Parte06

Operações na forma trigonométrica

Sejam os dois números complexos \(z_{1}=\rho_{1}(\text{cos}\,\theta_{1}+i\text{sen}\,\theta_{1})\) e \(z_{2}=\rho_{2}(\text{cos}\,\theta_{2}+i\text{sen}\,\theta_{2})\)

Multiplicação

\(\begin{array}{rcl} z_{1}\cdot z_{2} & = & \rho_{1}(\text{cos}\,\theta_{1}+i\cdot\text{sen}\,\theta_{1})\cdot \rho_{2}(\text{cos}\,\theta_{2}+i\cdot\text{sen}\,\theta_{2})=\\ &=& \rho_{1}\cdot\rho_{2}\cdot(\text{cos}\,\theta_{1}\cdot\text{cos}\,\theta_{2}+i\cdot\text{cos}\,\theta_{1}\cdot\text{sen}\,\theta_{2}+i\cdot\text{sen}\,\theta_{1}\cdot\text{cos}\,\theta_{2}+\cancel{i^2}^{\,\,(-1)}\cdot\text{sen}\,\theta_{1}\cdot\text{sen}\,\theta_{2})=\\\\ &=& \rho_{1}\cdot\rho_{2}[(\underbrace{\text{cos}\,\theta_{1}\cdot\text{cos}\,\theta_{2}-\text{sen}\,\theta_{1}\cdot\text{sen}\,\theta_{2}}_{\text{cos}(\theta_{1}+\theta_{2})})]+i\cdot(\underbrace{\text{sen}\,\theta_{1}\cdot\text{cos}\,\theta_{2}+\text{sen}\,\theta_{2}\cdot\text{cos}\,\theta_{1}}_{\text{sen}(\theta_{1}+\theta_{2})})] \end{array}\)

Portanto:

\(\boxed{z_{1}\cdot z_{2}=\rho_{1}\cdot\rho_{2}\cdot[\text{cos}(\theta_{1}+\theta_{2})+i\cdot\text{sen}(\theta_{1}+\theta_{2})]}\)

Exercício resolvido

Dados \(z_{1}=3\left(\text{cos}\,\dfrac{\pi}{3}+i\cdot\text{sen}\,\dfrac{\pi}{3}\right)\) e \(z_{2}=4\left(\text{cos}\,\dfrac{4\pi}{3}+i\cdot\text{sen}\,\dfrac{4\pi}{3}\right)\). Calcule, para \(\theta\in[0;\,2\pi[\):

a) \(z_{1}\cdot z_{2}\Rightarrow z_{1}\cdot z_{2}=3\cdot 4\left[\text{cos}\left(\dfrac{\pi}{3}+\dfrac{4\pi}{3}\right)+i\cdot\text{sen}\left(\dfrac{\pi}{3}+\dfrac{4\pi}{3}\right)\right]\therefore\)

\(\boxed{z_{1}\cdot z_{2}=12\cdot\left[\text{cos}\left(\dfrac{5\pi}{3}\right)+i\cdot\text{sen}\left(\dfrac{5\pi}{3}\right)\right]}\)

b) \(z_{1}^2\Rightarrow z_{1}^2=z_{1}\cdot z_{1}=3\cdot 3\left[\text{cos}\left(\dfrac{\pi}{3}+\dfrac{\pi}{3}\right)+i\cdot\text{sen}\left(\dfrac{\pi}{3}+\dfrac{\pi}{3}\right)\right]\therefore\)

\(\boxed{z_{1}^2=9\cdot\left[\text{cos}\left(\dfrac{2\pi}{3}\right)+i\cdot\text{sen}\left(\dfrac{2\pi}{3}\right)\right]}\)

c) \(z_{2}^2\Rightarrow z_{2}^2=z_{2}\cdot z_{2}=4\cdot 4\left[\text{cos}\left(\dfrac{4\pi}{3}+\dfrac{4\pi}{3}\right)+i\cdot\text{sen}\left(\dfrac{4\pi}{3}+\dfrac{4\pi}{3}\right)\right]\therefore\)

\(\boxed{z_{2}^2=16\cdot\left[\text{cos}\left(\dfrac{2\pi}{3}\right)+i\cdot\text{sen}\left(\dfrac{2\pi}{3}\right)\right]}^{(*)}\)

(*)Observe que \(\dfrac{8\pi}{3}\)rad é um arco de segunda volta, portanto, devemos buscar um correspondente dentro da primeira volta, como especifica a questão, e esse arco é \(\dfrac{2\pi}{3}\)rad.

Divisão

\(\begin{array}{rcl} \dfrac{z_{1}}{z_{2}} & = & \dfrac{\rho_{1}(\text{cos}\,\theta_{1}+i\cdot\text{sen}\,\theta_{1})}{\rho_{2}(\text{cos}\,\theta_{2}+i\cdot\text{sen}\,\theta_{2})}\times\dfrac{(\text{cos}\,\theta_{2}-i\cdot\text{sen}\,\theta_{2})}{(\text{cos}\,\theta_{2}-i\cdot\text{sen}\,\theta_{2})}=\\\\ &=& \dfrac{\rho_{1}}{\rho_{2}}\cdot\dfrac{\text{cos}\,\theta_{1}\cdot\text{cos}\,\theta_{2}-i\cdot\text{cos}\,\theta_{1}\cdot\text{sen}\,\theta_{2}+i\cdot\text{sen}\,\theta_{1}\cdot\text{cos}\,\theta_{2}-i^{2}\cdot\text{sen}\,\theta_{1}\cdot\text{sen}\,\theta_{2}}{[\text{cos}^{2}\,\theta_{2}-(i\cdot\text{sen}\,\theta_{2})^{2}]}=\\\\ &=& \dfrac{\rho_{1}}{\rho_{2}}\cdot\dfrac{(\text{cos}\,\theta_{1}\cdot\text{cos}\,\theta_{2}+\text{sen}\,\theta_{1}\cdot\text{sen}\,\theta_{2})+i\cdot(\text{sen}\,\theta_{1}\cdot\text{cos}\,\theta_{2}-\text{sen}\,\theta_{2}\cdot\text{cos}\,\theta_{1})}{\cancel{\text{cos}^{2}\,\theta_{2}+\text{sen}^{2}\,\theta_{2}}^{\,\,(1)}} \end{array}\)

Portanto:

\(\boxed{\dfrac{z_{1}}{z_{2}} =\dfrac{\rho_{1}}{\rho_{2}}\cdot\Big[\text{cos}(\theta_{1}-\theta_{2})+i\cdot\text{sen}(\theta_{1}-\theta_{2})\Big]}\)

Exercício resolvido

Dados \(z_{1}=3\left(\text{cos}\,\dfrac{\pi}{3}+i\cdot\text{sen}\,\dfrac{\pi}{3}\right)\) e \(z_{2}=4\left(\text{cos}\,\dfrac{4\pi}{3}+i\cdot\text{sen}\,\dfrac{4\pi}{3}\right)\). Calcule, para \(\theta\in[0;\,2\pi[\):

a) \(\dfrac{z_{2}}{z_{1}}\)

\(\dfrac{z_{2}}{z_{1}}=\dfrac{4}{3}\left[\text{cos}\left(\dfrac{4\pi}{3}-\dfrac{\pi}{3}\right)+i\cdot\text{sen}\left(\dfrac{4\pi}{3}-\dfrac{\pi}{3}\right)\right]\to\boxed{\dfrac{z_{1}}{z_{2}}=\dfrac{4}{3}\cdot\left(\text{cos}\,\pi+i\cdot\text{sen}\,\pi\right)}\)

b) \(\dfrac{z_{1}}{z_{2}}\)

\(\dfrac{z_{1}}{z_{2}}=\dfrac{3}{4}\left[\text{cos}\left(\dfrac{\pi}{3}-\dfrac{4\pi}{3}\right)+i\cdot\text{sen}\left(\dfrac{\pi}{3}-\dfrac{4\pi}{3}\right)\right]\to\boxed{\dfrac{z_{1}}{z_{2}}=\dfrac{3}{4}\cdot\left[\text{cos}\,\pi+i\cdot\text{sen}\,\pi\right]}^{(*)}\)

(*)Observe que \(-\pi\) rad, resultado de \(\left(\dfrac{\pi}{3}-\dfrac{4\pi}{3}\right)\) rad, é um arco de primeira volta negativa, isto é, \(\theta\in\,]-2\pi;\,0].\) Nesse caso, devemos buscar um correspondente dentro da primeira volta positiva, como especifica a questão, por isso o arco da solução é \(\pi\) rad.