Question

Escreva na forma trigonometria os seguintes números 3+4i÷3-3i

Answer 1

• Passando para a forma trigonometrica, temos que $\large\displaystyle\text{ \begin{aligned} z= \frac{5}{3\sqrt{2} } (\cos \left( 98.13^o \right) + i\sin \left( 98.13^o \right) )\end{aligned} }$.

Primeiramente irei racionalizar os números complexos 3+4i÷3-3i.

$\large\displaystyle\begin{aligned} \frac{3+4i}{3-3i}\cdot \frac{3+3i}{3+3i} \Leftrightarrow \end{aligned}$

$\large\displaystyle\begin{aligned} \frac{(3+4i)\cdot (3+3i)}{(3-3i)\cdot (3+3i)} \Leftrightarrow \end{aligned}$

$\large\displaystyle\begin{aligned} \frac{9 + 9i + 12i -12}{9 +9i -9i+9} \Leftrightarrow \end{aligned}$

$\large\displaystyle\begin{aligned} \frac{-3 }{18} + \frac{21 }{18} i\Leftrightarrow \end{aligned}$

$\large\displaystyle\text{ \begin{aligned} \Leftrightarrow \underline{\boxed{\frac{-1 }{6} + \frac{7 }{6} i}}\end{aligned} }$

Agora devemos aplicar na fórmula $\large\displaystyle\begin{aligned} z= p(\cos \theta + i\sin \theta )\ ;\ p=\sqrt{a^2 + b^2} \end{aligned}$ . Logo:

$\large\displaystyle\begin{aligned} z= p(\cos \theta + i\sin \theta )\ ;\ p=\sqrt{a^2 + b^2} \end{aligned}$

$\large\displaystyle\text{ \begin{aligned} z= p(\cos \theta + i\sin \theta )\ ;\ p=\sqrt{\left(-\frac{1}{6} \right)^2 + \left( \frac{7}{6} \right)^2} \end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} z= p(\cos \theta + i\sin \theta )\ ;\ p=\sqrt{\left(\frac{1}{36} \right) + \left( \frac{49}{36} \right)} \end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} z= p(\cos \theta + i\sin \theta )\ ;\ p=\sqrt{ \left( \frac{25}{18} \right)} \end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} z= p(\cos \theta + i\sin \theta )\ ;\ p= \frac{5}{3\sqrt{2}} \end{aligned} }$

Para finalizar, devemos achar o argumento ( $\theta$ ). Para isso devemos utilizar a seguinte fórmula:

$\large\displaystyle\begin{aligned} \cos \theta = \frac{a}{p} \ ;\ \sin \theta = \frac{b}{p}\end{aligned}$

• Aplicando na sua questão:

$\large\displaystyle\text{ \begin{aligned} \cos \theta = \frac{-\frac{1}{6} }{\frac{5}{3\sqrt{2}} } \ ;\ \sin \theta = \frac{\frac{7}{6}}{\frac{5}{3\sqrt{2}}}\end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} \cos \theta = -\frac{\sqrt{2}}{10} \ ;\ \sin \theta = \frac{7\sqrt{2}}{10} \end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} \underline{\boxed{\cos \theta =\frac{\sqrt{ 2}}{10} \ ;\ \sin \theta = \frac{7\sqrt{2} }{10}}} \end{aligned} }$

• Ficando então:

$\large\displaystyle\begin{aligned} z= p(\cos \theta + i\sin \theta )\end{aligned}$

$\large\displaystyle\text{ \begin{aligned} \therefore \boxed{\boxed{\green{z= \frac{5}{3\sqrt{2} } (\cos \left( 98.13^o \right) + i\sin \left( 98.13^o\right) )}}}\end{aligned} }$

Veja mais sobre:

Números complexos.

$\blue{\square}$ brainly.com.br/tarefa/22693420