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# Numeros Complexos Forma Algebrica e Trigonometrica-Gabarito-Lista_02

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```CÁLCULO DIFERENCIAL E INTEGRAL IV
NÚMEROS COMPLEXOS
FORMA ALGÉBRICA E TRIGONOMÉTRICA \u2013 LISTA 02

Profº Georges Cherry Rodrigues

LISTA DE EXERCÍCIOS - GABARITO

1) Represente os seguintes números no plano:

a) P1 = 2+3i b) P2 = 4-i c) P3 = -3-4i d) P4 = -1+2i e) P5 = -2i

Solução. Representando cada número complexo como pontos no plano Argand-Gauss, temos:

2) Determine o módulo e o argumento dos seguintes complexos:

a) 4+3i b) 2-2i c) 3+i d) 3 e)2i f) a+bi

Solução. Aplicando a fórmula do módulo e identificando os valores de cosseno e seno, temos:

a)
525916)3()4(34 22 \uf03d\uf03d\uf02b\uf03d\uf02b\uf03d\uf02b i
; º8,36
5
3
5
4
cos
:)( \uf040\uf0de
\uf0ef
\uf0ef
\uf0ee
\uf0ef\uf0ef
\uf0ed
\uf0ec
\uf03d
\uf03d
\uf071
\uf071
\uf071
sen
zArg
b)
22844)2()2(22 22 \uf03d\uf03d\uf02b\uf03d\uf02d\uf02b\uf03d\uf02d i
; º315
2
2
22
2
2
2
22
2
cos
:)( \uf03d\uf0de
\uf0ef
\uf0ef
\uf0ee
\uf0ef
\uf0ef
\uf0ed
\uf0ec
\uf02d\uf03d
\uf02d
\uf03d
\uf03d\uf03d
\uf071
\uf071
\uf071
sen
zArg
c)
1019)1()3(3 22 \uf03d\uf02b\uf03d\uf02b\uf03d\uf02b i
; º4,18
10
1
10
3
cos
:)( \uf040\uf0de
\uf0ef
\uf0ef
\uf0ee
\uf0ef
\uf0ef
\uf0ed
\uf0ec
\uf03d
\uf03d
\uf071
\uf071
\uf071
sen
zArg
d)
39)3(3 2 \uf03d\uf03d\uf03d
; º0
0
3
0
1
3
3
cos
:)( \uf03d\uf0de
\uf0ef
\uf0ef
\uf0ee
\uf0ef\uf0ef
\uf0ed
\uf0ec
\uf03d\uf03d
\uf03d\uf03d
\uf071
\uf071
\uf071
sen
zArg
e)
24)2(2 2 \uf03d\uf03d\uf03di
; º90
1
2
2
0
2
0
cos
:)( \uf03d\uf0de
\uf0ef
\uf0ef
\uf0ee
\uf0ef\uf0ef
\uf0ed
\uf0ec
\uf03d\uf03d
\uf03d\uf03d
\uf071
\uf071
\uf071
sen
zArg
e)
2222 )()( bababia \uf02b\uf03d\uf02b\uf03d\uf02b
;
a
b
arctg
ba
b
sen
ba
a
zArg \uf03d\uf0de
\uf0ef
\uf0ef
\uf0ee
\uf0ef
\uf0ef
\uf0ed
\uf0ec
\uf02b
\uf03d
\uf02b
\uf03d
\uf071
\uf071
\uf071
22
22
cos
:)(
3) Obtenha o produto w =
z z z1 2 3. .
onde:

a)
z i
z i
z i
1
2
3
16 160 160
5 325 325
308 308
\uf03d \uf02b
\uf03d \uf02b
\uf03d \uf02b
(cos sen )
(cos sen )
cos sen
\uf06f \uf06f
\uf06f \uf06f
\uf06f \uf06f
b)
)43sen43(cos6
)31sen31(cos4
)14sen14(cos3
3
2
1
\uf06f\uf06f
\uf06f\uf06f
\uf06f\uf06f
iz
iz
iz
\uf02b\uf03d
\uf02b\uf03d
\uf02b\uf03d

Solução. Aplicando a fórmula do produto entre complexos na forma trigonométrica e encontrando a
primeira determinação positiva, se necessário, temos:

a)
)7373.(cos80)793793.(cos80..
)]308º325º160()308º325º160)[cos(1)(5).(16(..
321
321
\uf06f\uf06f\uf06f\uf06f
\uf06f\uf06f
isenisenzzz
isenzzz
\uf02b\uf03d\uf02b\uf03d
\uf02b\uf02b\uf02b\uf02b\uf02b\uf03d

b)
)8888.(cos72..
)]43º31º14()43º31º14)[cos(6)(4).(3(..
321
321
\uf06f\uf06f
\uf06f\uf06f
isenzzz
isenzzz
\uf02b\uf03d
\uf02b\uf02b\uf02b\uf02b\uf02b\uf03d

4) Sendo z=
2
4 4
(cos sen )
\uf070 \uf070
\uf02b i
, determine z
2
, z
3
e z
4
.
Solução. Aplicando a fórmula da potência na forma trigonométrica dos complexos, temos:

i)
\uf028 \uf029 )
22
(cos2)
4
.2
4
.2.(cos2)
44
(cos2
2
2
2 \uf070\uf070\uf070\uf070\uf070\uf070 isenisenisenz \uf02b\uf03d\uf02b\uf03d\uf0fa
\uf0fb
\uf0f9
\uf0ea
\uf0eb
\uf0e9
\uf02b\uf03d

ii)
\uf028 \uf029 )
4
3
4
3
(cos22)
4
.3
4
.3.(cos2)
44
(cos2
3
3
3 \uf070\uf070\uf070\uf070\uf070\uf070 isenisenisenz \uf02b\uf03d\uf02b\uf03d\uf0fa
\uf0fb
\uf0f9
\uf0ea
\uf0eb
\uf0e9
\uf02b\uf03d

iii)
\uf028 \uf029 )(cos4)
4
.4
4
.4.(cos2)
44
(cos2
4
4
4 \uf070\uf070\uf070\uf070\uf070\uf070 isenisenisenz \uf02b\uf03d\uf02b\uf03d\uf0fa
\uf0fb
\uf0f9
\uf0ea
\uf0eb
\uf0e9
\uf02b\uf03d

OBS: Repare que
)
44
(cos2
\uf070\uf070
isenz \uf02b\uf03d
iz \uf02b\uf03d1
. O cálculo de
z
2
, z
3
e z
4
poderia ser feito nesta forma. No caso deste exercício, representando os resultados na forma
algébrica teríamos:
i)
iiisenz 2)0.(2)
22
(cos22 \uf03d\uf02b\uf03d\uf02b\uf03d
\uf070\uf070
ii)
iiisenz 22)
2
2
2
2
(22)
4
3
4
3
(cos223 \uf02b\uf02d\uf03d\uf02b\uf02d\uf03d\uf02b\uf03d
\uf070\uf070.
iii)
4)01(4)(cos44 \uf02d\uf03d\uf02b\uf02d\uf03d\uf02b\uf03d \uf070\uf070 isenz
5) Determine o módulo e o argumento do número
z 4
para os complexos:

a) z = 3(cos125\uf0b0+isen125\uf0b0) b) z = 2(cos300º + isen300º)

Solução. Aplicando a fórmula da potência na forma trigonométrica dos complexos e encontrando a
primeira determinação positiva, se necessário, temos:

a)
\uf05b \uf05d \uf028 \uf029
\uf0ee
\uf0ed
\uf0ec
\uf03d
\uf03d
\uf0de\uf02b\uf03d
\uf02b\uf03d\uf02b\uf03d\uf02b\uf03d
º140
81
)º140º140(cos81
)º500º500(cos81)]º125).(4()º125).(4.[cos(3)º125º125(cos3
4
444
\uf071
\uf072
isenz
isenisenisenz

b)
\uf05b \uf05d \uf028 \uf029
\uf0ee
\uf0ed
\uf0ec
\uf03d
\uf03d
\uf0de\uf02b\uf03d
\uf02b\uf03d\uf02b\uf03d\uf02b\uf03d
º120
16
)º120º120(cos16
)º1200º1200(cos16)]º300).(4()º300).(4.[cos(2)º300º300(cos2
4
444
\uf071
\uf072
isenz
isenisenisenz

6) Calcule as potências, dando a resposta na forma algébrica ou trigonométrica.

a)
( )1 3 8\uf02d i
b)
( )3 6\uf02b i

Solução. Escreve-se os números entre parênteses na forma trigonométrica e calculam-se as potências.

a) Calculando módulo e argumento:
-
2431)3()1(31 22 \uf03d\uf03d\uf02b\uf03d\uf02b\uf03d\uf02b i
; º300
2
3
2
1
cos
:)( \uf03d\uf0de
\uf0ef
\uf0ef
\uf0ee
\uf0ef\uf0ef
\uf0ed
\uf0ec
\uf02d\uf03d
\uf03d
\uf071
\uf071
\uf071
sen
zArg
-
\uf05b \uf05d \uf028 \uf029
\uf028 \uf029
\uf028 \uf029 ébricaAiii
ou
ricaTrigonométiseni
isenisenisen
lg3128128
2
3
2
1
.25631
º240º240(cos25631
)º2400º2400(cos256)]º300).(8()º300).(8.[cos(2)º300º300(cos2
8
8
88
\uf0ae\uf02d\uf02d\uf03d\uf0f7
\uf0f7
\uf0f8
\uf0f6
\uf0e7
\uf0e7
\uf0e8
\uf0e6
\uf02d\uf02d\uf03d\uf02d
\uf0ae\uf02b\uf03d\uf02d\uf0de
\uf02b\uf03d\uf02b\uf03d\uf02b

b) Calculando módulo e argumento:
-
2413)1()3(3 22 \uf03d\uf03d\uf02b\uf03d\uf02b\uf03d\uf02b i
; º30
2
1
2
3
cos
:)( \uf03d\uf0de
\uf0ef
\uf0ef
\uf0ee
\uf0ef\uf0ef
\uf0ed
\uf0ec
\uf03d
\uf03d
\uf071
\uf071
\uf071
sen
zArg
-
\uf05b \uf05d \uf028 \uf029
\uf028 \uf029
\uf028 \uf029 \uf028 \uf029 ébricaAii
ou
ricaTrigonométiseni
isenisenisen
lg640.1.643
)º180º180(cos643
)º180º180(cos64)]º30).(6()º30).(6.[cos(2)º30º30(cos2
6
6
66
\uf0ae\uf02d\uf03d\uf02b\uf02d\uf03d\uf02b
\uf0ae\uf02b\uf03d\uf02b\uf0de
\uf02b\uf03d\uf02b\uf03d\uf02b

OBS: A opção de representar o número na forma trigonométrica antes de calcular a potência agilizou os
cálculos. A escolha em todo caso é de cada um. Experimente calcular somente na forma algébrica e
verifique se houve mais ou menos dificuldade.

7) Dado o número complexo z = cos 45\uf0b0 + isen 45º calcule: w = z + z2 + z3 + z4 + z5 + z6

Solução. Observe que o valor de \u201cw\u201d representa uma soma de seis termos da progressão geométrica de
razão \u201cz\u201d. Temos:
i)
)1.(
2
2
º45º45cos iisenz \uf02b\uf03d\uf02b\uf03d
ii)
iiiiiz \uf03d\uf02d\uf02b\uf0f7
\uf0f8
\uf0f6
\uf0e7
\uf0e8
\uf0e6
\uf03d\uf02b\uf02b\uf0f7
\uf0f8
\uf0f6
\uf0e7
\uf0e8
\uf0e6
\uf03d\uf02b
\uf0f7
\uf0f7
\uf0f8
\uf0f6
\uf0e7
\uf0e7
\uf0e8
\uf0e6
\uf03d )121.(
4
2
)21.(
4
2
)1.(
2
2 22
2
2

iii)
)1(
2
2
)).(1(
2
2
. 23 iiizzz \uf02b\uf02d\uf03d\uf02b\uf03d\uf03d
iv)
1)).((. 2224 \uf02d\uf03d\uf03d\uf03d\uf03d iiizzz

v)
)1(
2
2
)1).(1(
2
2
. 45 iizzz \uf02d\uf02d\uf03d\uf02d\uf02b\uf03d\uf03d
vi)
\uf028 \uf029 \uf028 \uf029 iizz \uf02d\uf03d\uf03d\uf03d 3326

Logo,
2
2
2
2
1
)(
2
2
2
2
)1(
2
2
2
2
)(
2
2
2
2
iw
iiiiiw
\uf02b
\uf0f7
\uf0f7
\uf0f8
\uf0f6
\uf0e7
\uf0e7
\uf0e8
\uf0e6
\uf02d\uf02d\uf03d
\uf02d\uf02b
\uf0f7
\uf0f7
\uf0f8
\uf0f6
\uf0e7
\uf0e7
\uf0e8
\uf0e6
\uf02d\uf02d\uf02b\uf02d\uf02b
\uf0f7
\uf0f7
\uf0f8
\uf0f6
\uf0e7
\uf0e7
\uf0e8
\uf0e6
\uf02b\uf02d\uf02b\uf02b
\uf0f7
\uf0f7
\uf0f8
\uf0f6
\uf0e7
\uf0e7
\uf0e8
\uf0e6
\uf02b\uf03d

8) Escreva as expressões abaixo na forma
bia \uf02b
:
a)
iiii )36()4( \uf02b\uf02d\uf02b\uf02d
b) \uf028 \uf029
\uf028 \uf0292
2
3
2
i
i
\uf02b
\uf02d
c)
i
i
54
3
\uf02b
\uf02d

a)
iiiiiiiiii 67634364)36()4( 2 \uf02d\uf03d\uf02d\uf02b\uf03d\uf02d\uf02d\uf02b\uf02d\uf03d\uf02b\uf02d\uf02b\uf02d

b) \uf028 \uf029
\uf028 \uf029
\uf028 \uf029
\uf028 \uf029
\uf028 \uf029
\uf028 \uf029
\uf028 \uf029
\uf028 \uf029
\uf028 \uf029
2100
255025
10
55
.
3
326
.
3
3
.
3
2
3
2 2
22
22
22
2
2
iiii
i
iii
i
i
i
i
i
i
\uf02d\uf03d
\uf02b\uf02d
\uf03d\uf0fa
\uf0fb
\uf0f9
\uf0ea
\uf0eb
\uf0e9 \uf02d
\uf03d\uf0fa
\uf0fb
\uf0f9
\uf0ea
\uf0eb
\uf0e9
\uf02d
\uf02b\uf02d\uf02d
\uf03d\uf0fa
\uf0fb
\uf0f9
\uf0ea
\uf0eb
\uf0e9
\uf02d
\uf02d
\uf02b
\uf02d
\uf03d
\uf02b
\uf02d
c) \uf028 \uf029
\uf028 \uf029 \uf028 \uf029 41
197
2516
197
54
541512
54
54
.
54
3
22
2 ii
i
iii
i
i
i
i \uf02d
\uf03d
\uf02b
\uf02d
\uf03d
\uf02d
\uf02b\uf02d\uf02d
\uf03d
\uf02d
\uf02d
\uf02b
\uf02d

09) Escrevendo o complexo
31
1
i
i
z
\uf02b
\uf02d
\uf03d
, calcule os valores do módulo e do argumento.

Solução. Escrevendo o numerador e o denominador na forma complexa e dividindo, temos:

i)
4
7
2
1
2
1
cos
211)1()1(1 22
\uf070\uf071
\uf071
\uf071
\uf03d\uf0de
\uf0ef
\uf0ef
\uf0ee
\uf0ef
\uf0ef
\uf0ed
\uf0ec
\uf02d\uf03d
\uf03d
\uf0de\uf03d\uf02b\uf03d\uf02d\uf02b\uf03d\uf02d```