Zadatak 141 (Branka, gimnazija) - halapa.com · broj y imaginarni dio kompleksnog broja z. Pišemo:...

1

Zadatak 141 (Branka, gimnazija)

Pojednostavnite: 1 3 1 3.i i+ ⋅ + − ⋅

. 2 . 3 . 5 . 6A B C D

Rješenje 141 Ponovimo!

Kompleksan broj je broj oblika ,z x y i= + ⋅

gdje su x i y realni brojevi, a i imaginarna jedinica. Broj x zove se realni dio kompleksnog broja z, a

broj y imaginarni dio kompleksnog broja z. Pišemo:

I .Re m,x z y z= =

Standardni ili algebarski oblik kompleksnog broja je oblika

,z x y i= + ⋅

gdje su x i y realni brojevi.

Za kompleksne brojeve x + y · i i x – y · i kažemo da su kompleksno konjugirani jedan drugome.

Simbol konjugiranja jest povlaka iznad broja koji se konjugira:

.z x y i z x y i= + ⋅ ⇒ = − ⋅

( ) ( ) ( ) ( ), , ,22 2 2 2

2 .2

a b a a b b a a a b a b a b a b a b+ = + ⋅ ⋅ + = ⋅ = ⋅ − = − ⋅ +

( ) ( ) ( ),2 2

.2

1 ,n n n

a b i a b i a b i a b a b+ ⋅ ⋅ − ⋅ = + = − ⋅ = ⋅

Stavimo da je

1 3 1 3.x i i= + ⋅ + − ⋅

Nakon kvadriranja i sređivanja dobije se:

1 3 1 3 12

3 1 /3x i i x i i= + ⋅ + − ⋅ ⇒ = + ⋅ + − ⋅ ⇒

( )2

21 3 1 3x i i⇒ = + ⋅ + − ⋅ ⇒

( ) ( )2 2

21 3 2 1 3 1 3 1 3x i i i i⇒ = + ⋅ + ⋅ + ⋅ ⋅ − ⋅ + − ⋅ ⇒

( ) ( )21 3 2 1 3 1 3 1 3x i i i i⇒ = + ⋅ + ⋅ + ⋅ ⋅ − ⋅ + − ⋅ ⇒

( ) ( )21 2 1 3 31 13 3x i ii i⇒ = + ⋅ + ⋅+ ⋅ −⋅ ⋅ + ⋅− ⇒

( ) ( ) ( )22 2 2

2 2 1 3 1 3 2 2 1 3x i i x i⇒ = + ⋅ + ⋅ ⋅ − ⋅ ⇒ = + ⋅ − ⋅ ⇒

( ) ( )22 2 2 2

2 2 1 3 2 2 1 1 3 2 2 1 3x i x x⇒ = + ⋅ − ⋅ ⇒ = + ⋅ − − ⋅ ⇒ = + ⋅ + ⇒

2 2 2 2 22 2 4 2 2 2 2 4 6 6 6./x x x x x x⇒ = + ⋅ ⇒ = + ⋅ ⇒ = + ⇒ = ⇒ = ⇒ =

Odgovor je pod D.

Vježba 141

Pojednostavnite: 1 3 1 3.i i+ ⋅ − − ⋅

. 2 . 3 . 5 . 6A i B i C i D i⋅ ⋅ ⋅ ⋅

2

Rezultat: A.

Zadatak 142 (Iva, gimnazija)

Izračunaj: ( ) ( )4 4

1 1 .i i+ − −

. 1 . . 0 . 1A B i C D −

Rješenje 142

Ponovimo!

( ) ( ) ( )2 22 2

, , ,2

.2 1

2 2mn n m

a a a b a a b b a b a a b b i i⋅

= + = + ⋅ ⋅ + − = − ⋅ ⋅ + =

( ) ( ) ( )22 2 2 2

, ,1 .n n

i a b a b a b a a⋅ ⋅

= − − = − ⋅ + − =

Zakon distribucije množenja prema zbrajanju.

( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

1.inačica

( ) ( ) ( )( ) ( )( ) ( ) ( )2 2 2 24 4 2 2 2 2

1 1 1 1 1 2 1 2i i i i i i i i+ − − = + − − = + ⋅ + − − ⋅ + =

( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 2

1 2 1 1 2 1 2 2 21 1 21 1i i i i i i−= + ⋅ − − − ⋅ − = + ⋅ − − ⋅ = ⋅ − − ⋅− =

( ) ( ) ( ) ( )2 2

2 22 2

2 2 0.i ii i ⋅ − ⋅= ⋅ − ⋅ = =

Odgovor je pod C.

2.inačica

( ) ( ) ( )( ) ( )( ) ( ) ( )( ) ( ) ( )( )2 2

4 4 2 2 2 2 2 21 1 1 1 1 1 1 1i i i i i i i i+ − − = + − − = + − − ⋅ + + − =

( ) ( )( ) ( ) ( )( ) ( )2 21 1 1 1 1 2 1 2i i i i i i i i= + − − ⋅ + + − ⋅ + ⋅ + + − ⋅ + =

( ) ( ) ( ) ( ) ( ) ( )1 1 1 2 1 1 2 11 1 1 1 1 2 1 1 2 1 1 1i i i i i i i ii ii i= + − + ⋅ + + − ⋅ + ⋅ − + − ⋅ − = + + ⋅− + − + ⋅ − + − ⋅⋅ −+ =

2 2 0 0.i= ⋅ ⋅ ⋅ =

Odgovor je pod C.

Vježba 142

Izračunaj: ( ) ( )4 4

1 1 .i i− − +

. 1 . . 0 . 1A B i C D −

Rezultat: C.

Zadatak 143 (Željko, srednja škola)

Nađite apsolutnu vrijednost kompleksnog broja 1 3

2 .z i ii

= ⋅ + +

Rješenje 143

Ponovimo!

, , , .1 ,2 3

1

n a c a d b c a c a ci i i n

b d b d b d b d

⋅ + ⋅ ⋅= − = − = + = ⋅ =

⋅ ⋅




I .Re m,x z y z= =

3

Zapis z x y i= + ⋅ zovemo algebarski ili standardni prikaz kompleksnog broja.

Modul ili apsolutna vrijednost kompleksnog broja z = x + y · i definira se formulom

2 2.z x y= +

1.inačica

1 1 1 1 13 32 2 2

1

iz i i z i i z i i z i z

i i i i i= ⋅ + + ⇒ = ⋅ + + ⇒ = ⋅ + − ⇒ = + ⇒ = + ⇒

21 1 1 0

0 0.i

z z z z zi i i

+ − +⇒ = ⇒ = ⇒ = ⇒ = ⇒ =

2.inačica

1 1 13 32 2 2 2

2

iz i i z i i z i i z i i

i i i i

i

i

−= ⋅ + + ⇒ = ⋅ + + ⇒ = ⋅ + ⋅ − ⇒ = ⋅ + − ⇒

−

−

−

( )2 2 2

11

i iz i i z i i z i i i

− −⇒ = ⋅ + − ⇒ = ⋅ + − ⇒ = ⋅ − − ⇒

− −

0 0.z z⇒ = ⇒ =

Vježba 143

Nađite apsolutnu vrijednost kompleksnog broja 1 3

3 2 .z i ii

= ⋅ + + ⋅

Rezultat: 0.

Zadatak 144 (Ana, gimnazija)

Nađite vrijednost izraza: ( ) ( )

( ) ( )

3 31 1

.2 2

1 1

i i

i i

+ − −

+ − −

. . 2 . 1 . 2A i B i C D⋅

Rješenje 144

Ponovimo!

( ) ( )3 33 2 2 3 3 2

3 3 .2

,3

3 3a b a a b a b b a b a a b a b b+ = + ⋅ ⋅ + ⋅ ⋅ + − = − ⋅ ⋅ + ⋅ ⋅ −

( ) ( )2 22 2 2 2 1 2 3

2 2, , , ,1 .a b a a b b a b a a b b i i i i i+ = + ⋅ ⋅ + − = − ⋅ ⋅ + = = − = −

( ) ( ) ( ) ( )1 3 3 2

, ,2 2 2

.a a a b a b a a b b a b a b a b= − = − ⋅ + ⋅ + − = − ⋅ +

( ) ( )2 2

.a b i a b i a b+ ⋅ ⋅ − ⋅ = +

Skratiti razlomak znači brojnik i nazivnik tog razlomka podijeliti istim brojem različitim od nule i

jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

1.inačica

4

( ) ( )

( ) ( )

( )( )

3 2 1 1 2 3 3 2 1 1 2 33 3 1 3 1 3 1 1 3 1 3 11 1

2 2 2 2 2 21 1 1 2 1 1 2 1

i i i i i ii i

i i i i i i

+ ⋅ ⋅ + ⋅ ⋅ + − − ⋅ ⋅ + ⋅ ⋅ −+ − −= =

+ − − + ⋅ ⋅ + − − ⋅ ⋅ +

( ) ( ) ( )( )( )

( )1 3 3 1 1 3 3 1 1 3 3 1 3 3

1 2 1 1 2 1 1 2 1 1 2 1

i i i i i i i i

i i i i

+ ⋅ + ⋅ − − − − ⋅ + ⋅ − − − + ⋅ − − − − ⋅ − += = =

+ ⋅ − − − ⋅ − + ⋅ − − + ⋅ +

1 3 3 1 3 3 3 3 3 3 41.

1 2 1 1 2 1 2 2 2 2 4

1 3 1 3 4

1 1 1 1 4

i i i i i i i i i i i i i

i i i ii i i i

i+ ⋅ − − − + ⋅ + − + ⋅ − + ⋅ −− − + ⋅

− − +

⋅ − + ⋅ − ⋅= = = = = =

+ ⋅ − − + ⋅ + + ⋅ + ⋅ ⋅ + ⋅ ⋅ ⋅

Odgovor je pod C.

2.inačica

( ) ( )

( ) ( )

( ) ( )( ) ( ) ( ) ( ) ( )( )( ) ( )( ) ( ) ( )( )

2 23 3 1 1 1 1 1 1

1 1

2 2 1 1 1 11 1

i i i i i ii i

i i i ii i

+ − − ⋅ + + + ⋅ − + −+ − −

= =+ − − ⋅ + + −+ − −

( ) ( )( ) ( )

( ) ( )

( ) ( )

2 2 2 21 1 1 2 1 1 1 2 1 1 1 2 1 1 1 1 2 1

1 1 1 1 1 1 1 1

i i i i i i i i i i

i i i i i i i i

+ − + ⋅ + ⋅ + + + + − ⋅ + + − + ⋅ + ⋅ − + + + − ⋅ −= = =

+ − + ⋅ + + − + − + ⋅ + + −

( ) ( )

( ) ( )

1 1 2 21.

1 1 2

1 1 1 2 1 1 2 1

1 1 2 22

2 2i i i

i i

i i i

i ii i

+ + ⋅ + + ⋅ ⋅= = = =

+ + ⋅

− +

−+ ⋅ ⋅

⋅ − + − ⋅ − ⋅ ⋅

− + ⋅ ⋅

Odgovor je pod C.

Vježba 144

Nađite vrijednost izraza: ( ) ( )

( ) ( )

2 21 1

.3 3

1 1

i i

i i

+ − −

+ − −

. . 2 . 1 . 2A i B i C D⋅

Rezultat: C.

Zadatak 145 (MB, gimnazija)

Odredite apsolutnu vrijednost broja 2 2

2 cos 2 sin .7 7

z iπ π⋅ ⋅

= ⋅ + ⋅ ⋅

Rješenje 145

Ponovimo!

( )2 2cos sin 1 , .

n n na b a bα α+ = ⋅ = ⋅




I .Re m,x z y z= =


,z x y i= + ⋅



2 2.z x y= +


5

( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

Kompleksan broj z = x + y · i napisan u trigonometrijskom obliku glasi

( )co sin ,sz z iα α= ⋅ + ⋅

gdje je │z│ apsolutna vrijednost ili modul kompleksnog broja (udaljenost kompleksnog broja od

ishodišta kompleksne ravnine), α argument kompleksnog broja.

1.inačica

( ) ( )cos sincos sin

2 22 22 cos sin2 cos 2 sin

7 77 7

z z iz z i

z iz i

α αα α

π ππ π

= ⋅ + ⋅= ⋅ + ⋅

⇒ ⇒⋅ ⋅⋅ ⋅= ⋅ + ⋅= ⋅ + ⋅ ⋅

( )cos sin

2.2 2cos sin

7 72

z i

zi

z

z

α α

π π

= ⋅ + ⋅

⇒ ⇒ =⋅ ⋅= ⋅ + ⋅

2.inačica

realni dio

2 2 2 22 cos 2 sin 2 cos

imaginarni

2 si7

o

n

i

7 7

d

7z i z i

π π π π⋅ ⋅ ⋅ ⋅= ⋅ + ⋅ ⋅ ⇒ = ⋅ + ⋅ ⋅ ⇒

��

2 2 2 22 2 2 22 2

2 cos 2 sin 2 cos 2 sin7 7 7 7

z zπ π π π⋅ ⋅ ⋅ ⋅

⇒ = ⋅ + ⋅ ⇒ = ⋅ + ⋅ ⇒

2 2 2 22 2 2 24 cos 4 sin 4 cos sin 4 1

7 7 7 7z z z

π π π π⋅ ⋅ ⋅ ⋅⇒ = ⋅ + ⋅ ⇒ = ⋅ + ⇒ = ⋅ ⇒

4 2.z z⇒ = ⇒ =

Vježba 145

Odredite apsolutnu vrijednost broja 3 cos 3 sin .7 7

z iπ π

= ⋅ + ⋅ ⋅

Rezultat: 3.

Zadatak 146 (Ante, srednja škola)

Izračunaj ( ) ( )4 6

1 1 .i i− ⋅ +

. 16 . 32 . . 0A i B i C i D⋅ ⋅

Rješenje 146

Ponovimo!

( ) ( ) ( )2 22 2 2 2

2 2, , .mn n m

a a a b a a b b a b a a b b⋅

= + = + ⋅ ⋅ + − = − ⋅ ⋅ +

( ) , ,2 3

,1 .n n n n m n m

a b a b i i i a a a+

⋅ = ⋅ = − = − ⋅ =

1.inačica

( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( )2 3 2 34 6 2 2 2 32 2

1 1 1 1 1 2 1 2 1 2 1 1 2 1i i i i i i i i i i− ⋅ + = − ⋅ + = − ⋅ + ⋅ + ⋅ + = − ⋅ − ⋅ + ⋅ − =

( ) ( ) ( ) ( ) ( ) ( ) ( )2 3 2 3 2 2 3 3

2 2 2 2 2 2 4 1 8 32 .1 1 1 1i i i i i i i i= − ⋅ ⋅ + ⋅ = − ⋅ ⋅ ⋅ = − ⋅ ⋅ ⋅ = ⋅ − ⋅ ⋅ − = ⋅− −

Odgovor je pod B.

6

2.inačica

( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )444 6 4 4 2 2 22 2

1 1 1 1 1 1 1 1 1 1 1i i i i i i i i i− ⋅ + = − ⋅ + ⋅ + = − ⋅ + ⋅ + = + ⋅ + =

( ) ( ) ( ) ( )4 2 4

1 1 1 2 2 1 2 1 1 1 16 2 16 2 32 .i i i i i i= + ⋅ + ⋅ + = ⋅ + ⋅ − = ⋅ + = ⋅ ⋅ = ⋅−⋅

Odgovor je pod B.

Vježba 146

Izračunaj ( ) ( )4 6

1 1 .i i− ⋅ +

. 16 . 32 . . 0A i B i C i D⋅ ⋅

Rezultat: B.

Zadatak 147 (MS, gimnazija)

33

Ako je 2 11 , onda je 4. Dokazati.z i z z= + ⋅ + =

Rješenje 147

Ponovimo!




I .Re m,x z y z= =


,z x y i= + ⋅




.z x y i z x y i= + ⋅ ⇒ = − ⋅

( )2, , .

2z x y i m n mn n n nz z x y a a a b a b

z x y i

= + ⋅ ⇒ ⋅ = + = ⋅ = ⋅

= − ⋅

( )32 2 3 2 3

1, , , .a b a b a b a b a a i i i⋅ + ⋅ = ⋅ ⋅ + = = − = −

( ) ( )3 33 2 2 3 3 2

3 3 .2

,3

3 3a b a a b a b b a b a a b a b b+ = + ⋅ ⋅ + ⋅ ⋅ + − = − ⋅ ⋅ + ⋅ ⋅ −


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

1.inačica

Ako je 2 11 , onda je 2 11 .z i z i= + ⋅ = − ⋅

Transformiramo 2 11 i 2 11 .z i z i= + ⋅ = − ⋅

• ( ) ( )2 11 8 6 12 8 6 12 8 12 6z i z i i z i i z i i= + ⋅ ⇒ = − + ⋅ − ⇒ = − + ⋅ − ⇒ = + ⋅ − − ⇒

( )33 2 2 3

2 3 2 3 2 2 .z i i i z i⇒ = + ⋅ ⋅ + ⋅ ⋅ + ⇒ = +

• ( ) ( )2 11 8 6 12 8 6 12 8 12 6z i z i i z i i z i i= − ⋅ ⇒ = − + − ⋅ + ⇒ = − − ⋅ + ⇒ = − ⋅ − + ⇒

( )33 2 2 3

2 3 2 3 2 2 .z i i i z i⇒ = − ⋅ ⋅ + ⋅ ⋅ − ⇒ = −

Tada vrijedi:

( ) ( )3 33 333

2 2 2 2 2 2 4.z z i i i i i i+ = + + − = + + − −+ =+=

7

2.inačica

Ako je 2 11 , onda je 2 11 .z i z i= + ⋅ = − ⋅

( )kubiramo 3/

jednakos

33 3 3 33 3 3

4 4 4t

z z z z z z+ = ⇒ ⇒ + = ⇒ + = ⇒

( ) ( ) ( ) ( )2 33 2 3 3 3 33 3 3

3 3 4z z z z z z⇒ + ⋅ ⋅ + ⋅ ⋅ + = ⇒

3 32 23 332 233 3 64 3 3 64z z z z z z z z z z z z⇒ + ⋅ ⋅ + ⋅ ⋅ + = ⇒ + ⋅ ⋅ + ⋅ ⋅ + = ⇒

( )3 333 64z z z z z z⇒ + ⋅ ⋅ ⋅ + + = ⇒

( ) ( ) ( )3332 11 3 2 11 2 11 2 11 64i i i z z i⇒ + ⋅ + ⋅ + ⋅ ⋅ − ⋅ ⋅ + + − ⋅ = ⇒

( ) ( )2 2 3 33 3 3 32 3 2 2 64 2 3 4 111 11 21 2 6411i iz z z z+ ⋅ − ⋅⇒ + ⋅ ⋅ + + = ⇒ + ⋅ + ⋅ + + = ⇒+

( ) ( )33 333 3 34 3 125 64 4 3 5 64z z z z⇒ + ⋅ ⋅ + = ⇒ + ⋅ ⋅ + = ⇒

( ) ( ) ( )3 3 33 3 34 3 5 64 4 15 64 15 64 4z z z z z z⇒ + ⋅ ⋅ + = ⇒ + ⋅ + = ⇒ ⋅ + = − ⇒

( ) ( )pretpostav

3 33 315 60 15 60 15 4 60 60 60

ka

334

4

.z

zz

z z z+ =

⇒ ⋅ + = ⇒ ⇒ ⋅ + = ⇒ ⋅ = ⇒ = ��

Dobije se istinita jednakost. Dakle, početna pretpostavka je točna.

Vježba 147 33

Ako je 2 11 , onda je 4. Dokazati.z i z z= − ⋅ + =

Rezultat: Dokaz analogan.

Zadatak 148 (4A, TUPŠ)

Realan dio kompleksnog broja 6

1 2

b i

i

+ ⋅

− ⋅ jednak je 4. Koliki je realan broj b?

Rješenje 148

Ponovimo!

21, .

a c a ci

b d b d

⋅⋅ = = −

⋅




I .Re m,x z y z= =


,z x y i= + ⋅




.z x y i z x y i= + ⋅ ⇒ = − ⋅

8

Množenje zagrada

( ) ( ) .a b c d a c a d b c b d+ ⋅ + = ⋅ + ⋅ + ⋅ + ⋅

( ) ( )2 2

.z x y i

z z x y i x y i x yz x y i

= + ⋅ ⇒ ⋅ = + ⋅ ⋅ − ⋅ = +

= − ⋅


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

( ) ( )

( ) ( )

1 2

1

6 1 26 6

1 2 1 2 12 2 1 2

b i ib i b iz z z

i i i

i

ii

+ ⋅

+

+ ⋅ ⋅ + ⋅+ ⋅ + ⋅= ⇒ = ⋅ ⇒ = ⇒

− ⋅ − ⋅ − ⋅ + ⋅⋅ ⋅

( )26 12 2 16 12 2 6 12 2

2 2 1 4 51 2

i b i bi b i b i i b i bz z z

+ ⋅ + ⋅ + ⋅ ⋅ −+ ⋅ + ⋅ + ⋅ ⋅ + ⋅ + ⋅ − ⋅⇒ = ⇒ = ⇒ = ⇒

++

( ) ( )6 2

Re6 2 12 6 2 12 5

.125 5 5

Im5

bz

b b i b bz z i

bz

− ⋅=

− ⋅ + + ⋅ − ⋅ +⇒ = ⇒ = + ⋅ ⇒

+=

Iz uvjeta slijedi:

6 2Re 6 2 6 2

4 4 6 2 2/ 5 0 2 20 655 5

Re 4

bz b b

b b

z

− ⋅= − ⋅ − ⋅

⇒ = ⇒ = ⇒ − ⋅ = ⇒ − ⋅ −

=

⋅ = ⇒

( )2 14 2 1 72 ./ :4b b b⇒ − ⋅ = ⇒ − ⋅ = ⇒ = −−

Vježba 148

Realan dio kompleksnog broja 6

1 2

b i

i

+ ⋅

− ⋅ jednak je 2. Koliki je realan broj b?

Rezultat: b = – 2.

Zadatak 149 (Vesna, gimnazija)

Odredi realne brojeve x i y iz jednakosti 1

.1 1

x i y i

i i i

⋅ ⋅+ =

− +

Rješenje 149

Ponovimo!

2.1 1, ,

a c a ci i

b d b d

⋅= − = − ⋅ =

⋅




I .Re m,x z y z= =


,z x y i= + ⋅




.z x y i z x y i= + ⋅ ⇒ = − ⋅

Množenje zagrada

9

( ) ( ) .a b c d a c a d b c b d+ ⋅ + = ⋅ + ⋅ + ⋅ + ⋅

( ) ( )2 2

.z x y i


= + ⋅ ⇒ ⋅ = + ⋅ ⋅ − ⋅ = +

= − ⋅


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

Proširiti razlomak znači brojnik i nazivnik tog razlomka pomnožiti istim brojem različitim od nule i

jedinice

, 0 ., 1a a n

n nb b n

⋅= ≠ ≠

⋅

Dva su kompleksna broja jednaka ako i samo ako su im međusobno jednaki realni dijelovi i

međusobno jednaki imaginarni dijelovi, tj.

. ,a b i c d i a c b d+ ⋅ = + ⋅ ⇔ = =

( )

( ) ( )

( )

( ) ( )

1 11 1

21 1 1 1 1

1 1

1 11 1 1

x i i y i ix i y i x i y i i

i i i i i i i i i i i

i i i

i i i

⋅ ⋅ + ⋅ ⋅ −⋅ ⋅ ⋅ ⋅ −+ = ⇒ ⋅ + ⋅ = ⋅ ⇒ + = ⇒

− + − + − ⋅ + + ⋅+ −

+ −

−

−

− −

( )

( ) ( )2 21 1

2 2 2 2 1 1 1 1 1 11 1 1 1

x i x y i yx i x i y i y i i i⋅ + ⋅ − ⋅ − ⋅ −⋅ + ⋅ ⋅ − ⋅ − −⇒ + = ⇒ + = ⇒

− − + ++ +

22

/1

22 1 2 2

x i x y i y i x i x y i y ix i x y i y i

⋅ − ⋅ + − ⋅ − ⋅ + −⇒ + = ⇒ + = ⇒ ⋅ − + ⋅ + = − ⋅⋅ ⇒

( ) ( ) ( ) ( ) ( ) ( )2 2 0 2x y x i y i i x y x y i i x y x y i i⇒ − + + ⋅ + ⋅ = − ⋅ ⇒ − + + + ⋅ = − ⋅ ⇒ − + + + ⋅ = − ⋅ ⇒

02 2

2

jednakost metoda suprotnih

kompleksnih brojeva koeficijenata

x yy

x y

− + =⇒ ⇒ ⇒ ⇒ ⋅ = − ⇒

+ = −

/ : 22 2 1.y y⇒ ⋅ = − ⇒ = −

Računamo x.

21 2 2 1 1.

1

x yx x x

y

+ = −⇒ − = − ⇒ = − + ⇒ = −

= −

Za realne brojeve x i y vrijedi:

( ) ( ), 1, 1 .x y = − −

Vježba 149

Odredi realne brojeve x i y iz jednakosti 1

.1 1

x i y i

i i i

⋅ ⋅− =

+ −

Rezultat: ( ) ( ), 1, 1 .x y = − −

Zadatak 150 (Vesna, gimnazija)

Izračunaj: ( ) ( ) ( )2 2 2

1 1 2 1 3 .i i i− ⋅ − ⋅ ⋅ − ⋅

Rješenje 150

Ponovimo!

( )2

1 1, , .nn n

a b a b i i⋅ = ⋅ = − = −




10

I .Re m,x z y z= =


,z x y i= + ⋅




.z x y i z x y i= + ⋅ ⇒ = − ⋅

Množenje zagrada

( ) ( ) .a b c d a c a d b c b d+ ⋅ + = ⋅ + ⋅ + ⋅ + ⋅

( ) ( )2 2

.z x y i


= + ⋅ ⇒ ⋅ = + ⋅ ⋅ − ⋅ = +

= − ⋅


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )222 2 2 2

1 1 2 1 3 1 1 2 1 3 1 2 2 1 3i i i i i i i i i i− ⋅ − ⋅ ⋅ − ⋅ = − ⋅ − ⋅ ⋅ − ⋅ = − ⋅ − + ⋅ ⋅ − ⋅ =

( )( ) ( )( ) ( ) ( )( ) ( ) ( )( )2 2 2

1 2 2 1 1 3 1 2 2 1 3 1 3 1 3i i i i i i i i= − ⋅ − + ⋅ − ⋅ − ⋅ = − ⋅ − − ⋅ − ⋅ = − − ⋅ ⋅ − ⋅ =

( ) ( )( ) ( )( ) ( )( ) ( )22 2 22 2

1 3 1 3 1 3 1 9 10 100.i i= − + ⋅ ⋅ − ⋅ = − + = − + = − =

Vježba 150

Izračunaj: ( ) ( ) ( )4 4 4

1 1 2 1 3 .i i i− ⋅ − ⋅ ⋅ − ⋅

Rezultat: 10000.


Odredi realne brojeve x i y iz jednakosti ( ) ( )1 1 .i x i y i− ⋅ + + ⋅ =

Rješenje 151

Ponovimo!




I .Re m,x z y z= =


,z x y i= + ⋅


Dva su kompleksna broja jednaka ako i samo ako su im međusobno jednaki realni dijelovi i

međusobno jednaki imaginarni dijelovi, tj.

. ,a b i c d i a c b d+ ⋅ = + ⋅ ⇔ = =


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

( ) ( ) ( ) ( )1 1i x i y i x x i y y i i x y x i y i i x y x y i i− ⋅ + + ⋅ = ⇒ − ⋅ + + ⋅ = ⇒ + − ⋅ + ⋅ = ⇒ + + − + ⋅ = ⇒

11

( ) ( )

jednakostmetoda suprotnih

0 1 kompleksnihkoeficijenat

0

1 abrojeva

x yx y x y i i

x y

+ =⇒ + + − + ⋅ = + ⋅ ⇒ ⇒ ⇒ ⇒

− + =

12 1 /1 .22 :

2y y y⇒ ⋅ = ⇒ ⋅ = ⇒ =

Računamo x.

01 1

0 .12 2

2

x y

x xy

+ =

⇒ + = ⇒ = −=

Rješenje je:

( )1 1

, , .2 2

x y = −

Vježba 151

Odredi realne brojeve x i y iz jednakosti ( ) ( )1 1 .i x i y i+ ⋅ + − ⋅ =

Rezultat: ( )1 1

, , .2 2

x y = −


Odredi realni i imaginarni dio kompleksnog broja 4 2 2 6 2

.2

iz

+ ⋅ − ⋅ ⋅=

Rješenje 152

Ponovimo!

.a b a b

n n n

−− =




I .Re m,x z y z= =


,z x y i= + ⋅



( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

Skratiti razlomak znači brojnik i nazivnik tog razlomka podijeliti istim brojem različitim od nule i jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

( )4 2 2 6 24 2 2 6 2 4 2 2 6 2

2 2 2 2

iiz z z i

+ ⋅ − ⋅ ⋅+ ⋅ − ⋅ ⋅ + ⋅ ⋅= ⇒ = ⇒ = − ⋅ ⇒

( ) ( )2 2 2 2 26 2 22 2 3 2

2 2

2 6

2 2z i z i z i

⋅ + ⋅ +⋅ ⋅⇒ = − ⋅ ⇒ = − ⋅ ⇒ = + − ⋅ ⋅ ⇒

12

( )

( )

Re 2 2.

Im 3 2

z

z

= +⇒

= − ⋅

Vježba 152

Odredi realni i imaginarni dio kompleksnog broja 6 2 2 6 2

.2

iz

+ ⋅ ⋅ ⋅=

+

Rezultat: ( ) ( )Re 3 2 , Im 3 2.z z= + = ⋅


Pojednostavnite ( )8

1 .i−

Rješenje 153

Ponovimo!

( ) ( )2

,2 2

2 .1,2mn n m

a a a b a a b b i⋅

= − = − ⋅ ⋅ + = −

( )4

1, .n n n

a b a b i⋅ = ⋅ =




I .Re m,x z y z= =


,z x y i= + ⋅


( ) ( )( ) ( ) ( ) ( )4 48 2 4 42 2

1 1 1 2 1 1 2 1 21 1i i i i i i− = − = − ⋅ ⋅ + = − = − ⋅ −⋅ − =

( ) ( )4 4 4

2 2 16 1 16.i i= − ⋅ = − ⋅ = ⋅ =

Vježba 153

Pojednostavnite ( )8

1 .i+

Rezultat: 16.

Zadatak 154 (Nina, gimnazija)

Zadan je kompleksan broj z = 1 + 2 · i. Koliko je │z – 3│?

. 0 . 2 2 . 5 3 . 3 3A B C D⋅ − −

Rješenje 154

Ponovimo!

.a b a b⋅ = ⋅




I .Re m,x z y z= =


,z x y i= + ⋅


13


2 2.z x y= +

[ ] ( )2 2

3 1 2 3 2 2 2 2 4 21 4 42zz i ii− = = + ⋅ − = − + ⋅= + = − + = + = ⋅⋅ =

4 2 2 2.= ⋅ = ⋅

Vježba 154

Zadan je kompleksan broj z = 2 + 2 · i. Koliko je │z – 4│?

. 0 . 2 2 . 5 3 . 3 3A B C D⋅ − −

Rezultat: B.

Zadatak 155 (Ana, gimnazija)

Koji je od navedenih brojeva realan?

( ). 2 cos sin . 4 cos sin2 2

A i B iπ π

π π⋅ + ⋅ ⋅ + ⋅

. 6 cos sin . 8 cos sin3 3 4 4

C i D iπ π π π

⋅ + ⋅ ⋅ + ⋅

Rješenje 155

Ponovimo!

cos 1 sin 0 cos 0 sin 12

, , ,2

π ππ π= − = = =

, ,3 2 21

cos sin cos sin3 2 3 2 4 2 4 2

,π π π π

= = = =

Skup realnih brojeva R je unija skupa racionalnih brojeva Q i skupa iracionalnih brojeva I.

Imaginarni brojevi imaju oblik b · i gdje je b realni broj koji nije jednak nuli, a i je imaginarna jedinica za koju vrijedi

.1i = −


gdje su x i y realni brojevi, a i imaginarna jedinica. Broj x zove se realni dio kompleksnog broja z, a broj y imaginarni dio kompleksnog broja z. Pišemo:

I .Re m,x z y z= =


,z x y i= + ⋅

gdje su x i y realni brojevi. Zakon distribucije množenja prema zbrajanju.

( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

Preoblikujemo svaki od navedenih brojeva:

• ( ) ( ) ( )2 cos sin 2 1 0 2 1 realan broj0 2i iπ π⋅ + ⋅ = ⋅ − + ⋅ = ⋅ − + = −

14

• ( ) ( ) imaginaran bro4 cos sin 4 0 1 4 0 j42 2

i i i iπ π

⋅ + ⋅ = ⋅ + ⋅ = ⋅ + = ⋅

• 31

6 cos sin 6 3 3 33 3 2 2

kompleksan broji i iπ π

⋅ + ⋅ = ⋅ + ⋅ = + ⋅ ⋅

• 2 2

8 cos sin 8 4 2 4 2 .4 4 2

komple2

ksan broji i iπ π

⋅ + ⋅ = ⋅ + ⋅ = ⋅ + ⋅ ⋅

Odgovor je pod A.

Vježba 155

Koji je od navedenih brojeva imaginaran broj?

( ). 2 cos sin . 4 cos sin2 2

A i B iπ π

π π⋅ + ⋅ ⋅ + ⋅

. 6 cos sin . 8 cos sin3 3 4 4

C i D iπ π π π

⋅ + ⋅ ⋅ + ⋅

Rezultat: B.

Zadatak 156 (Darko, gimnazija)

Broj z prikazan je u kompleksnoj ravnini. Zapišite ga ili u trigonometrijskome ili u standardnome obliku.

Rješenje 156

Ponovimo!

3 31 10 0 0 0cos120 sin120 cos 60 sin 60, , , .

2,

2 2 2

b a ba

c c

⋅= − = = = ⋅ =


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

Kompleksan broj je broj oblika

15

,z x y i= + ⋅

gdje su x i y realni brojevi, a i imaginarna jedinica. Broj x zove se realni dio kompleksnog broja z, a broj y imaginarni dio kompleksnog broja z. Pišemo:

I .Re m,x z y z= =


,z x y i= + ⋅

gdje su x i y realni brojevi. Kompleksne brojeve predočujemo u koordinatnoj ravnini koju zovemo kompleksna ili Gaussova ravnina.

Trigonometrijski oblik kompleksnog broja

Svaki se kompleksan broj može prikazati u obliku

( ) ( )cos sin ili co i .s s nz r i z z iϕ ϕ ϕ ϕ= ⋅ + ⋅ = ⋅ + ⋅

Tu je r modul ili apsolutna vrijednost kompleksnog broja, r =│z│, mijenja se od 0 do + ∞, a argument

ili amplituda φ = arg z od – ∞ do + ∞. Smisao okretanja oko ishodišta O uzima se pozitivnim, ako

je protivan smislu okretanja kazaljke na satu.

Sa slike razabiremo da je modul kompleksnog broja z jednak polumjeru kružnice pa vrijedi:

16

( )( )

0314 , 120 0 0

4 cos120 sin120 42 2cos sin

rz i z i

z r i

ϕ

ϕ ϕ

= =⇒ = ⋅ + ⋅ ⇒ = ⋅ − + ⋅ ⇒

= ⋅ + ⋅

2 2 3 .z i⇒ = − + ⋅ ⋅

Trigonometrijski oblik kompleksnog broja glasi: ( )0 04 cos120 sin120 .z i= ⋅ + ⋅

Standardni oblik kompleksnog broja glasi: 2 2 3 .z i= − + ⋅ ⋅

Vježba 156

Broj z prikazan je u kompleksnoj ravnini. Zapišite ga ili u trigonometrijskome ili u

standardnome obliku.

Rezultat: ( )0 04 cos 60 sin 60 ili 2 2 3 .z i z i= ⋅ + ⋅ = + ⋅ ⋅

Zadatak 157 (Marija, Enisa, pedagoški fakultet)

( )20

Izračunaj 5 5 .i− ⋅

Rješenje 157

Ponovimo!


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +



broj y imaginarni dio kompleksnog broja z.

( ) ( ) ( ) ( )2 2

, , , .2 2 2

2mn n n n n m

a b a b a a a b a a b b a a⋅

⋅ = ⋅ = − = − ⋅ ⋅ + − =

21 , .

n m n mi a a a

+= − ⋅ =

Potencije imaginarne jedinice i

Za vrijedin N∈

4 4 1 4 2 4, , , .

31 1

n n n ni i i i i i

⋅ ⋅ + ⋅ + ⋅ += = = − = −

17

Preoblikujemo izraz:

( ) ( )( ) ( ) ( )( ) ( )10 102020 20 220 20 20 2

5 5 5 1 5 1 5 1 5 1 2i i i i i i− ⋅ = ⋅ − = ⋅ − = ⋅ − = ⋅ − ⋅ + =

( ) ( ) ( ) ( )10 10 10 1020 20 20 20 10 20 10 10

5 1 2 1 5 2 5 2 5 2 51 21i i i i i= ⋅ − ⋅ − = ⋅ − ⋅ = ⋅ − ⋅ = ⋅ − ⋅ = ⋅ ⋅− =

( ) ( )1010 : 4 2 20 10 2 20 10 20 10 2 10 10 10

5 2 5 2 1 5 22

5 2 25 2i=

= = ⋅ ⋅ = ⋅ ⋅ − = − ⋅ = − ⋅ = − ⋅ =

( )10 10

25 2 50 .= − ⋅ = −

Vježba 157

( )20

Izračunaj 2 2 .i− ⋅

Rezultat: 30

2 .−

Zadatak 158 (Kiki, gimnazija)

Neka je z = x + y · i. Odredi realni i imaginarni dio kompleksnog broja 1

.2

z

Rješenje 158

Ponovimo!




I .Re m,x z y z= =


,z x y i= + ⋅




.z x y i z x y i= + ⋅ ⇒ = − ⋅

( ) ( ) .2 2

x y i x y i x y+ ⋅ ⋅ − ⋅ = +


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

( ) ( )22 2 2

1 2, , .n n n

i a b a a b b a b a b= − + = + ⋅ ⋅ + ⋅ = ⋅

( ) ( )2 22 , , .

2 m a b a bn n ma b a a b b a a

n n n

−⋅− = − ⋅ ⋅ + = = −

, .

n na a a c a c

nb b d b db

⋅= ⋅ =

⋅

1.inačica

( ) ( )

1 1 1 1

2 2 2 2 2 2222z x x y i y ix y i x x y i y i

= = = =

+ ⋅ ⋅ ⋅ + ⋅+ ⋅ + ⋅ ⋅ ⋅ + ⋅

( )

1 1 1

2 2 2 2 2 22 1 2 2x x y i y x x y i y x y x y i

= = = =

+ ⋅ ⋅ ⋅ + ⋅ − + ⋅ ⋅ ⋅ − − + ⋅ ⋅ ⋅

18

( ) ( )( )( )

2 22

1 1

2 2 2 2 2 22 2 2

x y x y i

x y x y i x y x y i x y x y i

− − ⋅ ⋅ ⋅

= = ⋅ =

− + ⋅ ⋅ ⋅ − + ⋅ ⋅ ⋅ − − ⋅ ⋅ ⋅

( ) ( ) ( ) ( )

2 2 2 22 2

2 2 222 2 2 2 2 2 2 22 2 4

x y x y i x y x y i

x y x y x x y y x y

− − ⋅ ⋅ ⋅ − − ⋅ ⋅ ⋅= = =

− + ⋅ ⋅ − ⋅ ⋅ + + ⋅ ⋅

( )

2 2 2 2 2 22 2 2

4 2 2 4 2 2 4 2 2 4 22 22 4 2

x y x y i x y x y i x y x y i

x x y y x y x x y yx y

− − ⋅ ⋅ ⋅ − − ⋅ ⋅ ⋅ − − ⋅ ⋅ ⋅= = = =

− ⋅ ⋅ + + ⋅ ⋅ + ⋅ ⋅ ++

( ) ( )

2 22

.2 2

2 2 2 2

x y x yi

x y x y

− ⋅ ⋅= − ⋅

+ +

Tada je:

( ) ( )

2 21 1 2

Re , Im .2 2 2 2

2 2 2 2

x y x y

z zx y x y

− ⋅ ⋅= = −

+ +

2.inačica

( )

( )

222 221 1 1 1

2 2 2 22 2

x y ix y i x y i

z x y i x y i x y iz x yx y

− ⋅− ⋅ − ⋅= = = ⋅ = = =

+ ⋅ + ⋅ − ⋅ ++

( )

( ) ( )( )

( )

22 2 22 2 22 2 12

2 2 22 2 2 2 2 2

x x y i y i x x y i yx x y i y i

x y x y x y

− ⋅ ⋅ ⋅ + ⋅ − ⋅ ⋅ ⋅ + ⋅ −− ⋅ ⋅ ⋅ + ⋅= = = =

+ + +

( ) ( ) ( ) ( )

2 2 2 2 2 22 2 2

.2 2 2 2

2 2 2 2 2 2 2 2

x x y i y x y x y i x y x yi

x y x y x y x y

− ⋅ ⋅ ⋅ − − − ⋅ ⋅ ⋅ − ⋅ ⋅= = = − ⋅

+ + + +

Tada je:

( ) ( )

2 21 1 2

Re , Im .2 2 2 2

2 2 2 2

x y x y

z zx y x y

− ⋅ ⋅= = −

+ +

Vježba 158

Neka je z = x + y · i. Odredi realni i imaginarni dio kompleksnog broja 1

.2

z

Rezultat: 1 5 1 12

Re , Im .2 2169 169z z

= = −

Zadatak 159 (Ivan, veleučilište)

Neka je :f C C→ funkcija definirana s f(x) = a · x + b, gdje su ,a b R∈ konstante. Pokazati

da za svaki z C∈ vrijedi jednakost ( ) ( ).f z f z=

Rješenje 159

19

Ponovimo!




I .Re m,x z y z= =


,z x y i= + ⋅




.z x y i z x y i z x y i= + ⋅ ⇒ = + ⋅ ⇒ = − ⋅


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

Provjerimo jednakost ( ) ( )f z f z= tako da izračunamo svaku stranu posebno.

• ( ) ( ) ( ) ( ) ( )f x af z f x y i f x y i a x yx b i b= ⋅= + ⋅ = − ⋅ = = ⋅ − ⋅ ++ =

,a x a y i b a x b a y i= ⋅ − ⋅ ⋅ + = ⋅ + − ⋅ ⋅

• ( ) ( ) ( ) ( )f x af z f x y i a x y i b a x a y ix b b= + ⋅ = = ⋅ + ⋅ + = ⋅ += ⋅ + ⋅ ⋅ + =

.a x b a y i a x b a y i= ⋅ + + ⋅ ⋅ = ⋅ + − ⋅ ⋅

Zaključak:

( )

( )( ) ( ).

f z a x b a y if z f z

f z a x b a y i

= ⋅ + − ⋅ ⋅⇒ =

= ⋅ + − ⋅ ⋅

Vježba 159

Neka je :f C C→ funkcija definirana s f(x) = a · x2 + b · x + c, gdje su , ,a b c R∈

konstante. Pokazati da za svaki z C∈ vrijedi jednakost ( ) ( ).f z f z=

Rezultat: Dokaz analogan.

Zadatak 160 (Tessa, gimnazija)

Vrijednost potencije ( )8 5

4 3,

kk

i⋅ +

⋅ + gdje je i imaginarna jedinica, a k Z∈ jednaka je:

. 1 . . 1 .A B i C D i− −

Rješenje 160 Ponovimo!

( ),1

, .mn m n m n n m

a a a a a a a+ ⋅

= ⋅ = =

Potencije imaginarne jedinice

Neka je k prirodni broj. Tada za potencije imaginarne jedinice i vrijedi:

4 4 1 4 2 4, , , .

31 1

k k k ki i i i i i

⋅ ⋅ + ⋅ + ⋅ += = = − = −

Ako je eksponent potencije imaginarne jedinice djeljiv s 4, vrijednost potencije je 1. Ako je ostatak pri

dijeljenju eksponenta s 4 jednak 1, vrijednost potencije je i; ako je ostatak 2, vrijednost je – 1, a ako

je ostatak 3, vrijednost potencije je – i.

20

( ) ( ) ( ) ( ) ( ) ( )8 5 8 5 8 4 1 8 44 3

kk k kk

i i i i i i⋅ +

⋅ + ⋅ + ⋅ +⋅ += − = − ⋅ − = − ⋅ − =

( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )4 2 24 22 22 4 2

1 1 .

k kk

i i i i i i i i i

⋅ + ⋅ +⋅ +

= − ⋅ − = ⋅ − = ⋅ − = − ⋅ − = ⋅ − = −

Odgovor je pod D.

Vježba 160

Vrijednost potencije ( )8 5

4 1,

kk

i⋅ +

⋅ + gdje je i imaginarna jedinica, a k Z∈ jednaka je:

. 1 . . 1 .A B i C D i− −

Rezultat: B.

Zadatak 141 (Branka, gimnazija) - halapa.com · broj y imaginarni dio kompleksnog broja z. Pišemo:...

Documents

Transcript of Zadatak 141 (Branka, gimnazija) - halapa.com · broj y imaginarni dio kompleksnog broja z. Pišemo:...