Odredi kompleksan broj z = x + y · i za koji vrijedi z z i− = + ⋅1 2 . Rješenje 184 Ponovimo! Kompleksan broj je broj oblika z x y i= + ⋅ , gdje su x i y realni brojevi, a

1

Zadatak 181 (Dado, gimnazija)

Odredi ,x y R∈ tako da vrijedi 2 2 2 4.x y i y x i⋅ − ⋅ + ⋅ − ⋅ ⋅ =

Rješenje 181

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.

Zakon distribucije množenja prema zbrajanju.

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

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+ ⋅ = + ⋅ ⇔ = =

2 2 2 4 2 2 2 4x y i y x i x y y i x i⋅ − ⋅ + ⋅ − ⋅ ⋅ = ⇒ ⋅ + ⋅ − ⋅ − ⋅ ⋅ = ⇒

( ) ( ) ( ) ( )2 2 2 4 2 2 2 4 0x y y x i x y x y i i⇒ ⋅ + ⋅ + − − ⋅ ⋅ = ⇒ ⋅ + ⋅ + − ⋅ − ⋅ = + ⋅ ⇒

metoda suprotnih

koe

2 2 44.

2 0 ficijenata

x yy

x y

⋅ + ⋅ =⇒ ⇒ ⇒ =

− ⋅ − =

Računamo x.

2 2 42 2 4 4 2 8 4 2 4 8 2 4

4

x yx x x x

y

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

=

/ : 22 4 2.x x⇒ ⋅ = − ⇒ = −

Rješenje je:

( ) ( ), 2, 4 .x y = −

Vježba 181

Odredi ,x y R∈ tako da vrijedi 2 2 2 4 0.x y i y x i⋅ − ⋅ + ⋅ − ⋅ ⋅ − =

Rezultat: ( ) ( ), 2, 4 .x y = −

Zadatak 182 (Nikola, srednja škola)

Dokazati da je ( )4

1k

i⋅

+ realan broj, ako je k prirodan broj.

Rješenje 182

Ponovimo!




I .Re m,x z y z= =


,z x y i= + ⋅


2

( ) ( ) ( )2 2

,2 2

2 1, , .m nn n m n n

a a a b a a b b i a b a b⋅

= + = + ⋅ ⋅ + = − ⋅ = ⋅

( )2 2 2

2 .a b a a b b− = − ⋅ ⋅ +

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

1 1 1 1 2 1 2 1

k kk k

ki i i i i i

⋅+ = + = + = + ⋅ + = + ⋅ − =

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

2 2 21 1 4 1 4 .

k k k k ki i i= + ⋅ = ⋅ = ⋅ = ⋅ − = −−

( )Za slijedi 4 .k

k N R∈ − ∈

Vježba 182

Dokazati da je ( )4

1k

i⋅

− realan broj, ako je k prirodan broj.

Rezultat: Dokaz analogan, ( )4 .k

R− ∈

Zadatak 183 (Tomislav, gimnazija)

2

Ako je , onda iznosi1

iz z z

i

−= ⋅

−

2 13 3. . . 2 1 .

2 2 4A B C D

++

Rješenje 183

Ponovimo!




I .Re m,x z y z= =


,z x y i= + ⋅


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= + ⋅ ⇒ = − ⋅

Umnožak dvaju konjugirano kompleksnih brojeva realan je broj i nenegativan broj:

( ) ( ) .2 2

x y i x y i x y+ ⋅ ⋅ − ⋅ = +

( ) ( ) 2, , .

21,

2a c a c a b a ba b a b a b i

b d b d n n n

⋅ +⋅ = − ⋅ + = − = − = +

⋅

( ) ( ) ( )2 2 22 22 2

,2 2

2 22

, , .a a

a b a a b b a b a a b b a ab b

= + = + ⋅ ⋅ + − = − ⋅ ⋅ + =

Množenje zagrada

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

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

jedinice

3

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

Odredimo standardni ili algebarski oblik kompleksnog broja z:

( ) ( )

( ) ( )

22 12 2 2 21

2 21 1 1 1 1 1 1

i ii i i i iiz z z z

i i i i i

− ⋅ +− − + ⋅ − −+= ⇒ = ⋅ ⇒ = ⇒ = ⇒

− − + − ⋅ + +

( )2 2 1 2 2 1 2 1 2

1 1 2 2

i i i i i iz z z

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

+

( ) ( )2 1 2 1 2 1 2 1 2 1 2 1.

2 2 2 2 2

iz z i z i

+ + − ⋅ + − + −⇒ = ⇒ = + ⋅ ⇒ = − ⋅

Sada je:

2 22 1 2 1 2 1 2 1 2 1 2 1

2 2 2 2 2 2z z i i z z

+ − + − + −⋅ = + ⋅ ⋅ − ⋅ ⇒ ⋅ = + ⇒

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

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

2 2 4 42 2

z z z z+ − + ⋅ + − ⋅ +

⇒ ⋅ = + ⇒ ⋅ = + ⇒

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

4 4 4z z z z

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

2 2 2 22 1 2 1 2 1 2 1 6

4 4 4z z z z z z

+ + + + + +⇒ ⋅ = ⇒

+ ⋅ − ⋅⋅ = ⇒ ⋅ = ⇒

6

4

3.

2z z z z⇒ ⋅ = ⇒ ⋅ =

Odgovor je pod A.

Vježba 183

2

Ako je , onda iznosi1

iz z z

i

−= ⋅

+

2 13 3. . . 2 1 .

2 2 4A B C D

++

Rezultat: A.

Zadatak 184 (Cedric, Sean, Željko, Medox, Höhere Technische Lehranstalt)

Odredi kompleksan broj z = x + y · i za koji vrijedi 1 2 .z z i− = + ⋅

Rješenje 184

Ponovimo!




I .Re m,x z y z= =


,z x y i= + ⋅


4

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

2 2.z x y= +



. ,a b i c d i a c b d+ ⋅ = + ⋅ ⇔ = =

( ) ( )2 2 2

, .2

2a a a b a a b b= + = + ⋅ ⋅ +


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

[ ] ( )1 2 1 2z z i x y iz x y x y ii i− = + ⋅ ⇒ ⇒ + ⋅ − + ⋅ = ++ ⋅ ⋅= ⇒

jednakost

kompleksnih

br

2 2 2

o

21 2 1

jeva

2x y x y i i x y x y i i⇒ + − − ⋅ = + ⋅ ⇒ + − − ⋅ = + ⋅ ⇒ ⇒

( )

2 22 2 2 211 1

22 2

metoda

zamjene/ 1

x y xx y x x y x

yy y⋅ −

+ − =+ − = + = +⇒ ⇒ ⇒ ⇒ ⇒

− =− = = −

( ) 2/

22 2 22 1 4 1 4 1x x x x x x⇒ + − = + ⇒ + = + ⇒ + = + ⇒

( )2

22 2 24 1 4 1 2 4 1 2

2 2x x x x xx x x⇒ + = + ⇒ + = + ⋅ + ⇒ + = + ⋅ ⇒+

34 1 2 1 2 4 2 4 1 2 3 2 3 .2

2/ :x x x x x x⇒ = + ⋅ ⇒ + ⋅ = ⇒ ⋅ = − ⇒ ⋅ = ⇒ ⋅ = ⇒ =

Traženi kompleksan broj glasi:

.

2

322

2

3x

y i z i

y

z x=

=

= + ⋅ ⇒

−

⇒ = − ⋅

Vježba 184

Odredi kompleksan broj z = x + y · i za koji vrijedi 1 2 .z i z− − ⋅ =

Rezultat: 3

2 .2

z i= − ⋅

Zadatak 185 (4A, 4B, TUPŠ)

Neka je z = 3 + 2 · i. Koliko je ( )4

?i z z⋅ ⋅

Rješenje 185

Ponovimo!

( ) ( ), .2

1,mn n n n n m

a b a b a a i⋅

⋅ = ⋅ = = −




I .Re m,x z y z= =

5


,z x y i= + ⋅




.z x y i z x y i= + ⋅ ⇒ = − ⋅


( ) ( ) .2 2

x y i x y i x y+ ⋅ ⋅ − ⋅ = +

( ) ( ) ( )( ) ( )( ) ( )( )43 24 4 42 2

3 2 3 2 3 2 9 43 2

i z z i i i i iz i

z i⋅ ⋅ = = ⋅ + ⋅ ⋅ − ⋅ = ⋅ + =

= + ⋅

= − ⋅⋅ + =

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

13 13 13 13 13 1 13 1 13 28561.i i i i= ⋅ = ⋅ = ⋅ = ⋅ = ⋅ − = ⋅ = =

Vježba 185

Neka je z = 3 – 2 · i. Koliko je ( )4

?i z z⋅ ⋅

Rezultat: 28561.

Zadatak 186 (Tonka, gimnazija)

Dokaži da je ( )1000 500

1 2 .i+ =

Rješenje 186

Ponovimo!

( ) ( )2 2 2 2 4

2 1 1, , , , 1.mn n m

a a a b a a b b i i i⋅

= + = + ⋅ ⋅ + = − = =�

( ) .n n n

a b a b⋅ = ⋅




I .Re m,x z y z= =


,z x y i= + ⋅


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.

1.inačica

( ) ( )( ) ( ) ( ) ( )500 5001000 2 500 5002

1 1 1 2 11 2 2 11i i i i i i+ = + = + ⋅ + = + ⋅ −+ ⋅ − = =

( ) ( )125500 500 500 500 4 500 125 500 500

2 2 2 2 1 2 1 2 .i i i= ⋅ = ⋅ = ⋅ = ⋅ = ⋅ =

6

2.inačica

( ) ( )( ) ( ) ( ) ( )500 5001000 2 500 5002

1 1 1 2 11 2 2 11i i i i i i+ = + = + ⋅ + = + ⋅ −+ ⋅ − = =

( )

500 : 4 125

10

20

500 500 500 500 500 5002 2 2 2 1 2 .

0

i i i= ⋅ = ⋅ = = ⋅ = ⋅

=

=

�

Vježba 186

Dokaži da je ( )100 50

1 2 .i+ =

Rezultat: Dokaz analogan.

Zadatak 187 (Patrik, gimnazija)

Izračunati .a b i

b a i

+ ⋅

− ⋅

Rješenje 187

Ponovimo!

1 2, , , .1

a c a cn m n ma a a a a i

b d b d

⋅+= ⋅ = = − ⋅ =

⋅




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+ ⋅ ⋅ − ⋅ = +

Množenje zagrada

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


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅


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

( ) ( )( ) ( )

2 2 2

2 2

a b i b a ia b i a b i b a i a b a i b i a b i

b a i b a i b a i b a i b a i b a

+ ⋅ ⋅ + ⋅+ ⋅ + ⋅ + ⋅ ⋅ + ⋅ + ⋅ + ⋅ ⋅= ⋅ = = =

− ⋅ − ⋅ + ⋅ − ⋅ ⋅ + ⋅ +

7

( )2 2 2 2 2 21

2 2 2 2 2 2

a b a i b i a b a b a i b i a b a i b i

b a b a b a

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

⋅ − ⋅=

+ + +

( ) ( )2 22 2

.2 2 2 2

2 2

2 2

a b i ia i b

a bi

i

b a a bb a

+ ⋅ ⋅⋅ + ⋅

= =+

= =+

+

+

Vježba 187

Izračunati .a b i

b a i

+ ⋅

− + ⋅

Rezultat: .i−

Zadatak 188 (Maja, gimnazija)

Izračunati 1

.1

i

i

+

−

Rješenje 188

Ponovimo!

( )21 2 2 2

, 2, , .1n m n m

a a a a a i a b a a b b+

= ⋅ = = − + = + ⋅ ⋅ +




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+ ⋅ ⋅ − ⋅ = +

Množenje zagrada

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

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

jedinice

, 0 ., 1a a n

n nb b n

⋅= ≠ ≠

⋅


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅


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



8

. ,a b i c d i a c b d+ ⋅ = + ⋅ ⇔ = =

1.inačica

( ) ( )( ) ( )

( )2 2

1 1 11 1 1 1 2

2 21 1 1 1 1

proširiti

razloma 11 1k 1

i i ii i i i i

i i i i i

+ ⋅ + ++ + + + ⋅ += = ⋅ = = = =

− − + − ⋅ + ++

1 2 1 2 2.

2

1 1 2

2 22

i i i ii

−+ ⋅ − + ⋅ ⋅ ⋅= = = = =

2.inačica

Pretpostavimo da je zadani izraz jednak .x y i+ ⋅

Dalje slijedi:

( ) ( ) ( )/ 11 1

1 11 1

i ix y i x y i i x y i

i ii i

+ += + ⋅ ⇒ = + ⋅ ⇒⋅ − + = + ⋅ ⋅ − ⇒

− −

( )21 1 1i x x i y i y i i x x i y i y⇒ + = − ⋅ + ⋅ − ⋅ ⇒ + = − ⋅ + ⋅ − ⋅ − ⇒

( ) ( )

jednakost

kompleksnih

brojev

1 1

a

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

metoda suprotnih

ko

1 12

eficije1 ta2

n1 a

x y x yy

x y x y

= + + =⇒ ⇒ ⇒ ⇒ ⋅ = ⇒

= − + − + =

/ : 22 2 1.y y⇒ ⋅ = ⇒ =

Računamo x.

11 1 11 0.

1

yx x x

x y

=⇒ + = ⇒ = ⇒

+ =+ =

Rezultat glasi:

1 1 10 1 .

1 1

0

1 1

i i ix y i i i

i i i

x

y

+ + += + ⋅ ⇒ ⇒ = + ⋅ ⇒

−

==

=− −

Vježba 188

Izračunati 1

.1

i

i

−

+

Rezultat: .i−

Zadatak 189 (Dodo, gimnazija)

Koliki je argument φ u trigonometrijskome prikazu kompleksnog broja z = 5 · i?

2 3. . . .

3 2 3 2A B C D

π π π π⋅ ⋅

Rješenje 189

Ponovimo!


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= =


9

,z x y i= + ⋅

gdje su x i y realni brojevi. Svaki se kompleksan broj

z x y i= + ⋅

može prikazati u trigonometrijskom obliku

( )co sin ,sz r iϕ ϕ= ⋅ + ⋅

pri čemu je r modul kompleksnog broja

2 2,z r x y= = +

a kut φ argument kompleksnog broja za kojeg vrijedi

1i .0 2,li

y yarctg tg

x xϕ ϕ ϕ π

−= = ≤ < ⋅

Oznaka je g .ar zϕ =

Kompleksne brojeve predočujemo u koordinatnoj ravnini koju zovemo kompleksna ili Gaussova

ravnina.

1.inačica

05 0 5 .

5

xz i z i

y

== ⋅ ⇒ = + ⋅ ⇒

=

Računamo argument φ.

51 1.

0 2

y ytg tg tg

x x

πϕ ϕ ϕ ϕ

− −= ⇒ = ⇒ = ⇒ =

Odgovor je pod B.

2.inačica

Kompleksan broj z = 5 · i prikažemo u kompleksnoj ravnini i očitamo argument φ.

1

ϕϕϕϕ = ππππ

2

0

Im

Re

5 ⋅⋅⋅⋅ i

4 ⋅⋅⋅⋅ i

3 ⋅⋅⋅⋅ i

2 ⋅⋅⋅⋅ i

i

D

Odgovor je pod B

Vježba 189

Koliki je argument φ u trigonometrijskome prikazu kompleksnog broja z = 3 · i?

2 3. . . .

3 2 3 2A B C D

π π π π⋅ ⋅

Rezultat: B.

10

Zadatak 190 (Petra, gimnazija)

Nađi modul kompleksnog broja 1 1

.2 2

zi i

= −− +

Rješenje 190

Ponovimo!




I .Re m,x z y z= =


,z x y i= + ⋅


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= + ⋅ ⇒ = − ⋅


( ) ( ) .2 2

x y i x y i x y+ ⋅ ⋅ − ⋅ = +

2 1.,, ,0

a c a d b ca a a a a

b d b d

⋅ − ⋅− = = ≥ =

⋅

( ),2

.n m n m

a a a a a+

⋅ = =


2 2.z x y= +

Za apsolutnu vrijednost vrijedi:

111 2 1 2

2

.

2

,zz

z z z zz z

= ⋅ = ⋅


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

1.inačica

Kompleksan broj napišemo u standardnom obliku.

( )( ) ( )2 21 1 2 2

2 22 2 2 2 4 12

2

1

2i i i i i iz z z z

i i i i

+ − − + − + + += − ⇒ = ⇒ = ⇒ = ⇒

− + − ⋅ + +

−

+

2 2.

5 5

iz z i

⋅⇒ = ⇒ = ⋅

Modul iznosi:

0 22 2 22

0 02

5

22

5 55

z x y

x

z i z i zy

=

= ⋅ ⇒ = + ⋅ ⇒ ⇒ ⇒ = + ⇒==

+

22 2

.5 5

z z⇒ = ⇒ =

11

2.inačica

Preoblikujemo kompleksan broj.

( )( ) ( ) ( ) ( ) ( ) ( )2 21 1 2 2

2 2 2

2 2

2 2 2 2 2

i i i i i iz z z z

i i i i i i i i

+ − − + − −+ + += − ⇒ = ⇒ = ⇒ = ⇒

− + − ⋅ + − ⋅ + − ⋅ +

( ) ( )2

.2 2

iz

i i

⋅⇒ =

− ⋅ +

Modul iznosi:

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

22 2

2 2 2 2 2 2

ii iz z z

i i i i i i

⋅⋅ ⋅= ⇒ = ⇒ = ⇒

− ⋅ + − ⋅ + − ⋅ +

( )

22 2 2

22 2 22 2 4 1 4 12 1 2 1

iz z z

i i

⋅⇒ = ⇒ = ⇒ = ⇒

− ⋅ + + ⋅ ++ − ⋅ +

( )

2 2 2.

2 55 55

z z z⇒ = ⇒ = ⇒ =⋅

Vježba 190

Nađi modul kompleksnog broja 1 1

.2 2

zi i

= − −− + +

Rezultat: 2

.5

z =

Zadatak 191 (Petra, gimnazija)

Prikažite u trigonometrijskom obliku ( )sin 1 cos .z iα α= + ⋅ −

Rješenje 191

Ponovimo!




I .Re m,x z y z= =


,z x y i= + ⋅


Svaki se kompleksan broj z x y i= + ⋅




2 2,z r x y= = +


1i .0 2,li

y yarctg tg

x xϕ ϕ ϕ π

−= = ≤ < ⋅

Oznaka je

12

g .ar zϕ =

( )2 2 2 2 2 2

2 cos sin 1, , 1 cos 2 .sin2

a b a a b bα

α α α− = − ⋅ ⋅ + + = − = ⋅

( )2, 0 sin 2 2 sin c s, o, .a b a b a a a α α α⋅ = ⋅ = ≥ ⋅ = ⋅ ⋅

sin1

co, , , .

s

n m n ma a a a a tg tg tg

αα α β α β

α

+= ⋅ = = = ⇒ =


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


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

Računamo apsolutnu vrijednost zadanog kompleksnog broja.

( )sin

sin 1 cos1

2 2

cosz x

xz yi

y

αα α

α

== += + ⋅ − ⇒ ⇒ ⇒

= −

( )22 2 2

sin 1 cos sin 1 2 cos cosz zα α α α α⇒ = + − ⇒ = + − ⋅ + ⇒

2 2sin cos 1 2 cos 1 1 2 cos 2 2 cosz z zα α α α α⇒ = + + − ⋅ ⇒ = + − ⋅ ⇒ = − ⋅ ⇒

( ) 22 22 1 cos 2 2 sin 2 sin

2 2z z z

α αα⇒ = ⋅ − ⇒ = ⋅ ⋅ ⇒ = ⋅ ⇒

2 22 sin 2 sin .

2 2z z

α α⇒ = ⋅ ⇒ = ⋅

Računamo argument φ.

22 sin1 cos 2

sin2 sin cos

2

o

i

2

1 c s

s n

ytg tg tg

x

y

x

αα

ϕ ϕ ϕα αα

α

α

⋅ −= ⇒ ⇒ = ⇒ = ⇒

⋅ ⋅

= −

=

sin sin2 2 .

2 2cos cos

22

2 sin22 2

tg tg tg tg

α αα α

ϕ ϕ ϕα

ϕα α

⋅⇒ = ⇒ = ⇒ = ⇒ =

⋅ ⋅

Trigonometrijski oblik glasi:

( )2 sin

22 sin cos sin .

2 2 2o si

2

c s nz

z

z iz iϕ ϕ

α

α α α

αϕ

= ⋅

⇒ ⇒ = ⋅ ⋅

=

= ⋅⋅ + ⋅ +

Vježba 191

Prikažite u trigonometrijskom obliku ( )sin cos 1 .z iα α= − ⋅ −

Rezultat: 2 sin cos sin .2 2 2

z iα α α

= ⋅ ⋅ + ⋅

13

Zadatak 192 (Dubravko, gimnazija)

Izračunati:

21 1

, 1.1 1

a a aa i i i a

a a a a

−+ ⋅ + − ⋅ − − + >

− −

Rješenje 192

Ponovimo!




I .Re m,x z y z= =


,z x y i= + ⋅



2 2.z x y= +


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

( ), ,1

, .

n n ma a n a c a d b cn n ma a nn

b b d b db

⋅ + ⋅⋅= = = + =

⋅

, , .aa c a d b c a

a b a bb d b d b b

⋅ − ⋅− = = ⋅ = ⋅

⋅

02., 0 0 , 0,a a a a

a= ≥ = ≠

2 21 1 1 1

11 1 1 1

a a a a a aa i i i a i i i

a a a a a a a a

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

− − − −

222 2 21 12 2

11 1

a a aa

a a a a

−= + + + − − − + =

− −

( )

( )

( )( )

222 2

1 121

2 2 2 21 1

aa aa

a a a a

−= + + + − + =

− −

( )

( ) ( )

22 2 41 1 1

2 2 2 21 11 1

aa a a

a a a a

−= + + + − + =

− −

( ) ( )

( )

( )

( )

2 2 24 4 41 1 1

2 2 221 1

a a a a a a

a a a a

+ − − + + −= + − =

⋅ − −

14

( ) ( )

( )

( )

( )

2 2 24 4 41 1 1

2 2 221 1

a a a a a a

a a a a

+ − + − + −= + − =

⋅ − −

( ) ( )

( )

( )

( )

2 2 24 4 41 1 1

2 2 221 1

a a a a a a

a a a a

+ − + − + −= + − =

⋅ − −

( ) ( )

( )

( )2 2 24 4 4

1 1 1

22 11

a a a a a a

a aa a

+ − + − + −= + − =

−⋅ −

( ) ( )( )

( )2 2 24 4 4

1 1 1

1 1

a a a a a a

a a a a

+ − + − + −= + − =

⋅ − −

( ) ( )( )

( )2 2 24 4 4

1 1

1 1

1

a a

a a a a a a

a a= + − =

⋅

+ − + + −

−

−

−

( )( )

( )( )

1 1 1 1 12 24 41 1

1 1 1

a aa a a a

a a a a a a

− + −= + − ⋅ + − = + − ⋅ =

⋅ − − ⋅ −

( )( )

( )( )

( )02 2 24 4 4

1 1 1 01

01

.1 1

a a a a a aa a a a

a a= + − ⋅ = + − ⋅ = + − ⋅ =

⋅ − −

−

⋅

− +

Vježba 192

Izračunati:

21 1

, 1.1 1

a a aa i i i a

a a a a

−+ ⋅ + + ⋅ − + >

− −

Rezultat: 0.

Zadatak 193 (Ana, medicinska škola)

Zadan je kompleksan broj ( )2

, gdje je .a

z a i a Ri

= + + ∈ Zapišite ga u standardnom obliku

( ), , .z x y i x y R= + ⋅ ∈

Rješenje 193

Ponovimo!

( )1 12 2 2 2

2 1, , , .a

a b a a b b i i ai b b

+ = + ⋅ ⋅ + = − = − = ⋅




I .Re m,x z y z= =


,z x y i= + ⋅


( ) ( )12 2 2 2

2 2 1a

z a i z a a i i a z a a i a ii i

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

15

2 22 1 1 .z a a i a i z a a i⇒ = + ⋅ ⋅ − − ⋅ ⇒ = − + ⋅

Vježba 193

Zadan je kompleksan broj ( )2

, gdje je .a

z a i a Ri

= + − ∈ Zapišite ga u standardnom obliku

( ), , .z x y i x y R= + ⋅ ∈

Rezultat: 2

1 3 .z a a i= − + ⋅ ⋅

Zadatak 194 (Ana, medicinska škola)

Odredite apsolutnu vrijednost broja 2 2

2 cos 2 sin .7 7

z iπ π⋅ ⋅

= ⋅ + ⋅ ⋅

Rješenje 194

Ponovimo!

( ) 2 2cos sin 1, .

n n na b a b x x⋅ = ⋅ + =




I .Re m,x z y z= =


,z x y i= + ⋅



2 2.z x y= +


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

22 cos

72 2

22 cos 2 sin2 si7 7

2

n7

2z x y i x

z iy

z x y

π

π ππ

⋅= + ⋅ = ⋅

⇒ ⇒ ⇒⋅ ⋅⋅= ⋅

=+ ⋅ ⋅

=

+

⋅

2 22 2 2 22 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

7 7 7 7z z

π π π π⋅ ⋅ ⋅ ⋅⇒ = ⋅ + ⋅ ⇒ = ⋅ + ⇒

4 1 4 2.z z z⇒ = ⋅ ⇒ = ⇒ =

Vježba 194

Odredite apsolutnu vrijednost broja 2 2

3 cos 3 sin .7 7

z iπ π⋅ ⋅

= ⋅ + ⋅ ⋅

Rezultat: 3.

16

Zadatak 195 (Darko, gimnazija)

Koliko ima kompleksnih brojeva za koje vrijede jednakosti 2, 4 1?z i z i− = − ⋅ =

. 0 . 1 . 2 . 4A B C D

Rješenje 195

Ponovimo!

( ) ( )2 2 2 2

2 0, 0, .n

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




I .Re m,x z y z= =


,z x y i= + ⋅



2 2.z x y= +


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

Neka je zadan kompleksan broj .z x y i= + ⋅

Tada vrijedi:

[ ]( )

( )

1 22 2

4 1 4 1 4 1z x y

x y iz i x y i i

z i x y i i x y ii

+ − ⋅ =− = + ⋅ − =⇒ ⇒ ⇒ ⇒

− ⋅ = + ⋅= +

− ⋅ = + ⋅ =⋅

−

( )

( )

( )

( )

( )

( )

2

222 2

2 2 1 22 21 2 1 2

22 22 2 22 24 1

/

2/4 1

4 1

x yx y x y

x y x yx y

+ − =+ − = + − =

⇒ ⇒ ⇒ ⇒

+ − = + − =+ − =

( )

( )

22 2 2 2 21 4 2 1 4 2 4 1

2 2 2 2 228 16 1 8 1 164 1

x y x y y x y y

x y y x y yx y

+ − = + − ⋅ + = + − ⋅ = −⇒ ⇒ ⇒ ⇒

+ − ⋅ + = + − ⋅ = −+ − =

( )

metoda suprotnih

koeficijenat

2 22 22 32 3

2 22 28 18 /5 11 a5

x y yx y y

x y yx y y

+ − ⋅ =+ − ⋅ =⇒ ⇒ ⇒ ⇒

+ − ⋅ = −+ − ⋅ −⋅ = −

2 22 3

6 18 6 18 3.2 2

8 15

/ : 6x y y

y y y

x y y

+ − ⋅ =⇒ ⇒ ⋅ = ⇒ ⋅ = ⇒ =

− − + ⋅ =

Računamo x.

2 22 2 2 22 3

3 2 3 3 9 6 3 3 33

x y yx x x

y

+ − ⋅ =⇒ + − ⋅ = ⇒ + − = ⇒ + = ⇒

=

17

2 20 .3 03x x x⇒ = ⇒ = ⇒ =+

Kompleksan broj je

0 , 30 3 3 .

x yz i z i

z x y i

= =⇒ = + ⋅ ⇒ = ⋅

= + ⋅

Dakle, postoji jedan kompleksan broj.

Odgovor je pod B.

Vježba 195

Koliko ima kompleksnih brojeva za koje vrijede jednakosti 1, 3 1?z i z i− = − ⋅ =

. 0 . 1 . 2 . 4A B C D

Rezultat: B.

Zadatak 196 (4A, 4B, TUPŠ)

Napišite 5 kao umnožak nekih dvaju kompleksnih brojeva kojima su i realni i imaginarni

dijelovi različiti od 0.

Rješenje 196

Ponovimo!

( ) ,2

, , .a b a b a c a c a b a b

a an n n b d b d n n n

+ ⋅ −= + = ⋅ = = −

⋅

1 21, , .

n m n ma a a a a i

+= ⋅ = = −




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+ ⋅ ⋅ − ⋅ = +

Množenje zagrada

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



. ,a b i c d i a c b d+ ⋅ = + ⋅ ⇔ = =


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

1.inačica

Broj 5 možemo prikazati kao umnožak odgovarajućeg kompleksnog broja i pripadnog mu konjugirano

kompleksnog. Primjeri:

18

• ( ) ( ) ( ) ( )2 2

2 3 2 3 2 3 2 3 5i i+ ⋅ ⋅ − ⋅ = + = + =

• ( ) ( ) 2 22 2 2 1 4 1 5i i+ ⋅ − = + = + =

•

2 27 3 7 3 7 3 7 3 7 3 10

52 2 2 2 2

10

22 2 2 2 2i i

++ ⋅ ⋅ − ⋅ = + = + = = = =

itd.

2.inačica

Na početku možemo uzeti bilo koji kompleksan broj koji zadovoljava uvjete zadatka. Na primjer,

1 .z i= −

Sada moramo odrediti kompleksan broj w x y i= + ⋅

tako da vrijedi:

( ) ( )5 1 5.z w i x y i⋅ = ⇒ − ⋅ + ⋅ =

Izračunat ćemo x i y na dva načina.

1. način rada

( ) ( ) ( ) ( )1

/1

51 5 1 5

1i x y i i x y i x y i

ii− ⋅ + ⋅ = ⇒ − ⋅ + ⋅ = ⇒ + ⋅ =

− −⋅ ⇒

( )( ) ( )

5 15 1 5 5

2 21 1 1 1 1 1

ii ix y i x y i x y i

i i i i

⋅ ++ + ⋅⇒ + ⋅ = ⋅ ⇒ + ⋅ = ⇒ + ⋅ = ⇒

− + − ⋅ + +

5 5 5 5 5 5 5 5.

1 1 2 2 2 2 2

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

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

+

Dakle, dva kompleksa broja

5 51 ,

2 2z i w i= − = + ⋅

pomnoženi daju umnožak 5.

2. način rada

( ) ( ) ( )21 5 5 1 5i x y i x y i x i y i x y i x i y− ⋅ + ⋅ = ⇒ + ⋅ − ⋅ − ⋅ = ⇒ + ⋅ − ⋅ − ⋅ − = ⇒

( ) ( ) ( ) ( )5 5 5 0x y i x i y x y x y i x y x y i i⇒ + ⋅ − ⋅ + = ⇒ + + − + ⋅ = ⇒ + + − + ⋅ = + ⋅ ⇒

52 5

0

jednakost metoda suprotnih

kompleksnih brojeva koeficijenata

x yy

x y

+ =⇒ ⇒ ⇒ ⇒ ⋅ = ⇒

− + =

/ : 25

2 5 .2

y y⇒ ⋅ = ⇒ =

Računamo x.

( )/ 1

05 5 5 5

0 .52 2 2 2

2

x y

x x x xy

− + =

⇒ − + = ⇒ ⋅ −− = − ⇒ − = − ⇒ ==

Dakle, kompleksan broj w glasi:

5 5.

2 2w i= + ⋅

Vježba 196

Napišite 7 kao umnožak nekih dvaju kompleksnih brojeva kojima su i realni i imaginarni

dijelovi različiti od 0.

19

Rezultat: 5 2 , 5 2

.7 71 ,

2 2

z i w i

z i w i

= + ⋅ = − ⋅

= + = − ⋅

Zadatak 197 (Lana, gimnazija)

Odredite realan broj b ako je ( ) ( )4 2 1 10 .i b i i− ⋅ ⋅ − + ⋅ = ⋅

Rješenje 197

Ponovimo!

.2

1i = −




I .Re m,x z y z= =


,z x y i= + ⋅


Množenje zagrada

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



. ,a b i c d i a c b d+ ⋅ = + ⋅ ⇔ = =


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

( ) ( ) 24 2 1 10 4 4 2 2 10i b i i b i i b i i− ⋅ ⋅ − + ⋅ = ⋅ ⇒ − + ⋅ ⋅ + ⋅ − ⋅ ⋅ = ⋅ ⇒

( )4 4 2 2 1 10 4 4 2 2 10b i i b i b i i b i⇒ − + ⋅ ⋅ + ⋅ − ⋅ ⋅ − = ⋅ ⇒ − + ⋅ ⋅ + ⋅ + ⋅ = ⋅ ⇒

( ) ( ) ( ) ( )4 2 4 2 10 4 2 4 2 0 10b b i i i b b i i⇒ − + ⋅ + ⋅ ⋅ + ⋅ = ⋅ ⇒ − + ⋅ + ⋅ + ⋅ = + ⋅ ⇒

4 2 0 2 4 2 4 2 4 22.

4 2 10 4 10 2 4

/ : 2

/8 :4 48 2

b b b b bb

b b b b b

− + ⋅ = ⋅ = ⋅ = ⋅ = =⇒ ⇒ ⇒ ⇒ ⇒ ⇒ =

⋅ + = ⋅ = − ⋅ = ⋅ = =

Vježba 197

Odredite realan broj b ako je ( ) ( )4 2 1 10 .i b i i− + ⋅ ⋅ − ⋅ = ⋅

Rezultat: b = 2.

Zadatak 198 (Tea, gimnazija)

Koliki je argument φ u trigonometrijskome zapisu kompleksnog broja

cos sin ?3 3

z i iπ π

= ⋅ + ⋅

Rješenje 198

Ponovimo!



20


I .Re m,x z y z= =


,z x y i= + ⋅


Svaki se kompleksan broj z x y i= + ⋅




2 2,z r x y= = +


1i .0 2,li

y yarctg tg

x xϕ ϕ ϕ π

−= = ≤ < ⋅

Oznaka je g .ar zϕ =

Kompleksne brojeve predočujemo u koordinatnoj ravnini koju zovemo kompleksna ili Gaussova

ravnina.


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


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

Trokut je dio ravnine omeđen s tri dužine. Te dužine zovemo stranice trokuta.

Pravokutni trokuti imaju jedan pravi kut (kut od 90º). Stranice koje zatvaraju pravi kut zovu se katete,

a najdulja stranica je hipotenuza pravokutnog trokuta.

Pitagorin poučak

Trokut je pravokutan ako i samo ako je kvadrat nad hipotenuzom jednak zbroju kvadrata nad

katetama.

2 2 2 2 2 2 2 2 2, , .c a b a c b b c a= + = − = −

Tangens šiljastog kuta pravokutnog trokuta jednak je omjeru duljine katete nasuprot tog kuta i duljine

katete uz taj kut.

311 21 cos s, , , , in

3 2 3 2.

n m n ma a a a a i

π π+= ⋅ = = − = =

( )2 3

, , , , .6 2

a

a d a c a d b c a c a cb a a tgc b c b d b d b d b d

d

π⋅ ⋅ + ⋅ ⋅= + = ⋅ = = =

⋅ ⋅ ⋅

, .1

n a c a d b cn

b d b d

⋅ − ⋅= − =

⋅

Preoblikujemo kompleksan broj tako da ga zapišemo u standardnom obliku.

21

1cos

3 2

3sin

3

31cos sin

3

2

3 2 2z i i z i i

π

π

π

π= ⋅ + ⋅ ⇒ ⇒ = ⋅ + ⋅ ⇒

=

=

( )3 3 3 31 1 1 12

12 2 2 2 2 2 2 2

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

3Re

2 .1

Im2

z

z

= −

⇒

=

Kompleksan broj z nalazi se u drugom kvadrantu kompleksne ravnine.

Re

Im

z ϕϕϕϕ

ππππ

ααααC

y

x

αααα

ππππ

ϕϕϕϕz

Im

Re

Sa slika vidi se:

racionalizacija2

nazivnika

2

1 1

12

3 3 3

2

ytg tg tg tg

xα α α α= ⇒ = ⇒ = ⇒ = ⇒ ⇒

( )

3 3 3 31 1.

2 3 3 63 33

tg tg tg tgπ

α α α α α−

⇒ = ⋅ ⇒ = ⇒ = ⇒ = ⇒ =

Argument φ iznosi:

6 5.

6 1 6 6 6

π π π π π πϕ π α ϕ π ϕ ϕ ϕ

⋅ − ⋅= − ⇒ = − ⇒ = − ⇒ = ⇒ =

Vježba 198

Koliki je argument φ u trigonometrijskome zapisu kompleksnog broja

sin cos ?3 3

z i iπ π

= ⋅ ⋅ +

Rezultat: 5

.6

πϕ

⋅=

22

Zadatak 199 (Patrik, srednja škola)

Jednakost ( ) ( ) ( ) ( )1 1 2 1 1a i i i b i+ ⋅ ⋅ − ⋅ = + ⋅ − ⋅ ispunjena je ako je:

. 3 . 1 . 4 . 2A a b B a b C a b D a b+ = + = − + = − + =

Rješenje 199

Ponovimo!

1 21, , .

n m n ma a a a a i

+= ⋅ = = −

Množenje zagrada

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




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⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +



. ,a b i c d i a c b d+ ⋅ = + ⋅ ⇔ = =

( ) ( ) ( ) ( ) 2 21 1 2 1 1 1 2 2 1a i i i b i i a i a i b i i b i+ ⋅ ⋅ − ⋅ = + ⋅ − ⋅ ⇒ − ⋅ + ⋅ − ⋅ ⋅ = − ⋅ + − ⋅ ⇒

2 21 1

2 22 2 2 2i a i a i b i i b i i a i a i b i i b i⇒ − ⋅ + ⋅ − ⋅ ⋅ = − ⋅ + − ⋅ ⇒ − ⋅ + ⋅ − ⋅ ⋅ = − ⋅ + − ⋅ ⇒

( ) ( )2 2 1 1 2 2i a i a b i i b i a i a b i i b⇒ − ⋅ + ⋅ − ⋅ ⋅ − = − ⋅ + − ⋅ − ⇒ − ⋅ + ⋅ + ⋅ = − ⋅ + + ⇒

( ) ( )jednakost kompleksnih

brojeva

22 2 1

2 1

a ba a i b b i

a b

⋅ =⇒ ⋅ + − + ⋅ = + − + ⋅ ⇒ ⇒ ⇒

− + = − +

metoda suprotnih

koeficijenata

2 0 2 03 3

1 2 3

a b a ba

a b a b

⋅ − = ⋅ − =⇒ ⇒ ⇒ ⇒ ⋅ = ⇒

+ = + + =

/ : 33 3 1.a a⇒ ⋅ = ⇒ =

Računamo b.

31 3 3 1 2.

1

a bb b b

a

+ =⇒ + = ⇒ = − ⇒ =

=

Vidi se:

1 2 3.a b a b+ = + ⇒ + =

Odgovor je pod A.

Vježba 199 Jednakost ( ) ( ) ( ) ( )1 2 1 1 1i a i b i i− + ⋅ ⋅ + ⋅ = − + ⋅ ⋅ + ispunjena je ako je:

. 3 . 1 . 4 . 2A a b B a b C a b D a b+ = + = − + = − + =

Rezultat: A.

23

Zadatak 200 (Lara, gimnazija)

( )5

Iz 1 slijedi :i a b i− = + ⋅

. 1 . 0 . 1 . 0A a b B a b C a b D a b− = + = + = − − =

Rješenje 200

Ponovimo!

( ), ,1 2

1, .mn m n m n n m

a a a a a a a i+ ⋅

= ⋅ = = = −

( ) ( )2 .2 2

,2 n n n

a b a a b b a b a b− = − ⋅ ⋅ + ⋅ = ⋅


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




I .Re m,x z y z= =


,z x y i= + ⋅




. ,a b i c d i a c b d+ ⋅ = + ⋅ ⇔ = =

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

5 4 1 21 1 1 1 1i a b i i i a b i i i a b i− = + ⋅ ⇒ − ⋅ − = + ⋅ ⇒ − ⋅ − = + ⋅ ⇒

( ) ( ) ( ) ( )2 22

1 2 1 1 2 1 1i i i a b i i i a b i⇒ − ⋅ + ⋅ − = + ⋅ ⇒ − ⋅ − ⋅ − = + ⋅ ⇒

( ) ( ) ( ) ( ) ( )1 12 2 2

2 1 2 1 4 1i i a b i i i a b i i i a b i⇒ − ⋅ ⋅ − = + ⋅ ⇒ − ⋅ ⋅ − = + ⋅ ⇒ ⋅ ⋅ − = + ⋅− ⇒

( ) ( ) ( )4 1 1 4 1 4 4i a b i i a b i i a b i⇒ ⋅ − ⋅ − = + ⋅ ⇒ − ⋅ − = + ⋅ ⇒ − + ⋅ = + ⋅ ⇒

jednakost kompleksnih

br

4 4.

4 4ojeva

a a

b b

− = = −⇒ ⇒ ⇒

= =

Tada je

.44 044a b a b a b+ = − + ⇒ + = ⇒ + =− +

Odgovor je pod B.

Vježba 200

( )5

Iz 1 0 slijedi :i a b i− − − ⋅ =

. 1 . 0 . 1 . 0A a b B a b C a b D a b− = + = + = − − =

Rezultat: B.

Documents

Odredi kompleksan broj z = x + y · i za koji vrijedi z z i− = + ⋅1 2 . Rješenje 184 Ponovimo! Kompleksan broj je broj oblika z x y i= + ⋅ , gdje su x i y realni brojevi, a