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Escreva $z$ na forma trigonométrica 0

Escreva $z$ na forma trigonométrica

Números complexos: Infinito 12 A - Parte 3 Pág. 141 Ex. 47

Enunciado

Escreva $z$ na forma trigonométrica:

  1. $z = 1 – i\sqrt 3 $
     
  2. $z =  – 1 + i$
     
  3. $z =  – 5$
     
  4. $z = 3i$
     
  5. $z = \frac{1}{3} + \frac{1}{3}i$
     
  6. $z =  – \sqrt 2  – \sqrt 6 i$
     
  7. $z = \frac{4}{{1 – i\sqrt 3 }}$
     
  8. $z = \frac{2}{{\sqrt 6  – i\sqrt 2 }}$

Resolução >> Resolução

  1.  
    $$\begin{array}{*{20}{l}}
      z& = &{1 – i\sqrt 3 } \\
      {}& = &{2\left( {\frac{1}{2} – \frac{{\sqrt 3 }}{2}i} \right)} \\
      {}& = &{2\operatorname{cis} \left( { – \frac{\pi }{3}} \right)}
    \end{array}$$
     
  2.  
    $$\begin{array}{*{20}{l}}
      z& = &{ – 1 + i} \\
      {}& = &{\sqrt 2 \left( { – \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2}i} \right)} \\
      {}& = &{\sqrt 2 \operatorname{cis} \left( {\frac{{3\pi }}{4}} \right)}
    \end{array}$$
     
  3.  
    $$\begin{array}{*{20}{l}}
      z& = &{ – 5} \\
      {}& = &{5\operatorname{cis} \pi }
    \end{array}$$

     

  4.  
    $$\begin{array}{*{20}{l}}
      z& = &{3i} \\
      {}& = &{3\operatorname{cis} \frac{\pi }{2}}
    \end{array}$$
     
  5.  
    $$\begin{array}{*{20}{l}}
      z& = &{\frac{1}{3} + \frac{1}{3}i} \\
      {}& = &{\frac{{\sqrt 2 }}{3}\left( {\frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2}i} \right)} \\
      {}& = &{\frac{{\sqrt 2 }}{3}\operatorname{cis} \frac{\pi }{4}}
    \end{array}$$
     
  6.  
    $$\begin{array}{*{20}{l}}
      z& = &{ – \sqrt 2  – \sqrt 6 i} \\
      {}& = &{2\sqrt 2 \left( { – \frac{1}{2} – \frac{{\sqrt 3 }}{2}i} \right)} \\
      {}& = &{2\sqrt 2 \operatorname{cis} \left( {\frac{{4\pi }}{3}} \right)}
    \end{array}$$
     
  7.  
    $$\begin{array}{*{20}{l}}
      z& = &{\frac{4}{{1 – i\sqrt 3 }}} \\
      {}& = &{\frac{{4\operatorname{cis} \left( 0 \right)}}{{2\left( {\frac{1}{2} – \frac{{\sqrt 3 }}{2}i} \right)}}} \\
      {}& = &{\frac{{4\operatorname{cis} \left( 0 \right)}}{{2\operatorname{cis} \left( { – \frac{\pi }{3}} \right)}}} \\
      {}& = &{2\operatorname{cis} \left( {0 + \frac{\pi }{3}} \right)} \\
      {}& = &{2\operatorname{cis} \frac{\pi }{3}}
    \end{array}$$
     
  8.  
    $$\begin{array}{*{20}{l}}
      z& = &{\frac{2}{{\sqrt 6  – i\sqrt 2 }}} \\
      {}& = &{\frac{{2\operatorname{cis} \left( 0 \right)}}{{2\sqrt 2 \left( {\frac{{\sqrt 3 }}{2} – \frac{1}{2}i} \right)}}} \\
      {}& = &{\frac{{2\operatorname{cis} \left( 0 \right)}}{{2\sqrt 2 \operatorname{cis} \left( { – \frac{\pi }{6}} \right)}}} \\
      {}& = &{\frac{2}{{2\sqrt 2 }}\operatorname{cis} \frac{\pi }{6}} \\
      {}& = &{\frac{{\sqrt 2 }}{2}\operatorname{cis} \frac{\pi }{6}}
    \end{array}$$
<< Enunciado
Determine na forma trigonométrica 0

Determine na forma trigonométrica

Números complexos: Infinito 12 A - Parte 3 Pág. 140 Ex. 46

Enunciado

Sendo $$\begin{array}{*{20}{c}}
  {z = 2\operatorname{cis} \frac{\pi }{3}}&{\text{e}}&{w = 3\operatorname{cis} \frac{\pi }{2}}
\end{array}$$ determine na forma trigonométrica:

  1. $zw$
     
  2. $\frac{z}{w}$
     
  3. ${z^3}$

Resolução >> Resolução

  1.  
    $$\begin{array}{*{20}{l}}
      {zw}& = &{\left( {2\operatorname{cis} \frac{\pi }{3}} \right) \times \left( {3\operatorname{cis} \frac{\pi }{2}} \right)} \\
      {}& = &{\left( {2 \times 3} \right)\operatorname{cis} \left( {\frac{\pi }{3} + \frac{\pi }{2}} \right)} \\
      {}& = &{6\operatorname{cis} \frac{{5\pi }}{6}}
    \end{array}$$
     
  2.  
    $$\begin{array}{*{20}{l}}
      {\frac{z}{w}}& = &{\frac{{2\operatorname{cis} \frac{\pi }{3}}}{{3\operatorname{cis} \frac{\pi }{2}}}} \\
      {}& = &{\frac{2}{3}\operatorname{cis} \left( {\frac{\pi }{3} – \frac{\pi }{2}} \right)} \\
      {}& = &{\frac{2}{3}\operatorname{cis} \left( { – \frac{\pi }{6}} \right)}
    \end{array}$$
     
  3.  
    $$\begin{array}{*{20}{l}}
      {{z^3}}& = &{{{\left( {2\operatorname{cis} \frac{\pi }{3}} \right)}^3}} \\
      {}& = &{{2^3}\operatorname{cis} \left( {3 \times \frac{\pi }{3}} \right)} \\
      {}& = &{8\operatorname{cis} \pi }
    \end{array}$$
<< Enunciado
Calcule 0

Calcule

Números complexos: Infinito 12 A - Parte 3 Pág. 99 Ex. 60

Enunciado

Calcule:

  1.  
    $${\left( { – 1 – \sqrt 3 i} \right)^6}$$
     
  2.  
    $${\left( {\frac{{2 + 2i}}{{2 – 2i}}} \right)^4}$$
     
  3.  
    $${\left[ {3\operatorname{cis} \left( { – \frac{{4\pi }}{3}} \right)} \right]^5}$$

Resolução >> Resolução

Forma trigonométrica da potência (Fórmula de Moivre):

Se $z = \rho \operatorname{cis} \theta $ é um número complexo não nulo, então $${z^n} = {\rho ^n}\operatorname{cis} \left( {n\theta } \right)$$

 

  1.  
    $$\begin{array}{*{20}{l}}
      {{{\left( { – 1 – \sqrt 3 i} \right)}^6}}& = &{{{\left[ {2\left( { – \frac{1}{2} – \frac{{\sqrt 3 }}{2}i} \right)} \right]}^6}} \\
      {}& = &{{{\left( {2\operatorname{cis} \frac{{4\pi }}{3}} \right)}^6}} \\
      {}& = &{{2^6}\operatorname{cis} \left( {6 \times \frac{{4\pi }}{3}} \right)} \\
      {}& = &{64\operatorname{cis} \left( {8\pi } \right)} \\
      {}& = &{64\operatorname{cis} \left( 0 \right)}
    \end{array}$$
     
  2.  
    $$\begin{array}{*{20}{l}}
      {{{\left( {\frac{{2 + 2i}}{{2 – 2i}}} \right)}^4}}& = &{{{\left( {\frac{{2\sqrt 2 \operatorname{cis} \frac{\pi }{4}}}{{2\sqrt 2 \operatorname{cis} \left( { – \frac{\pi }{4}} \right)}}} \right)}^4}} \\
      {}& = &{{{\left( {\frac{{2\sqrt 2 }}{{2\sqrt 2 }}\operatorname{cis} \left( {\frac{\pi }{4} + \frac{\pi }{4}} \right)} \right)}^4}} \\
      {}& = &{{1^4}\operatorname{cis} \left( {4 \times \frac{\pi }{2}} \right)} \\
      {}& = &{\operatorname{cis} \left( {2\pi } \right)} \\
      {}& = &{\operatorname{cis} \left( 0 \right)}
    \end{array}$$
     
  3.  
    $$\begin{array}{*{20}{l}}
      {{{\left[ {3\operatorname{cis} \left( { – \frac{{4\pi }}{3}} \right)} \right]}^5}}& = &{{3^5}\operatorname{cis} \left( {5 \times \left( { – \frac{{4\pi }}{3}} \right)} \right)} \\
      {}& = &{243\operatorname{cis} \left( { – \frac{{20\pi }}{3} + 8\pi } \right)} \\
      {}& = &{243\operatorname{cis} \frac{{4\pi }}{3}}
    \end{array}$$
<< Enunciado
Calcule 0

Calcule

Números complexos: Infinito 12 A - Parte 3 Pág. 97 Ex. 59

Enunciado

Calcule:

  1.  
    $$\frac{{2\operatorname{cis} \frac{\pi }{3}}}{{4\operatorname{cis} \left( {\frac{{2\pi }}{3}} \right)}}$$
     
  2.  
    $$\frac{{ – 2}}{{\operatorname{cis} \left( { – \theta } \right)}}$$
     
  3.  
    $$\frac{{ – \operatorname{cis} \frac{\pi }{6}}}{{2\operatorname{cis} \theta }}$$
     
  4.  
    $$\left( {2\operatorname{cis} \frac{{5\pi }}{6}} \right) \times \left[ {3\operatorname{cis} \left( { – \frac{{2\pi }}{3}} \right)} \right]$$

Resolução >> Resolução

Forma trigonométrica do quociente:

Se ${z_1} = {\rho _1}\operatorname{cis} \left( {{\theta _1}} \right)$ e ${z_2} = {\rho _2}\operatorname{cis} \left( {{\theta _2}} \right)$ são dois números complexos não nulos, então $$\frac{{{z_1}}}{{{z_2}}} = \frac{{{\rho _1}}}{{{\rho _2}}}\operatorname{cis} \left( {{\theta _1} – {\theta _2}} \right)$$

 

  1.  
    $$\begin{array}{*{20}{l}}
      {\frac{{2\operatorname{cis} \frac{\pi }{3}}}{{4\operatorname{cis} \left( {\frac{{2\pi }}{3}} \right)}}}& = &{\frac{2}{4}\operatorname{cis} \left( {\frac{\pi }{3} – \frac{{2\pi }}{3}} \right)} \\
      {}& = &{\frac{1}{2}\operatorname{cis} \left( { – \frac{\pi }{3}} \right)} \\
      {}& = &{\frac{1}{2}\operatorname{cis} \frac{{5\pi }}{3}}
    \end{array}$$
     
  2.  
    $$\begin{array}{*{20}{l}}
      {\frac{{ – 2}}{{\operatorname{cis} \left( { – \theta } \right)}}}& = &{\frac{{2\operatorname{cis} \pi }}{{\operatorname{cis} \left( { – \theta } \right)}}} \\
      {}& = &{\frac{2}{1}\operatorname{cis} \left( {\pi  + \theta } \right)} \\
      {}& = &{2\operatorname{cis} \left( {\pi  + \theta } \right)}
    \end{array}$$
     
  3.  
    $$\begin{array}{*{20}{l}}
      {\frac{{ – \operatorname{cis} \frac{\pi }{6}}}{{2\operatorname{cis} \theta }}}& = &{\frac{{\operatorname{cis} \left( {\pi  + \frac{\pi }{6}} \right)}}{{2\operatorname{cis} \theta }}} \\
      {}& = &{\frac{1}{2}\operatorname{cis} \left( {\pi  + \frac{\pi }{6} – \theta } \right)} \\
      {}& = &{\frac{1}{2}\operatorname{cis} \left( {\frac{{7\pi }}{6} – \theta } \right)}
    \end{array}$$
     
  4.  
    $$\begin{array}{*{20}{l}}
      {\left( {2\operatorname{cis} \frac{{5\pi }}{6}} \right) \times \left[ {3\operatorname{cis} \left( { – \frac{{2\pi }}{3}} \right)} \right]}& = &{\left( {2 \times 3} \right)\operatorname{cis} \left( {\frac{{5\pi }}{6} + \left( { – \frac{{2\pi }}{3}} \right)} \right)} \\
      {}& = &{6\operatorname{cis} \frac{\pi }{6}}
    \end{array}$$
<< Enunciado
Represente na forma trigonométrica 0

Represente na forma trigonométrica

Números complexos: Infinito 12 A - Parte 3 Pág. 96 Ex. 57

Enunciado

Represente na forma trigonométrica:

  1.  $z =  – 3\operatorname{cis} \theta $
     
  2. $z = 2\cos \theta  – 2i\operatorname{sen} \theta $
     
  3. $z =  – \cos \theta  – i\operatorname{sen} \theta $
     
  4. $z = \frac{1}{{2\operatorname{cis} \left( {\frac{\pi }{3} – \theta } \right)}}$

Resolução >> Resolução

  1.  
    $$\begin{array}{*{20}{l}}
      z& = &{ – 3\operatorname{cis} \theta } \\
      {}& = &{3\operatorname{cis} \left( {\pi  + \theta } \right)}
    \end{array}$$
     
  2.  
    $$\begin{array}{*{20}{l}}
      z& = &{2\cos \theta  – 2i\operatorname{sen} \theta } \\
      {}& = &{2\cos \left( { – \theta } \right) + 2i\operatorname{sen} \left( { – \theta } \right)} \\
      {}& = &{2\operatorname{cis} \left( { – \theta } \right)}
    \end{array}$$
     
  3.  
    $$\begin{array}{*{20}{l}}
      z& = &{ – \cos \theta  – i\operatorname{sen} \theta } \\
      {}& = &{\cos \left( {\pi  + \theta } \right) + i\operatorname{sen} \left( {\pi  + \theta } \right)} \\
      {}& = &{\operatorname{cis} \left( {\pi  + \theta } \right)}
    \end{array}$$
     
  4.  
    $$\begin{array}{*{20}{l}}
      z& = &{\frac{1}{{2\operatorname{cis} \left( {\frac{\pi }{3} – \theta } \right)}}} \\
      {}& = &{\frac{1}{2}\operatorname{cis} \left( { – \frac{\pi }{3} + \theta } \right)} \\
      {}& = &{\frac{1}{2}\operatorname{cis} \left( {\theta  – \frac{\pi }{3}} \right)}
    \end{array}$$
<< Enunciado
Represente na forma trigonométrica o simétrico e o inverso dos números complexos 0

Represente na forma trigonométrica o simétrico e o inverso dos números complexos

Números complexos: Infinito 12 A - Parte 3 Pág. 96 Ex. 56

Enunciado

Represente na forma trigonométrica o simétrico e o inverso de cada um dos seguintes números complexos:

  1. $z =  – 3 + 3i$
     
  2. $z = 2\operatorname{cis} \left( { – \pi } \right)$
     
  3. $z = 2,3\operatorname{cis} \left( {\frac{{5\pi }}{4}} \right)$

Resolução >> Resolução

Forma trigonométrica do simétrico:

Se $z = \rho \operatorname{cis} \theta $, então $ – z = \rho \operatorname{cis} \left( {\pi  + \theta } \right)$.

Forma trigonométrica do produto de dois números complexos 0

Forma trigonométrica do produto de dois números complexos

Números complexos

Forma trigonométrica do produto:

Se ${z_1} = {\rho _1}\operatorname{cis} {\theta _1}$ e ${z_2} = {\rho _2}\operatorname{cis} {\theta _2}$ são dois complexos não nulos, então $${z_1}.{z_2} = {\rho _1}{\rho _2}\operatorname{cis} \left( {{\theta _1} + {\theta _2}} \right)$$

 

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Calcule o produto na forma trigonométrica 0

Calcule o produto na forma trigonométrica

Números complexos: Infinito 12 A - Parte 3 Pág. 94 Ex. 54

Enunciado

Sendo $$\begin{array}{*{20}{l}}
  {{z_1} = 3\operatorname{cis} \left( { – \frac{\pi }{3}} \right)}&{\text{;}}&{{z_2} = \sqrt 2 \operatorname{cis} \frac{\pi }{6}}&{\text{e}}&{{z_3} = \operatorname{cis} \frac{{5\pi }}{6}}
\end{array}$$ calcule:

  1. ${z_1}.{z_2}$
     
  2. ${z_2}.{z_3}$
     
  3. ${z_1}.{z_2}.{z_3}$

Resolução >> Resolução

Forma trigonométrica do produto:

Se ${z_1} = {\rho _1}\operatorname{cis} {\theta _1}$ e ${z_2} = {\rho _2}\operatorname{cis} {\theta _2}$ são dois complexos não nulos, então $${z_1}.{z_2} = {\rho _1}{\rho _2}\operatorname{cis} \left( {{\theta _1} + {\theta _2}} \right)$$

  1.   
    $$\begin{array}{*{20}{l}}
      {{z_1}.{z_2}}& = &{3\operatorname{cis} \left( { – \frac{\pi }{3}} \right) \times \sqrt 2 \operatorname{cis} \frac{\pi }{6}} \\
      {}& = &{3\sqrt 2 \operatorname{cis} \left( { – \frac{\pi }{3} + \frac{\pi }{6}} \right)} \\
      {}& = &{3\sqrt 2 \operatorname{cis} \left( { – \frac{\pi }{6}} \right)}
    \end{array}$$
     
  2.  
    $$\begin{array}{*{20}{l}}
      {{z_2}.{z_3}}& = &{\sqrt 2 \operatorname{cis} \frac{\pi }{6} \times \operatorname{cis} \frac{{5\pi }}{6}} \\
      {}& = &{\sqrt 2 \operatorname{cis} \left( {\frac{\pi }{6} + \frac{{5\pi }}{6}} \right)} \\
      {}& = &{\sqrt 2 \operatorname{cis} \pi }
    \end{array}$$
     
  3.  
    $$\begin{array}{*{20}{l}}
      {{z_1}.{z_2}.{z_3}}& = &{3\operatorname{cis} \left( { – \frac{\pi }{3}} \right) \times \sqrt 2 \operatorname{cis} \frac{\pi }{6} \times \operatorname{cis} \frac{{5\pi }}{6}} \\
      {}& = &{3\sqrt 2 \operatorname{cis} \left( { – \frac{\pi }{3} + \frac{\pi }{6} + \frac{{5\pi }}{6}} \right)} \\
      {}& = &{3\sqrt 2 \operatorname{cis} \frac{{2\pi }}{3}}
    \end{array}$$
<< Enunciado
Represente, na forma trigonométrica, os conjugados dos números complexos 0

Represente, na forma trigonométrica, os conjugados dos números complexos

Números complexos: Infinito 12 A - Parte 3 Pág. 93 Ex. 53

Enunciado

Represente, na forma trigonométrica, os conjugados dos números complexos:

  1. $z =  – 3$
     
  2. $z = 2i$
     
  3. $z = 2\operatorname{cis} \left( { – \frac{\pi }{3}} \right)$

Resolução >> Resolução

Forma trigonométrica do conjugado:

Se $z = \rho \operatorname{cis} \theta $, então $\overline z  = \rho \operatorname{cis} \left( { – \theta } \right)$.

Represente na forma algébrica os números complexos 0

Represente na forma algébrica os números complexos

Números complexos: Infinito 12 A - Parte 3 Pág. 92 Ex. 52

Enunciado

Represente na forma algébrica os números complexos:

  1. $z = 5\operatorname{cis} \pi $
     
  2. $z = 3\operatorname{cis} \frac{\pi }{2}$
     
  3. $z = \sqrt 2 \operatorname{cis} \frac{{7\pi }}{4}$
     
  4. $z = \operatorname{cis} \frac{{7\pi }}{6}$
     
  5. $z = \sqrt 3 \operatorname{cis} \left( { – \frac{\pi }{3}} \right)$

Resolução >> Resolução

  1.  
    $$\begin{array}{*{20}{l}}
      z& = &{5\operatorname{cis} \pi } \\
      {}& = &{5\left( {\cos \pi  + i\operatorname{sen} \pi } \right)} \\
      {}& = &{5\left( { – 1 + 0i} \right)} \\
      {}& = &{ – 5}
    \end{array}$$
     
  2.  
    $$\begin{array}{*{20}{l}}
      z& = &{3\operatorname{cis} \frac{\pi }{2}} \\
      {}& = &{3\left( {\cos \frac{\pi }{2} + i\operatorname{sen} \frac{\pi }{2}} \right)} \\
      {}& = &{3\left( {0 + i} \right)} \\
      {}& = &{3i}
    \end{array}$$
     
  3.  
    $$\begin{array}{*{20}{l}}
      z& = &{\sqrt 2 \operatorname{cis} \frac{{7\pi }}{4}} \\
      {}& = &{\sqrt 2 \left( {\cos \frac{{7\pi }}{4} + i\operatorname{sen} \frac{{7\pi }}{4}} \right)} \\
      {}& = &{\sqrt 2 \left( {\cos \frac{\pi }{4} + i\operatorname{sen} \left( { – \frac{\pi }{4}} \right)} \right)} \\
      {}& = &{\sqrt 2 \left( {\frac{{\sqrt 2 }}{2} – \frac{{\sqrt 2 }}{2}i} \right)} \\
      {}& = &{1 – i}
    \end{array}$$
     
  4.  
    $$\begin{array}{*{20}{l}}
      z& = &{\operatorname{cis} \frac{{7\pi }}{4}} \\
      {}& = &{\cos \frac{{7\pi }}{6} + i\operatorname{sen} \frac{{7\pi }}{6}} \\
      {}& = &{ – \cos \frac{\pi }{6} – i\operatorname{sen} \frac{\pi }{6}} \\
      {}& = &{ – \frac{{\sqrt 3 }}{2} – \frac{1}{2}i}
    \end{array}$$
     
  5.  
    $$\begin{array}{*{20}{l}}
      z& = &{\sqrt 3 \operatorname{cis} \left( { – \frac{\pi }{3}} \right)} \\
      {}& = &{\sqrt 3 \left( {\cos \left( { – \frac{\pi }{3}} \right) + i\operatorname{sen} \left( { – \frac{\pi }{3}} \right)} \right)} \\
      {}& = &{\sqrt 3 \left( {\cos \frac{\pi }{3} – i\operatorname{sen} \frac{\pi }{3}} \right)} \\
      {}& = &{\sqrt 3 \left( {\frac{1}{2} – \frac{{\sqrt 3 }}{2}i} \right)} \\
      {}& = &{\frac{{\sqrt 3 }}{2} – \frac{3}{2}i}
    \end{array}$$
<< Enunciado