAQ = 45o

BCP = QCA = 30o

ABR = RAB = 15o

Chng minh rng QRP = 90o 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 vMC

MP=

3 ta c

pmbm = i v

cmpm = i

3

V th

p =c+

3b

1 +

3+

b c1 +

3i.

Tng t

q =c+

3a

1 +

3+

a c1 +

3i.

im B c c t im A bng cch quay quanh R mt gc = 150o v

th

b = a

(

3

2+

1

2i

).


30

S dng bin i i s ta cp

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 mtng trn

Xt bn im phn bit Mi (zi) , i {1, 2, 3, 4}

Mnh 2.1.7. Cc im M1,M2,M3 thng hng khi v ch khi

z3 z1z2 z1 R

.

Mnh 2.1.8. ng thng M1M2 vM3M4 vung gc khi v ch khi

z1 z2z3 z4 iR

.

Mnh 2.1.9. Bn im phn bit M1 (z1) ,M2 (z2) ,M3 (z3) ,M4 (z4)

thng hng hoc cng nm trn mt ng trn khi v ch khi

k =z3 z2z1 z2 :

z3 z4z1 z4 R

S k c gi l t s kp ca bn im

M1 (z1) ,M2 (z2) ,M3 (z3) ,M4 (z4) .

Ch 1) cc im M1 (z1) ,M2 (z2) ,M3 (z3) ,M4 (z4) thng hng khi v

ch khi

z3 z2z1 z2 R

vz3 z4z1 z4 R

.

2) Cc im M1 (z1) ,M2 (z2) ,M3 (z3) ,M4 (z4) nm trn mt ng trn

khi v ch khi

k =z3 z2z1 z2 :

z3 z4z1 z4 R

viz3 z2z1 z2 / R v

z3 z4z1 z4 R.


31

2.1.3 Tam gic ng dng

Xt su im A1 (a1) , A2 (a2) , A3 (a3) , B1 (b1) , B2 (b2) , B3 (b3) trong

mt phng phc. Ta ni rng tam gic A1A2A3 v B1B2B3 ng dng

vi nhau nu gc Ak bng gc Bk , k {1, 2, 3} .

Mnh 2.1.10. Tam gic A1A2A3 v B1B2B3 ng dng v c cng

hng khi v ch khi:

a2 a1a3 a1 =

b2 b1b3 b1 .

Mnh 2.1.11. Tam gic A1A2A3 v B1B2B3 ng dng v c ngc

hng khi v ch khi:

a2 a1a3 a1 =

b2 b1b3 b1

.

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 ab a =

e bc b =

f ca 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 ABCvABC c cng trng tm.Gii T

MC

AB=NA

BC=PB

CA.

Ta c tan(C AB) = tan(ABC) = tan(CBA). V th tam gicAC B,BAC,CBA ng dng vi nhau . p dng kt qu bi trn tac li gii .

Bi ton 3 Cho tam gic ABO u vi tm S, tam gic u khc

ABO c cng hng vi tam giac ABO v S 6= A, S 6= B . Gi M vN ln lt l trung im AB v AB. Chng minh rng tam gic SBMv SAN ng dng.

( 30th IMO-Shortlist)

Gii

Gi R l bn knh ng trn ngoi tip tam gic ABO v

= cos2pi

3+ i sin

2pi

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,R2.

Xt R + z l ta ca B khi ta A l R z. Ta trungim M,N l

zM =zB + zA

2=R2 +R z

2=R(2 + ) z

2=R z

2=(R + z)

2.

v

zN =zA + zB

2=R+R + z

2=R (+ 1) + z

2=R2 + z

2=z R

2=R z2 .

Khi ta c

zB zSzM zS =

zA zSzN zS

R + z(R+z)

2

=R zRz2

. = 1 ||2 = 1.

Vy gic SBM v SAN ng dng v ngc hng.


33

2.1.4 Tam gic u

Mnh 2.1.12. Gi s z1, z2, z3 l ta cc nh ca tam gic

A1A2A3. Khi cc khng nh sau tng ng:

1) A1A2A3 l tam gic u;

2) |z1 z2| = |z2 z3| = |z3 z1| ;3) z21 + z

22 + z

23 = z1z2 + z2z3 + z3z1;

4)z2 z1z3 z2 =

z3 z2z1 z2 ;

5)1

z z1 +1

z z2 +1

z z3 = 0 vi z =z1 + z2 + z3

3;

6)(z1 + z2 +

2z3) (z1 +

2z2 + z3)

= 0 vi = cos2pi3 + i sin2pi3 ;

7)

1 1 1z1 z2 z3z2 z3 z1 = 0.

Mnh 2.1.13. Gi s z1, z2, z3 l ta cc nh ca tam gic c

hng dng A1A2A3. Khi cc khng nh sau tng ng:


2) z3 z1 = (z2 z1) vi = cospi3

+ i sinpi

3;

3) z2 z1 = (z3 z1) vi = cos5pi3

+ i sin5pi

3;

4) z1 + z2 + 2z3 = 0 vi = cos

2pi

3+ i sin

2pi

3.

Mnh 2.1.14. Gi s z1, z2, z3 l ta cc nh ca tam gic c

hng m A1A2A3. Khi cc khng nh sau tng ng :


2) z3 z1 = (z2 z1) vi = cos5pi3

+ i sin5pi

3;

3) z2 z1 = (z3 z1) vi = cospi3

+ i sinpi

3;

4) z1 + 2z2 + z3 = 0 vi = cos

2pi

3+ i sin

2pi

3.

Mnh 2.1.15. Gi s z1, z2, z3 l ta cc nh ca tam gic u

A1A2A3. Khi cc khng nh sau tng ng :


2) z1z2 = z2z3 = z3z1;


34

3) z21 = z2z3 v z22 = z1z3.

Bi ton 1 V pha ngoi tam gic ABC dng ba tam gic u c

hng dng AC B,BAC,CBA. Chng minh rng cc trng tm caba 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+ b2 = 0 , b+ a+ c2 = 0 , c+ b+ a2 = 0 (1)

Vi a, b, c ,l ta cc im A, B, C . Trng tm cc tam gicAC B,BAC,CBA 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+ b2 = 0. Tht vy

(c + a+ b2

)= (a+ b+ c) + (a + b+ c) + (a+ b + c) 2

=(b+ a+ c2

)+(a+ c+ b2

)+(c+ b+ a2

)= 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 A0, B0, C0 l trng tm ca cc n gic trn cc cnh

BC,CA,AB S o cc gc AC0B , BA0C , AB0C l2pi

nK hiu a , b , c , a0 , b0 , c0 l ta cc im A,B,C,A0, B0, C0. S

dng cc cng thc php quay ta c

a = c0 + (b c0) ;b = a0 + (c a0) ;c = b0 + (a b0) ;


35

Suy ra

a0 =b c1 , b0 =

c a1 , c0 =

a b1 .

Tam gic A0B0C0 u khi v ch khi

a20 + b20 + c

20 = a0b0 + b0c0 + c0a0

Thay cc gi tr a0 , b0 , c0 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 2pin

=2pi

3ta 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, R v z = x+ yi C.Mnh 2.1.16. Cho hai ng thng d1 v d2 c phng trnh

1.z + 1.z + 1 = 0 v 2.z + 2.z + 2 = 0

ng thng d1 v d2 l:

1) song song khi v ch khi11

=22

;

2) vung gc khi v ch khi11

+22

= 0;

3) Ct nhau khi v ch khi116= 22

;


36

Phng trnh ng thng xc nh bi hai im: Phng trnh

mt ng thng c xc nh bi hai im

P1 (z1) , P2 (z2) l z1 z1 1z2 z2 1z z 1 = 0.

Din tch tam gic

nh l 2.1.17. Din tch tam gic A1A2A3 vi ta cc nh z1, z2, z3bng modun ca s

i

4

z1 z1 1z2 z2 1z3 z3 1 .

H qu Din tch tam gic nh hng A1A2A3 vi ta cc nh

z1, z2, z3 l

area [A1A2A3] =1

2Im (z1z2 + z2z3 + z3z1) .

Bi ton 1 Cho tam gic A1A2A3 v cc im M1,M2,M3 nm trn

ng thng A2A3, A3A1, A1A2. Gi s rng M1,M2,M3 chia on thng

[A2A3] , [A3A1] , [A1A2] theo t s 1, 2, 3. Chng minh rng :

area [M1M2M3]

area [A1A2A3]=

1 123(1 1) (1 2) (1 3) .

Gii Ta cc im M1,M2,M3 l:

m1 =a2 1a1

1 1 ,m2 =a3 2a1

1 2 ,m3 =a1 3a2

1 3p dng cng thc (2) ta c


37

area [M1M2M3] =1

2(m1m2 +m2m3 +m3m1)

=1

2

((a2 1a3) (a3 2a1)

(1 1) (1 2) +(a3 2a1) (a1 3a2)

(1 2) (1 3)+

(a1 3a2) (a2 1a3)(1 3) (1 1)

)=

1 123(1 1) (1 2) (1 3)area [A1A2A3]

Ch Cng thc dng (3) ta suy ra nh l Menelaus :Cc im

M1,M2,M3 thng hng khi v ch khi 123 = 1 tng ng vi:

M1A2M1A3

.M2A3M2A1

.M3A1M3A2

= 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,pi

2

)sao cho

a+ b cos + c sin = 0. Chng minh rng

1 < area [ABC] 6 1 +

2

2.

Gii Ta c

1 = |a|2 = |b cos + c sin|2 = (b cos + c sin) (b cos + c sin)= |b|2 cos2 + |c|2 sin2 + (bc+ bc) sincos= 1 +

b2 + c2

bcsincos.

T trn ta c b2 + c2 = 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] = 12

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

2sin +

2

2cos

)]=

1

2

(1 +

2 sin( +

pi

4

))Ta thy rng

pi

4< +

pi

4 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,R1, R2, R3 ln lt l

bn knh ng trn ngoi tip tam gic ABC,GBC,GCA,GAB.

Chng minh rng R1 +R2 +R3 > 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

cospi

11+ cos

3pi

11+ cos

5pi

11+ cos

7pi

11+ cos

9pi

11=

1

2.

Gii: t z = cospi

11+ i sin

pi

11ta c

z + z3 + z5 + z7 + z9 =z11 zz2 1 =

1 zz2 1 =

1

1 z .

Ta chng minh kt qu sau.

Nu z = cos t+ i sin t v z 6= 1 th Re(

1

1 z)

=1

2Tht vy

1

1 z =1

1 (cos t+ i sin t ) =1

(1 cos t) i sin t=

1

2 sin2 t2 2i sin t2cos t2=

1

2 sin t2(sint2 i cos t2)

=sin t2 + i cos

t2

2 sin t2=

1

2+ i

cos t22 sin t2

.

Suy ra Re

(1

1 z)

=1

2.


46

Bi ton 2 Tnh tch

P = cos20o.cos40o.cos80o

Gii:t z = cos20o + i sin 20o ta c z9 = 1v z = cos20o - i sin 20ov

cos20o =z2 + 1

2z, cos40o =

z4 + 1

2z2, cos80o =

z8 + 1

2z4.

nn

P =

(z2 + 1

) (z4 + 1

) (z8 + 1

)8z7

=

(z2 1) (z2 + 1) (z4 + 1) (z8 + 1)

8z7 (z2 1)=

z16 18 (z9 z7) =

z7 18 (1 z7) =

1

8

.

Cch gii 2

sin 20o.P = sin 20o.cos20o.cos40o.cos80o

=1

2sin 40o.cos40o.cos80o

=1

4sin 80o.cos80o =

1

8sin 160o =

1

8sin 20o.

Vy P =1

8.

Bi ton 3 Cho x, y, z l cc s thc tha mn sinx+ 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 z1 = cosx + i sinx , z2 = cosy + i sin y , z3 = cosz + i sin z , ta

c z1 + z2 + z3 = 0 v |z1| = |z2| = |z3| = 1 . T yz21 + z

22 + z

23 = (z1 + z2 + z3)

2 2 (z1z2 + z2z3 + z3z1)= 2z1z2z3

(1

z1+

1

z2+

1

z3

)= 2z1z2z3 (z1 + z2 + z3)

= 2z1z2z3 (z1 + z2 + z3) = 0

Suy ra ( cos2x + cos2y + cos2z) + i (sin 2x+ sin 2y + sin 2z) = 0 .

T y ta c iu phi chng minh.


47


cos210o + cos250o + cos270o =3

2.

Gii t z = cos10o + i sin 10o ta c z9 = i

cos10o =z2 + 1

2z, cos50o =

z10 + 1

2z5, cos70o =

z14 + 1

2z7.

ng thc cn chng minh tng ng vi(z2 + 1

2z

)2+

(z10 + 1

2z5

)2+

(z14 + 1

2z7

)2=

3

2.

hay

z16 + 2z14 + z12 + z24 + 2z14 + z4 + z28 + 2z14 + 1 = 6z14

z28 + z24 + z16 + z12 + z4 + 1 = 0.S dng z8 = 1 ta thu c

z16 + z12 z10 z6 + z4 + 1 = 0 (z4 + 1) (z12 z6 + 1) = 0

(z4 + 1

) (z18 + 1

)z6 + 1

= 0.

Khng nh c chng minh.

Bi ton 5 Gii phng trnh

cosx+ cos2x cos3x = 1.

Gii t z = cosx+ i sinx ta c

cosx =z2 + 1

2z, cos2x =

z4 + 1

2z2, cos3x =

z6 + 1

2z3.

phng trnh tr thnh


48

z2 + 1

2z+z4 + 1

2z2 z

6 + 1

2z3= 1

z4 + z2 + z5 + z z6 1 2z3 0 (z6 z5 z4 + z3)+ (z3 z2 z + 1) = 0 (z3 + 1) (z3 z2 z + 1) = 0 (z3 + 1) (z 1)2 (z + 1) = 0

.

v z = 1 z = 1 , z3 = 1 nn

x {2kpi |k Z} , x {pi + 2kpi |k Z} , x {pi + 2kpi

3|k Z

}.

hp cc nghim ta c

x {kpi |k Z} , x {

2k + 1

3pi |k Z

}.

Bi ton 6 Tnh tng S =nk=1

qk. cos kx v T =nk=1

qk. sin kx .

Gii Ta c

1 + S + iT =nk=1

qk (cos kx+ i sin kx ) =nk=1

qk (cosx+ i sinx)k

=1 qn+1 (cosx+ i sinx)n+1

1 q cosx iq sinx=

1 qn+1 [cos (n+ 1)x+ i sin (n+ 1)x]1 q cosx iq sinx

=

[1 qn+1 cos (n+ 1)x iqn+1 sin (n+ 1)x] [1 q cosx+ iq sinx]

q2 2q cosx+ 1 .

v th

1 + S =qn+2 cosnx qn+1 cos (n+ 1)x q cosx+ 1

q2 2q cosx+ 1T =

qn+2 sinnx qn+1 sin (n+ 1)x+ q sinxq2 2q cosx+ 1 .


49

Ch Khi q = 1 ta c h thc quen thuc

nk=1

cos kx =sin nx2 cos

(n+1)x2

sin x2v

nk=1

sin kx =sin nx2 sin

(n+1)x2

sin x2.

Tht vy

nk=1

cos kx =cosnx cos (n+ 1)x (1 cosx)

2 (1 cosx)

=2 sin x2 sin

(2n+1)x2 2 sin2 x2

4 sin2 x2

=sin (2n+1)x2 sin x2

2 sin x2=

sin nx2 cos(n+1)x

2

sin x2.

vnk=1

sin kx =sinnx sin (n+ 1)x+ sinx

2 (1 cosx)

=2 sin x2cos

x2 2 sin x2cos (2n+1)x2

4 sin2 x2

=cosx2 cos (2n+1)x2

2 sin x2=

sin nx2 sin(n+1)x

2

sin x2.

Bi ton 7 Cho cc im A1, A2, ...., A10, c xp theo th t v cch

u nhau trn ng trn bn knh R. Chng minh rng

A1A4 A1A2 = R.

Gii t z = cospi

10+ i sin

pi

10khng mt tnh tng qut ta cho R = 1.

Ta cn ch ra 2 sin3pi

10- sin

pi

10= 1.

Trong trng hp tng qut z = cos a+ i sin a th sin a =z2 1

2iz. ng

thc cn chng minh trn tr thnh

2z6 12iz3

z2 12iz

= 1 z6 z4 + z2 1 = 2iz3

. V z5 = i nn z8 z6 + z4 z2 + 1 = 0. ng thc ny ng v(z8 z6 + z4 z2 + 1) (z2 + 1) = z10 + 1 = 0 v z2 + 1 6= 0


50


cospi

7 cos 2pi

7+ cos

3pi

7=

1

2. (

5thIMO)

Gii t z = cospi7 + i sinpi7 ta c z

7 + 1 = 0 Do z 6= 1 v z7 + 1 =(z + 1)

(z6 z5 + z4 z3 + z2 z + 1) = 0 nn z6 z5 + z4 z3+ =

0 z (z2 z + 1) = 11 z3 .

Ta c cospi

7 cos 2pi

7+ cos

3pi

7= Re

(z3 z2 + z) .

Ta chng minh kt qu sau.

Nu z = cos t+ i sin t v z 6= 1 th Re(

1

1 z)

=1

2Tht vy

1

1 z =1

1 (cos t+ i sin t ) =1

(1 cos t) i sin t=

1

2 sin2 t2 2i sin t2cos t2=

1

2 sin t2(sint2 i cos t2)

=sin t2 + i cos

t2

2 sin t2=

1

2+ i

cos t22 sin t2

.

Suy ra Re

(1

1 z)

=1

2.

Bi ton 9 Chng minh rng:trung bnh cng ca cc s k sin ko

(k = 2, 4, . . . , 180) l cot 1o.

(1996 USA Mathematical Olympiad)

Gii: t z = cos t+ i sin t ta c ng thc

z + 2z2 + ...+ nzn = (z + ...+ zn) +(z2 + ...+ zn

)+ ...+ zn

=1

z 1[(zn+1 z)+ (zn+1 z2)+ ...+ (zn+1 zn)]

=nzn+1

z 1 zn+1 z(z 1)2 ,


51

t y ta thu c hai ng thc

nk=1

k cos kt =(n+ 1) sin (2n+1)t2

2 sin t2 1 cos (n+ 1) t

4 sin2 t2. (1)

vnk=1

k sin kt =sin (n+ 1) t

4 sin2 t2 n cos

(2n+1)t2

2 sin t2. (2)

S dng h thc (2) ta thu c

2 sin 2o + 4 sin 4o + ....+ 178 sin 178o = 2 (sin 2o + 2 sin 2.2o + ...+ 89 sin 89.2o)

= 2

(sin 90.2o

4 sin2 1o 90 cos 179

o

2 sin 1o

)= 90 cos 179

o

sin 1o= 90 cot 1o.

Suy ra

1

90(2 sin 2o + 4 sin 4o + ....+ 178 sin 178o + 180 sin 180o) = cot 1o.

Khng nh c chng minh.

Bi ton 10 Cho n l s nguyn dng. Tm s thc ao v

akl , k, l = 1, n , k > l tha mn

sin2 nx

sin2 x= a0 +

16l6k6n

aklcos2 (k l)x;

vi mi s thc x 6= mpi , m Z .

Gii: S dng ng nht thc

S1 =nj=1

cos2jx =sinnx cos (n+ 1)x

sinx

S2 =nj=1

sin 2jx =sinnx sin (n+ 1)x

sinx

Ta thu c

S21 + S22 =

(sinnx

sin

)2.


52

mt khc

S21 + S22 = (cos2x+ cos4x+ ...+ cos2nx)

2 +

(sin 2x+ sin 4x+ ...+ sin 2nx)2

= n+

16l6k6n(cos2kx cos 2lx+ sin 2kx sin 2lx)

= n+

16l6k6ncos2 (k l)x;

.

nh vy a0 = n , akl = 2 , 1 6 l < k 6 n.

Bi ton 11 Tnh cc tng sau vi = 30o

i) 1 +cos

cos+cos (2)

cos2+cos (3)

cos3+ ...+

cos ((n 1) )cosn1

;

ii) coscos + cos2cos (2) + cos3cos (3) + ...+ cosncos (n).


1 + cos2n(pin

)+ cos2n

(2pin

)+ ...+ cos2n

((n1)pi

n

)= n.4n (2 + Cn2n)

vi mi s nguyn dng n > 2 .Bi ton 13 Cho s nguyn p > 0. Tm cc s thc a0, a1, ..., ap viap 6= 0 sao cho

cos 2p = a0 + a1 sin2 + ...+ ap.

(sin2

)pvi 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. tx = u,

y = v. H cho tr thnh

u

(1 +

1

u2 + v2

)=

23

v

(1 1

u2 + v2

)=

4

27

u+

1

u2 + v2=

23

(1)

v 1u2 + v2

=4

27

(2)


53

ly phng trnh th (2) nhn vi i sau cng vi phng trnh (1) ta

c u+ iv +u ivu2 + v2

=23

+ i4

27

()t z = u+ iv th u2 + v2 = |z|2. Khi phng trnh (*) tr thnh

z +z

|z|2 =23

+ i4

27

z + 1z

=23

+ i4

27

z2 (

23

+ i4

27

)z + 1 = 0

z = 13 2

21+ i

(2

27

2

).

Suy ra

(u, v) =

(13 2

21,2

27

2

).

H cho c hai nghim

(x, y) =

( 13 2

21

)2,

(2

27

2

)2 = (1121 4

3

7,22

7 8

7

).

Bi ton 2 Gii h phng trnh{x3 3xy2 = 13x2y y3 =

3

Gii {x3 3xy2 = 13x2y y3 =

3

{x3 3xy2 = 13x2 (iy) + i3y3 = i

3

{x3 + 3x (iy)2 = 1

3x2 (iy) + (iy)3 = i

3

{

(x+ iy)3 = 1 i

3 (1)

3x2 (iy) + (iy)3 = i

3 (2) .


54

t z = x+ iy t (1) ta c z3 = 1 i3 = 2 (cos (pi3)+ i sin (pi3))Suy ra

z = 3

2

(cospi3 + 2kpi

3+ i sin

pi3 + 2kpi3

)= 3

2

(cospi + 6kpi

9+ i sin

pi + 6kpi9

)vi k = 0, 1, 2.

T y ta c cc nghim ca h phng trnh cho l

(x, y) =

(3

2cospi + 6kpi

9,

3

2 sinpi + 6kpi

9

)vi k = 0, 1, 2.

2.2.3 Bi tp

Bi 1 Chng minh rng :

a) cos2 + cos22 + ...+ cos2n =sin (n+ 1) cosn

2 sin+n 1

2;

b) sin2 + sin2 2 + ...+ sin2 n =n+ 1

2 sin (n+ 1) cosn

2 sin;

.

Bi 2 Chng minh rng :

a) cos2pi

2n+ 1+ cos

4pi

2n+ 1+ cos

6pi

2n+ 1+ ...+ cos

2npi

2n+ 1= 1

2;

b) cospi

16=

2 +

2 + 2

2;

c) sin2pi

5=

10 + 2

5

2.

.

Bi 3 Chng minh rng vi mi n chn (n = 2m),ta u c:

cosn =

cosn C2ncosn2 sin2 + C4ncosn4 sin4 ...+ (1)mCnn sinn .Bi 4 Chng minh rng :


55

a) sinpi

2n+ 1sin

2pi

2n+ 1... sin

npi

2n+ 1=

2n+ 1

2n;

b) cospi

2n+ 1cos

2pi

2n+ 1...cos

npi

2n+ 1=

1

2n;

c) sinpi

2nsin

2pi

2n... sin

(n 1) pi2n

=

n

2n1;

d) cospi

2ncos

2pi

2ncos

(n 1)pi2n

=

n

2n1.

Bi 5 Gii phng trnha) x5 + x4 + x3 + x2 + x+ 1 = 0 ;

b) x5 + x4 + 2x3 + 3x2 + 4x+ 5 = 0 , 0 6= C .Bi 6 Chng minh rng vi mi n l (n = 2m+ 1),ta u c:

sinn =

C1ncosn1 sin C3ncosn3 sin3 + ...+ (1)m1Cn1n sinn3 cos3.

Bi 7 Gii h phng trnh :x+

3x yx2 + y2

= 3

y x+ 3yx2 + y2

= 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

3n1k=0

(1)k C2k+16n 3k.


56

Gii Ta c3n1k=0

(1)k C2k+16n 3k =3n1k=0

C2k+16n (3)k =3n1k=0

C2k+16n

(i

3)2k

=1

i

3

3n1k=0

C2k+16n

(i

3)2k+1

=1

i

3Im(

1 + i

3)6n

=1

i

3Im[2(cos

pi

3+ i sin

pi

3

)]6n=

1

i

3

[26n (cos2pin+ i sin 2pin)

]= 0.

Bi ton 2 Tnh tng

Sn =nk=0

Ckn cos k

vi [0, pi] .Gii Ta c t z = cos + i sin v Tn =

nk=0

Ckn sin k. Ta c

Sn + iTn =nk=0

Ckn (cos k + i sin k) =nk=0

Ckn (cos + i sin)k

=nk=0

Cknzk = (1 + z)k . (1)

Dng ta cc ca s phc 1 + z l

1 + cos + i sin = 2 cos2

2+ 2i sin

2cos

2

= 2cos

2

(cos

2+ i sin

2

).

V [0, pi] ng thc (1) tr thnhSn + iTn =

(2cos

2

)n (cos

n

2+ i sin

n

2

).

T y ta thu c

Sn =(

2cos

2

)ncos

n

2

Tn =(

2cos

2

)nsin

n

2.


57

Biton 3 Chng minh ng thc(C0n C2n + C4n ....

)2+(C1n C3n + C5n ....

)2= 2n.

Gii t xn = C0n C2n + C4n .... v yn = C1n C3n + C5n ....ta c

(1 + i)n = xn + iynLy modun hai v ta thu c |xn + iyn| = |(1 + i)n| = |1 + i|n = 2niu ny tng ng vi x2n + y

2n = 2

n. Khng nh c chng minh.

Ch Ta c th xy dng cng thc tnh xn, yn nh sau

(1 + i)n =(

2(cos

pi

4+ i sin

pi

4

))n= 2

n2

(cos

npi

4+ i sin

npi

4

).

Suy ra xn = 2n2 cos

npi

4v yn = 2

n2 sin

npi

4.

Biton 4 Chng minh ng thc

Cmn + Cm+pn + C

m+2pn + ... =

2n

p

p1k=0

(cos

kpi

p

)ncos

(n 2m) kpip

.

Gii Gi 0, 1, ..., p1 l cc cn bc p ca n v. Khi

p1k=0

mk (1 + k)n =

nk=0

Ckn(km0 + ...+

kmp1

). (1)

Ta c kt qu quen thuc sau:

km0 + ...+ kmp1 =

{p , p | (k m)0 , p 6 | (k m) .(2)

Xt

mk (1 + k)n

=

(cos

2mkpi

p i sin 2mkpi

p

)(2 cos

kpi

p

)n(cos

2nkpi

p+ i sin

2nkpi

p

)= 2n

(cos

kpi

p

)n(cos

(n 2m) kpip

+ i sin(n 2m) kpi

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 :p1k=0

(cos

kpi

p

)nsin

(n 2m) kpip

= 0

Bi ton 5 Cho an, bn, cn l cc s nguyn vian = C

0n + C

3n + C

6n + ;

bn = C1n + C

4n + C

7n + ;

cn = C2n + C

5n + C

8n + ;

Chng minh rng

1) a3n + b3n + c

3n 3anbncn = 2n;

2)a2n + b2n + c

2n anbn bncn cnan = 1.

3) Hai trong ba s nguyn an, bn, cn 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 = an+bn+cn , (1 + )n = an+bn+cn

2 ,(1 + 2

)n= an+bn

2+cn

v th

a3n + b3n + c

3n 3anbncn = (an + bn + cn)

(an + bn+ cn

2) (an + bn

2 + cn2n)

= 2n (1 + )n(1 + 2

)n= 2n

(2)n ()n = 2n2) S dng ng thc quen thuc

x3 + y3 + z3 3xyz = (x+ y + z) (x2 + y2 + z2 xy yz zx) .t h thc trn ta c

a2n + b2n + c

2n anbn bncn cnan = 1

3)a2n + b2n + c

2n anbn bncn cnan = 1

(an bn)2 + (bn cn)2 + (cn an)2 = 2t ng thc ny suy ra:hai trong ba s an, bn, cn bng nhau v s th

ba bng 1.

Ch T bi ton 5 ta thu c

an =1

3

[2n + cos

npi

3+ (1)n cos2npi

3

]=

1

3

(2n + 2cos

npi

3

)bn =

1

3

[2n + cos

(n 2) pi3

+ (1)n cos(2n 4)pi3

]=

1

3

(2n + 2cos

(n 2)pi3

)cn =

1

3

[2n + cos

(n 4)pi3

+ (1)n cos(2n 8)pi3

]=

1

3

(2n + 2cos

(n 4)pi3

).


59

D thy rng

an = bn khi v ch khi n 1 mod (3) ;bn = cn khi v ch khi n 2 mod (3) ;cn = an 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} vchia ht cho 3?

(Romanian Mathematical Regional Contest Trian Lalescu)

Gii:Gi xn, yn, zn 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 tmcc gi tr xn .

t = cos2pi

3+ i sin

2pi

3, ta thy rng xn + yn + zn = 4

n v

xn + yn + 2zn =

j1+j2+j3+j4=n

2j1+3j2+7j3+9j4

=(2 + 3 + 7 + 9

)n= 1.

hay xn 1 + yn + 2zn = 0. Ta thu c xn 1 = yn = zn = k. Khi 3k = xn + yn + zn 1 = 4n 1 dn nk = 13 (4

n 1). Suy ra xn = k + 1 = 13 (4n + 2) .

Bi ton 7 Cho s nguyn t n v cc s nguyn dng a1, a2, ..., am .

Gi f (k) l s cc b m s (c1, c2, ..., cm) tha mn iu kin 0 6 ci 6 aiv c1 + c2 + ...+ cm k (modm). Chng minh rngf (0) = f (1) = ... = f (n 1) khi v ch khi n |aj vi j no thuc tp{1, 2, ...,m} .

Gii Xt = cos2pin + i sin2pin . rng ng thc sau lun ng

nk=1

(X +X2 + ...+Xak

)=

16ck6ak

Xc1+...+cn

v

f (0)+f (1)+...+f (n 1)n1 = 16ck6ak

Xc1+...+cn =mk=1

( + 2 + ...+ ak

)


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

mk=1

( + 2 + ...+ ak

)= 0

tc l + 2 + ...+ ak = 0 vi j no thuc tp {1, 2, ...,m}

Bi ton 8 Cho s nguyn t n v cc s nguyn dng a1, a2, ..., am.

Gi f (k) l s cc b m s (c1, c2, ..., cm) tha mn iu kin 0 6 ci 6 aiv c1 + c2 + ...+ cm k (modm). Chng minh rngf (0) = f (1) = ... = f (n 1) khi v ch khi n |aj vi j no thuc tp{1, 2, ...,m} . (

36thIMO)

Gii Trng hp p = 2 l tm thng. Xt p > 3 v = cos2pi3 + i sin2pi3 .

K hiu xj l s cc tp con vi tnh cht |B| = p v m (B) j (modp).Khi

n1j=0

xjj =

BA,|B|=p

m(B) =

16c1

61

vi t [0, pi ]Bi ton 11 Chng minh ng thc sau

1)C0n + C4n + C

8n + ... =

1

4

(2n + 2

n2+1 + cos

npi

4

).

(Romanian Mathematicanl Olympiad Second Round 1981)

2)C0n +C5n +C

10n + ... =

1

5

2n +(

5 + 1)n

2n1cos

npi

5+

(5 1

)n2n1

cos2npi

5]

Bi 12 Cho cc s nguyn An, Bn, Cn xc nh nh sauAn = C

0n C3n + C6n ... ,

Bn = C1n + C4n C7n + ... ,Cn = C

2n C5n + C8n ... .

Chng minh cc ng thc sau

1) A2n +B2n + C

2n AnBn BnCn CnAn = 3n;

2) A2n + AnBn +B2n = 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


M unh ngha v tnh cht ca s phcnh nghaTnh cht s phcCc tnh cht lin quan n php cng Cc tnh cht lin quan n php nhn

Dng i s ca s phcnh ngha v tnh chtGii phng trnh bc hai ngha hnh hc ca cc s phc v modun ngha hnh hc ca cc php ton i s

Dng lng gic ca s phcTa cc trong mt phngTa cc ca s phc Cc php ton s phc trong ta cc ngha hnh hc ca php nhn Cn bc n ca n v

Bi tp

S dng s phc trong gii ton s cpS phc v cc bi ton hnh hcMt vi khi nim v tnh cht iu kin thng hng , vung gc v cng thuc mt ng trn Tam gic ng dng Tam gic u Hnh hc gii tch vi s phc Tch thc ca hai s phcBi tp

S phc v cc bi ton i s , lng gicCc bi ton lng gic Cc bi ton i s Bi tp

S phc v cc bi ton t hp

Kt lun Ti liu tham kho

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