Números Complexos: Exercício Resolvidos

Exercício 1

1. SEJA

$2z - i = \ overline z \ leftrightarrow 2 \ left ({x + yi} \ right) - i = x - 2x yi \ leftrightarrow + 2yi - i = x - yi \ leftrightarrow$

$\ Leftrightarrow 2x + \ left ({2a - 1} \ right) i = x - yi \ leftrightarrow$

$\ Leftrightarrow \ left \ {{\ begin {array} {* {20} {c}} {2x = x} \ \ {2a - 1 = - y} \ end {array}} \ right. \ Leftrightarrow \ left \ {{\ begin {array} {* {20} {c}} {x = 0} \ \ {3a = 1} \ end {array}} \ right. \ Leftrightarrow \ left \ {{\ begin {array} {* {20} {c}} {x = 0} \ \ {y = \ frac {1} {3}} \ end {array}} \ right.$

$S = \ \ left {{\ frac {1} {3} i} \ right \}$

2. ${2} ^ z + z = - 1 \ leftrightarrow {z} ^ 2 + z + 1 = 0 \ leftrightarrow$

$\ Leftrightarrow z = \ frac {{- 1 \ pm \ sqrt {{1} ^ 2 - 4 \ times 1 \ times 1}}} {{2 \ times 1}} \ leftrightarrow z = \ frac {{1 - \ pm \ sqrt {- 3}}} {2} \ leftrightarrow$

$\ Leftrightarrow z = \ frac {{- 1 \ pm \ sqrt 3 i}} {2} \ leftrightarrow z = - \ frac {1} {2} - \ frac {{\ sqrt 3}} {2} i \ vee z = - \ frac {1} {2} + \ frac {{\ sqrt 3}} {2} i$

$S = \ \ left {{- \ frac {1} {2} - \ frac {{\ sqrt 3}} {2} i, - \ frac {1} {2} + \ frac {{\ sqrt 3}} {2} i} \ right \}$

3. ${Z ^ 3} + 2z = 0 \ leftrightarrow z \ left ({{z ^ 2 + 2}} \ right) = 0 \ leftrightarrow z = 0 \ vee {z ^ 2} + 2 = 0 \ leftrightarrow z = 0 \ vee {z ^ 2} = - 2 \ leftrightarrow$

$\ Leftrightarrow z = 0 \ vee z = \ pm \ sqrt {- 2} \ leftrightarrow z = 0 \ vee z = \ pm \ sqrt 2 i$

$S = \ left \ {{- \ sqrt 2 i, 0, \ sqrt 2 i} \ right \}$

Exercício 2

$t = \ frac {1} {{2}} + i = \ frac {{2 - i}} {{\ left ({2 + i} \ right) \ left ({2 - i} \ right)}} = \ frac {{2 - i}} {{{2} ^ 2 - {i ^ 2}}} = \ frac {{2 - i}} {{4 - \ left ({- 1} \ right)} } = \ frac {{2 - i}} {5} = \ frac {2} {5} - \ frac {1} {5} i$

${Z ^ 2} = {\ left ({2 - 3i} \ right) ^ 2} = 4 - 12i + 9 {i} ^ 2 = 4 - 12i + 9 \ left ({- 1} \ right) = 4 - 12i - 9 = - 5 - 12i$

$\ Left ({1 - i} \ right) \ overline z = \ left ({1 - i} \ right) \ left ({2 + 3i} \ right) = 2 + 3i - 2i - 3 {i ^ 2} =$

$= 2 + i - 3 \ left ({- 1} \ right) = 2 + i + 3 = 5 + i$

$z + w = 2 - 3i + 2x - yi = \ left ({2 + 2x} \ right) + \ left ({- 3 - y} \ right) i$

Para Opaco z + w

SEJA UM imaginário puro um verdadeiro contraditório nula TEM Que Ser EO coeficiente da Parte Imaginária TEM Que Ser Diferente de zero. Logo:

$2 + 2x = 0 \ leftrightarrow 2x = - 2 \ leftrightarrow x = - 1$

$- 3 - y \ ne 0 \ leftrightarrow - y \ ne 3 \ leftrightarrow y \ ne - 3$

Exercício 3

SEJA

• $1 \ leqslant \ operatorname {Re} \ left (z \ right) <4 \ leftrightarrow 1 \ leqslant x <4$

• $\ Operatorname {Im} \ left ({z - i} \ right) \ geqslant 2 \ leftrightarrow \ operatorname {Im} \ left ({x + yi - i} \ right) \ geqslant 2 \ leftrightarrow \ operatorname {Im} \ Esquerda ({x + \ left ({y - 1} \ right) i} \ right) \ geqslant 2 \ leftrightarrow$

$\ Leftrightarrow y - 1 \ geqslant 2 \ leftrightarrow y \ geqslant 3$

SEJA

• $\ Left | {z + 2 - i} \ right | \ leqslant 2 \ leftrightarrow \ left | {z - \ left ({- 2 + i} \ right)} \ right | \ leqslant 2$

• $\ Operatorname {Im} \ left (z \ right) \ geqslant 1 \ leftrightarrow y \ geqslant 1$

$0 \ leqslant {\ text {arg}} \ left ({z + 1 - i} \ right) <\ frac {\ pi} {2} {\ text {}} \ wedge {\ text {}} \ left | {z + 1 - i} \ right | \ geqslant 2$

• $0 \ leqslant \ arg \ left ({z + 1 - i} \ right) <\ frac {\ pi} {2} \ leftrightarrow 0 \ leqslant \ arg \ left ({z - \ left ({- 1 + i} \ right)} \ right) <\ frac {\ pi} {2}$