Question

determine o modulo do seguinte numero complexo z= 1/2 + 1/3 i

Answer 1

z = 1/2 + (1/3)i 
........___________......_______ 
P = V(1/2)² + (1/3)² = V1/4+1/9 -----mmc = 36. Logo: 
........._______......____.......__ 
P = V(9+4)/36 = V13/36 = (V13)/6. 

Answer 2

✅ Tendo terminado resolução, concluímos que o módulo do referido número complexo é:

     $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf |z| = \frac{\sqrt{13}}{6}\:u\cdot c \:\:\:}}\end{gathered} }$

Sejam o número complexo:

       $\Large\displaystyle\begin{gathered} z= \frac{1}{2} + \frac{1}{3}i \end{gathered}$

Sabendo que todo número complexo pode ser escrito na forma:

        $\Large\displaystyle\begin{gathered} z = a + bi\end{gathered}$

Então o seu módulo pode ser escrito na forma:

        $\Large\displaystyle\text{ \begin{gathered} |z| = \sqrt{a^{2} + b^{2}}\end{gathered} }$

Então, temos:

        $\Large\displaystyle\text{ \begin{gathered} |z| = \sqrt{\Bigg(\frac{1}{2} \Bigg)^{2} + \Bigg(\frac{1}{3} \Bigg)^{2}}\end{gathered} }$

              $\Large\displaystyle\text{ \begin{gathered} = \sqrt{\frac{1^{2}}{2^{2}} + \frac{1^{2}}{3^{2}} }\end{gathered} }$

              $\Large\displaystyle\text{ \begin{gathered} = \sqrt{\frac{1}{4} + \frac{1}{9} }\end{gathered} }$

               $\Large\displaystyle\text{ \begin{gathered} = \sqrt{\frac{9 + 4}{36} }\end{gathered} }$

               $\Large\displaystyle\text{ \begin{gathered} = \sqrt{\frac{13}{36} }\end{gathered} }$

               $\Large\displaystyle\text{ \begin{gathered} = \frac{\sqrt{13}}{\sqrt{36}} \end{gathered} }$

               $\Large\displaystyle\text{ \begin{gathered} = \frac{\sqrt{13}}{6} \end{gathered} }$

✅ Portanto, o módulo do número complexo é:

         $\Large\displaystyle\text{ \begin{gathered}|z| = \frac{\sqrt{13}}{6}\:u\cdot c \end{gathered} }$

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