Determinar o módulo e o argumento de cada um dos seguintes números complexos:
Colocar na forma trigonométrica os complexos abaixo.
  1. Z=1+i
    R: Z=\sqrt{2}\left[\cos \left(\frac{\pi}{4}\right)+i \cdot \operatorname{sen}\left(\frac{\pi}{4}\right)\right]
  2. Z=-3+3 \sqrt{3} i
    R: Z=6\left[\cos \left(\frac{2 \pi}{3}\right)+i \cdot \operatorname{sen}\left(\frac{2 \pi}{3}\right)\right]
  3. Z=1+i \sqrt{3}
    R: Z=2\left[\cos \left(\frac{\pi}{3}\right)+i \cdot \operatorname{sen}\left(\frac{\pi}{3}\right)\right]
  4. Z=1-i \sqrt{3}
    R: Z=2\left[\cos \left(\frac{5 \pi}{3}\right)+i \cdot \operatorname{sen}\left(\frac{5 \pi}{3}\right)\right]
  5. Z=2+2 \sqrt{3} \cdot i
    R: Z=4\left[\cos \left(\frac{\pi}{3}\right)+i \cdot \operatorname{sen}\left(\frac{\pi}{3}\right)\right]
  6. Z=-2+2 \sqrt{3} \cdot i
    R: Z=4\left[\cos \left(\frac{2 \pi}{3}\right)+i \cdot \operatorname{sen}\left(\frac{2 \pi}{3}\right)\right]
  7. Z=-4 \sqrt{3}-4 i
    R: Z=8\left[\cos \left(\frac{7 \pi}{6}\right)+i \cdot \operatorname{sen}\left(\frac{7 \pi}{6}\right)\right]
  8. Z=-7-7 i
    R: Z=7 \sqrt{2}\left[\cos \left(\frac{5 \pi}{4}\right)+i \cdot \operatorname{sen}\left(\frac{5 \pi}{4}\right)\right]
  9. Z=-2-2 i
    R: Z=2 \sqrt{2}\left[\cos \left(\frac{5 \pi}{4}\right)+i \cdot \operatorname{sen}\left(\frac{5 \pi}{4}\right)\right]
  10. Z=-1+i
    R: Z=\sqrt{2}\left[\cos \left(\frac{3 \pi}{4}\right)+i \cdot \operatorname{sen}\left(\frac{3 \pi}{4}\right)\right]
  11. Z=\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2} i
    R: Z=\sqrt{2}\left[\cos \left(\frac{\pi}{6}\right)+i \cdot \operatorname{sen}\left(\frac{\pi}{6}\right)\right]
  12. Z=8 i
    R: Z=8\left[\cos \left(\frac{\pi}{2}\right)+i \cdot \operatorname{sen}\left(\frac{\pi}{2}\right)\right]
  13. Z=2 i
    R: Z=2\left[\cos \left(\frac{\pi}{2}\right)+i \cdot \operatorname{sen}\left(\frac{\pi}{2}\right)\right]
  14. Z=4
    R: Z=4[\cos (0)+i \cdot \operatorname{sen}(0)]

Passo 1

O enunciado pede, para cada um dos complexos abaixo, determinar o módulo e o argumento e escrever o número na forma trigonométrica. Os números fornecidos são:
  1. Z=1+i
  2. Z=-3+3\sqrt{3}\,i
  3. Z=1+i\sqrt{3}
  4. Z=1-i\sqrt{3}
  5. Z=2+2\sqrt{3}\,i
  6. Z=-2+2\sqrt{3}\,i
  7. Z=-4\sqrt{3}-4\,i
  8. Z=-7-7\,i
  9. Z=-2-2\,i
  10. Z=-1+i
  11. Z=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{2}}{2}\,i
  12. Z=8\,i
  13. Z=2\,i
  14. Z=4

Passo 2

Antes de resolver cada caso, vamos fixar a receita passo a passo (a “fórmula de bolso”) que vamos usar e explicar por que funciona. Para um número complexo Z=a+bi:
  • O módulo (distância ao origem) é dado por
|Z|=\sqrt{a^{2}+b^{2}}.
Explicação: pensamos em Z como o ponto (a,b) no plano; a distância ao origem segue o teorema de Pitágoras.
  • O argumento (ângulo com o eixo real positivo) é um ângulo \theta tal que
\cos\theta=\frac{a}{|Z|},\qquad \operatorname{sen}\theta=\frac{b}{|Z|}.
Praticamente, podemos calcular um ângulo auxiliar \alpha=\arctan\!\left(\frac{|b|}{|a|}\right) e depois ajustar pelo quadrante:
  • Se a>0 e b\ge0 (Q I): \theta=\alpha.
  • Se a<0 e b\ge0 (Q II): \theta=\pi-\alpha.
  • Se a<0 e b<0 (Q III): \theta=\pi+\alpha.
  • Se a>0 e b<0 (Q IV): \theta=2\pi-\alpha.
  • Casos especiais: se a=0 e b>0 então \theta=\tfrac{\pi}{2}; se a=0 e b<0 então \theta=\tfrac{3\pi}{2}; se b=0 com a>0 então \theta=0; se b=0 com a<0 então \theta=\pi.
  • Depois escrevemos na forma trigonométrica:
Z=|Z|\left[\cos(\theta)+i\cdot\operatorname{sen}(\theta)\right].
Vou usar ângulos notáveis conhecidos: \arctan(1)=\tfrac{\pi}{4}, \arctan(\sqrt{3})=\tfrac{\pi}{3}, \arctan(1/\sqrt{3})=\tfrac{\pi}{6}. Agora aplicamos isso em cada número, passo a passo.

Passo 3

Para 1) Z=1+i: temos a=1, b=1.
  • Módulo:
|Z|=\sqrt{1^{2}+1^{2}}=\sqrt{2}.
  • Ângulo: \alpha=\arctan\!\left(\frac{|b|}{|a|}\right)=\arctan(1)=\tfrac{\pi}{4}. Como a>0,b>0 (Q I), \theta=\tfrac{\pi}{4}.
  • Forma trigonométrica:
Z=\sqrt{2}\left[\cos\!\left(\tfrac{\pi}{4}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{\pi}{4}\right)\right].

Passo 4

Para 2) Z=-3+3\sqrt{3}\,i: a=-3, b=3\sqrt{3}.
  • Módulo:
|Z|=\sqrt{(-3)^{2}+(3\sqrt{3})^{2}}=\sqrt{9+27}=\sqrt{36}=6.
  • Ângulo: \alpha=\arctan\!\left(\frac{|b|}{|a|}\right)=\arctan\!\left(\frac{3\sqrt{3}}{3}\right)=\arctan(\sqrt{3})=\tfrac{\pi}{3}.
    Aqui a<0,b>0 (Q II), então \theta=\pi-\alpha=\pi-\tfrac{\pi}{3}=\tfrac{2\pi}{3}.
  • Forma trigonométrica:
Z=6\left[\cos\!\left(\tfrac{2\pi}{3}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{2\pi}{3}\right)\right].

Passo 5

Para 3) Z=1+i\sqrt{3}: a=1, b=\sqrt{3}.
  • Módulo:
|Z|=\sqrt{1^{2}+(\sqrt{3})^{2}}=\sqrt{1+3}=\sqrt{4}=2.
  • Ângulo: \alpha=\arctan\!\left(\frac{\sqrt{3}}{1}\right)=\arctan(\sqrt{3})=\tfrac{\pi}{3}. Como a>0,b>0 (Q I), \theta=\tfrac{\pi}{3}.
  • Forma trigonométrica:
Z=2\left[\cos\!\left(\tfrac{\pi}{3}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{\pi}{3}\right)\right].

Passo 6

Para 4) Z=1-i\sqrt{3}: a=1, b=-\sqrt{3}.
  • Módulo:
|Z|=\sqrt{1^{2}+(-\sqrt{3})^{2}}=\sqrt{1+3}=\sqrt{4}=2.
  • Ângulo: \alpha=\arctan\!\left(\frac{|b|}{|a|}\right)=\arctan(\sqrt{3})=\tfrac{\pi}{3}. Aqui a>0,b<0 (Q IV), então \theta=2\pi-\alpha=2\pi-\tfrac{\pi}{3}=\tfrac{5\pi}{3}.
  • Forma trigonométrica:
Z=2\left[\cos\!\left(\tfrac{5\pi}{3}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{5\pi}{3}\right)\right].

Passo 7

Para 5) Z=2+2\sqrt{3}\,i: a=2, b=2\sqrt{3}.
  • Módulo:
|Z|=\sqrt{2^{2}+(2\sqrt{3})^{2}}=\sqrt{4+12}=\sqrt{16}=4.
  • Ângulo: \alpha=\arctan\!\left(\frac{2\sqrt{3}}{2}\right)=\arctan(\sqrt{3})=\tfrac{\pi}{3}. Como a>0,b>0 (Q I), \theta=\tfrac{\pi}{3}.
  • Forma trigonométrica:
Z=4\left[\cos\!\left(\tfrac{\pi}{3}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{\pi}{3}\right)\right].

Passo 8

Para 6) Z=-2+2\sqrt{3}\,i: a=-2, b=2\sqrt{3}.
  • Módulo:
|Z|=\sqrt{(-2)^{2}+(2\sqrt{3})^{2}}=\sqrt{4+12}=\sqrt{16}=4.
  • Ângulo: \alpha=\arctan\!\left(\frac{2\sqrt{3}}{2}\right)=\arctan(\sqrt{3})=\tfrac{\pi}{3}. Aqui a<0,b>0 (Q II), então \theta=\pi-\alpha=\pi-\tfrac{\pi}{3}=\tfrac{2\pi}{3}.
  • Forma trigonométrica:
Z=4\left[\cos\!\left(\tfrac{2\pi}{3}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{2\pi}{3}\right)\right].

Passo 9

Para 7) Z=-4\sqrt{3}-4\,i: a=-4\sqrt{3}, b=-4.
  • Módulo:
|Z|=\sqrt{(-4\sqrt{3})^{2}+(-4)^{2}}=\sqrt{48+16}=\sqrt{64}=8.
  • Ângulo: \alpha=\arctan\!\left(\frac{|b|}{|a|}\right)=\arctan\!\left(\frac{4}{4\sqrt{3}}\right)=\arctan\!\left(\frac{1}{\sqrt{3}}\right)=\tfrac{\pi}{6}. Como a<0,b<0 (Q III), \theta=\pi+\alpha=\pi+\tfrac{\pi}{6}=\tfrac{7\pi}{6}.
  • Forma trigonométrica:
Z=8\left[\cos\!\left(\tfrac{7\pi}{6}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{7\pi}{6}\right)\right].

Passo 10

Para 8) Z=-7-7\,i: a=-7, b=-7.
  • Módulo:
|Z|=\sqrt{(-7)^{2}+(-7)^{2}}=\sqrt{49+49}=\sqrt{98}=7\sqrt{2}.
  • Ângulo: \alpha=\arctan\!\left(\frac{|b|}{|a|}\right)=\arctan(1)=\tfrac{\pi}{4}. Aqui a<0,b<0 (Q III), então \theta=\pi+\alpha=\pi+\tfrac{\pi}{4}=\tfrac{5\pi}{4}.
  • Forma trigonométrica:
Z=7\sqrt{2}\left[\cos\!\left(\tfrac{5\pi}{4}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{5\pi}{4}\right)\right].

Passo 11

Para 9) Z=-2-2\,i: a=-2, b=-2.
  • Módulo:
|Z|=\sqrt{(-2)^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}.
  • Ângulo: \alpha=\arctan(1)=\tfrac{\pi}{4}. a<0,b<0 (Q III), \theta=\pi+\tfrac{\pi}{4}=\tfrac{5\pi}{4}.
  • Forma trigonométrica:
Z=2\sqrt{2}\left[\cos\!\left(\tfrac{5\pi}{4}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{5\pi}{4}\right)\right].

Passo 12

Para 10) Z=-1+i: a=-1, b=1.
  • Módulo:
|Z|=\sqrt{(-1)^{2}+1^{2}}=\sqrt{1+1}=\sqrt{2}.
  • Ângulo: \alpha=\arctan\!\left(\frac{1}{1}\right)=\tfrac{\pi}{4}. Aqui a<0,b>0 (Q II), então \theta=\pi-\tfrac{\pi}{4}=\tfrac{3\pi}{4}.
  • Forma trigonométrica:
Z=\sqrt{2}\left[\cos\!\left(\tfrac{3\pi}{4}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{3\pi}{4}\right)\right].

Passo 13

Para 11) Z=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{2}}{2}\,i: a=\dfrac{\sqrt{6}}{2}, b=\dfrac{\sqrt{2}}{2}.
  • Módulo:
|Z|=\sqrt{\left(\dfrac{\sqrt{6}}{2}\right)^{2}+\left(\dfrac{\sqrt{2}}{2}\right)^{2}}=\sqrt{\dfrac{6}{4}+\dfrac{2}{4}}=\sqrt{\dfrac{8}{4}}=\sqrt{2}.
  • Ângulo: razão \dfrac{b}{a}=\dfrac{\sqrt{2}/2}{\sqrt{6}/2}=\dfrac{\sqrt{2}}{\sqrt{6}}=\dfrac{1}{\sqrt{3}}. Assim \alpha=\arctan\!\left(\dfrac{1}{\sqrt{3}}\right)=\tfrac{\pi}{6}. Como a>0,b>0 (Q I), \theta=\tfrac{\pi}{6}.
  • Forma trigonométrica:
Z=\sqrt{2}\left[\cos\!\left(\tfrac{\pi}{6}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{\pi}{6}\right)\right].

Passo 14

Para 12) Z=8\,i: a=0, b=8.
  • Módulo:
|Z|=\sqrt{0^{2}+8^{2}}=8.
  • Ângulo: ponto sobre o eixo imag. positivo, portanto \theta=\tfrac{\pi}{2}.
  • Forma trigonométrica:
Z=8\left[\cos\!\left(\tfrac{\pi}{2}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{\pi}{2}\right)\right].

Passo 15

Para 13) Z=2\,i: a=0, b=2.
  • Módulo:
|Z|=\sqrt{0^{2}+2^{2}}=2.
  • Ângulo: eixo imag. positivo, \theta=\tfrac{\pi}{2}.
  • Forma trigonométrica:
Z=2\left[\cos\!\left(\tfrac{\pi}{2}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{\pi}{2}\right)\right].

Passo 16

Para 14) Z=4: a=4, b=0.
  • Módulo:
|Z|=\sqrt{4^{2}+0^{2}}=4.
  • Ângulo: eixo real positivo, \theta=0.
  • Forma trigonométrica:
Z=4\left[\cos(0)+i\cdot\operatorname{sen}(0)\right].

Resposta

  1. Z=\sqrt{2}\left[\cos\!\left(\tfrac{\pi}{4}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{\pi}{4}\right)\right]
  2. Z=6\left[\cos\!\left(\tfrac{2\pi}{3}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{2\pi}{3}\right)\right]
  3. Z=2\left[\cos\!\left(\tfrac{\pi}{3}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{\pi}{3}\right)\right]
  4. Z=2\left[\cos\!\left(\tfrac{5\pi}{3}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{5\pi}{3}\right)\right]
  5. Z=4\left[\cos\!\left(\tfrac{\pi}{3}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{\pi}{3}\right)\right]
  6. Z=4\left[\cos\!\left(\tfrac{2\pi}{3}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{2\pi}{3}\right)\right]
  7. Z=8\left[\cos\!\left(\tfrac{7\pi}{6}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{7\pi}{6}\right)\right]
  8. Z=7\sqrt{2}\left[\cos\!\left(\tfrac{5\pi}{4}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{5\pi}{4}\right)\right]
  9. Z=2\sqrt{2}\left[\cos\!\left(\tfrac{5\pi}{4}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{5\pi}{4}\right)\right]
  10. Z=\sqrt{2}\left[\cos\!\left(\tfrac{3\pi}{4}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{3\pi}{4}\right)\right]
  11. Z=\sqrt{2}\left[\cos\!\left(\tfrac{\pi}{6}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{\pi}{6}\right)\right]
  12. Z=8\left[\cos\!\left(\tfrac{\pi}{2}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{\pi}{2}\right)\right]
  13. Z=2\left[\cos\!\left(\tfrac{\pi}{2}\right)+i\cdot\operatorname{sen}\!\left(\tfrac{\pi}{2}\right)\right]
  14. Z=4\left[\cos(0)+i\cdot\operatorname{sen}(0)\right]

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