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Luyện thi đại học - thpt môn Toán phần số phức
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Luyn thi i hc mn Ton phn S phc
A. CHUN B KIN THC
I. DNG I S CA S PHC .
1. Mt s phc l mt biu thc c dng a + bi, trong a, b l cc s thc v s i tho
mn i2 = -1. K hiu s phc l z v vit z = a + bi .
i c gi l n v o
a c gi l phn thc. K hiu Re(z) = a
b c gi l phn o ca s phc z = a + bi , k hiu Im(z) = b
Tp hp cc s phc k hiu l C.
*) Mt s lu :
- Mi s thc a dng u c xem nh l s phc vi phn o b = 0.
- S phc z = a + bi c a = 0 c gi l s thun o hay l s o.
- S 0 va l s thc va l s o.
2. Hai s phc bng nhau.
Cho z = a + bi v z = a + bi.
z = z '
'
a a
b b
3. Biu din hnh hc ca s phc.
Mi s phc c biu din bi mt im M(a;b) trn mt phng to Oxy.
Ngc li, mi im M(a;b) biu din mt s phc l z = a + bi .
4. Php cng v php tr cc s phc.
Cho hai s phc z = a + bi v z = a + bi. Ta nh ngha:
' ( ') ( ')
' ( ') ( ')
z z a a b b i
z z a a b b i
5. Php nhn s phc.
Cho hai s phc z = a + bi v z = a + bi. Ta nh ngha:
' ' ' ( ' ' )zz aa bb ab a b i
6. S phc lin hp.
Cho s phc z = a + bi. S phc z = a bi gi l s phc lin hp vi s phc
trn.
Vy z = a bi = a - bi
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Luyn thi i hc mn Ton phn S phc
Ch : 10) z = z z v z gi l hai s phc lin hp vi nhau.
20) z. z = a
2 + b
2
*) Tnh cht ca s phc lin hp:
(1): z z
(2): ' 'z z z z
(3): . ' . 'z z z z
(4): z. z = 2 2a b (z = a + bi )
7. Mun ca s phc.
Cho s phc z = a + bi . Ta k hiu z l mun ca s ph z, l s thc khng
m c xc nh nh sau:
- Nu M(a;b) biu din s phc z = a + bi, th z = OM
= 2 2a b
- Nu z = a + bi, th z = .z z = 2 2a b
8. Php chia s phc khc 0.
Cho s phc z = a + bi 0 (tc l a2+b
2 > 0 )
Ta nh ngha s nghch o z-1
ca s phc z 0 l s
z-1
= 22 2
1 1z z
a b z
Thng 'z
zca php chia s phc z cho s phc z 0 c xc nh nh sau:
1
2
' '..
z z zz z
z z
Vi cc php tnh cng, tr, nhn chia s phc ni trn n cng c y tnh
cht giao hon, phn phi, kt hp nh cc php cng, tr, nhn, chia s thc thng
thng.
II. DNG LNG GIC CA S PHC.
1. Cho s phc z 0. Gi M l mt im trong mt phng phc biu din s phc z. S
o (radian) ca mi gc lng gic tia u l Ox, tia cui OM c gi l mt acgumen
ca z.
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Luyn thi i hc mn Ton phn S phc
Nh vy nu l mt acgumen ca z, th mi acgumen u c dng:
+ 2k, k Z.
2. Dng lng gic ca s phc.
Xt s phc z = a + bi 0 (a, b R)
Gi r l mun ca z v l mt acgumen ca z.
Ta c: a = rcos , b = rsin
z = r(cos +isin), trong r > 0, c gi l dng lng gic ca s phc z 0.
z = a + bi (a, b R) gi l dng i s ca z.
3. Nhn v chia s phc di dng lng gic.
Nu z = r(cos +isin)
z' = r(cos +isin) (r 0, r 0)
th: z.z = r.r[cos( +) +isin( +)]
' '
os( ' ) isin( ' )z r
cz r
khi r > 0.
4. Cng thc Moivre.
[z = r(cos +isin)]n = rn(cos n +isin n)
5. Cn bc hai ca s phc di dng lng gic.
Cho s phc z = r(cos +isin) (r>0)
Khi z c hai cn bc hai l: os isin2 2
r c
v - os isin2 2
r c
= os isin
2 2r c
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Luyn thi i hc mn Ton phn S phc
B. MT S DNG TON C BN.
VN 1: DNG I S CA S PHC
Dng 1: Cc php tnh v s phc.
Phng php gii:
S dng cc cng thc cng , tr, nhn, chia v lu tha s phc.
Ch cho HS: Trong khi tnh ton v s phc ta cng c th s dng cc hng
ng thc ng nh nh trong s thc. Chng hn bnh phng ca tng hoc hiu, lp
phng ca tng hoc hiu 2 s phc
V d 1: Cho s phc z = 3 1
2 2i
Tnh cc s phc sau: z ; z2; ( z )
3; 1 + z + z
2
Gii:
a) V z = 3 1
2 2i z =
3 1
2 2i
b) Ta c z2 =
2
3 1
2 2i
= 2
3 1 3
4 4 2i i =
1 3
2 2i
( z )2 =
2
23 1 3 1 3 1 3
2 2 4 4 2 2 2i i i i
( z )3 =( z )
2 . z =
1 3 3 1 3 1 3 3
2 2 2 2 4 2 4 4i i i i i
Ta c: 1 + z + z2 =
3 1 1 3 3 3 1 31
2 2 2 2 2 2i i i
Nhn xt: Trong bi ton ny, tnh 3
z ta c th s dng hng ng thc nh trong s
thc.
V d 2: Tm s phc lin hp ca: 1
(1 )(3 2 )3
z i ii
Gii:
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Luyn thi i hc mn Ton phn S phc
Ta c : 3 3
5 5(3 )(3 ) 10
i iz i i
i i
Suy ra s phc lin hp ca z l: 53 9
10 10z i
V d 3: Tm m un ca s phc (1 )(2 )
1 2
i iz
i
Gii: Ta c : 5 1
15 5
iz i
Vy, m un ca z bng: 2
1 261
5 5z
V d 4: Tm cc s thc x, y tho mn:
3x + y + 5xi = 2y 1 +(x y)i
Gii:
Theo gi thit: 3x + y + 5xi = 2y 1 +(x y)i
(3x + y) + (5x)i = (2y 1) +(x y)i
3 2 1
5
x y y
x x y
Gii h ny ta c:
1
7
4
7
x
y
V d 5: Tnh:
i105
+ i23
+ i20
i34
Gii:
tnh ton bi ny, ta ch n nh ngha n v o t suy ra lu tha
ca n v o nh sau:
Ta c: i2 = -1; i
3 = -i; i
4 = i
3.i
= 1; i
5 = i; i
6 = -1
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Luyn thi i hc mn Ton phn S phc
Bng quy np d dng chng minh c: i4n
= 1; i4n+1
= i; i4n+2
= -1; i4n+3
= -i; n
N*
Vy in {-1;1;-i;i}, n N.
Nu n nguyn m, in = (i
-1)
-n =
1n
ni
i
.
Nh vy theo kt qu trn, ta d dng tnh c:
i105
+ i23
+ i20
i34
= i4.26+1
+ i4.5+3
+ i4.5
i4.8+2
= i i + 1 + 1 = 2
V d 6: Tnh s phc sau:
z = (1+i)15
Gii:
Ta c: (1 + i)2 = 1 + 2i 1 = 2i (1 + i)
14 = (2i)
7 = 128.i
7 = -128.i
z = (1+i)15
= (1+i)14
(1+i) = -128i (1+i) = -128 (-1 + i) = 128 128i.
V d 7: Tnh s phc sau: z = 16 8
1 1
1 1
i i
i i
Gii:
Ta c: 1 (1 )(1 ) 2
1 2 2
i i i ii
i
1
1
ii
i
. Vy
16 81 1
1 1
i i
i i
=i
16 +(-i)
8 = 2
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Luyn thi i hc mn Ton phn S phc
Dng 2: Cc bi ton chng minh.
Trong dng ny ta gp cc bi ton chng minh mt tnh cht, hoc mt ng thc
v s phc.
gii cc bi ton dng trn, ta p dng cc tnh cht ca cc php ton cng,
tr, nhn, chia, s phc lin hp, mun ca s phc c chng minh.
V d 8: Cho z1, z2 C.
CMR: E = 1 2 1 2.z z z z R
gii bi ton ny ta s dng mt tnh cht quan trng ca s phc lin hp
l:
z R z = z
Tht vy: Gi s z = x + yi z = x yi.
z = z x + yi = x yi y = 0 z = x R
Gii bi ton trn:
Ta c E = 1 2 1 2 1 2 1 2.z z z z z z z z = E E R
V d 9: Chng minh rng:
1) E1 = 7 7
2 5 2 5i i R
2) E2 = 19 7 20 5
9 7 6
n ni i
i i
R
Gii:
1) Ta c: 1E = 7 7 7 7 7 7
12 5 2 5 2 5 2 5 2 5 2 5i i i i i i E
E1R
2
19 7 (9 ) 20 5 (7 6 )19 7 20 52)
9 7 6 82 85
164 82 170 852 2
82 85
n nn n
n nn n
i i i ii iE
i i
i ii i
2 2E E E2 R
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V d 10: Cho z C.
CMR: 1
12
z hoc |z2 + 1| 1
Gii:
Ta chng minh bng phn chng: Gi s 2
11
2
1 1
z
z
t z = a+bi z2
= a2 b
2 + 2a + bi
2
11
2
1 1
z
z
2 2 2 2
2 2 2 2 22 2 2 2
1(1 ) 2( ) 4 1 0
2( ) 2( ) 0
(1 ) 4 1
a b a b a
a b a ba b a b
Cng hai bt ng thc trn ta c: (a2 + b
2 )
2 + (2a+1)
2 < 0 v l pcm
Luyn thi i hc mn Ton phn S phc
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