Question

Escreva na forma trigonométrica os números complexos
Z= 2+2i

Answer 1

Olá


Forma algébrica de um número complexo:

z = a + bi


Forma trigonométrica de um número complexo:

z = ρ.( cosθ + isenθ )



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z = 2 + 2i


Calculando o módulo de z (ρ)

ρ = 

ρ = $\sqrt{2^2+2^2}$

ρ = $\sqrt{8}$

ρ = $2 \sqrt{2}$



Calculando o cosθ


$\mathsf{cos\theta = \dfrac{a}{\rho} }\\\\\\\mathsf{cos\theta~=~\dfrac{\diagup\!\!\!\!2}{\diagup\!\!\!\!2\sqrt{2}}}\\\\\\\mathsf{cos\theta= \dfrac{1}{ \sqrt{2} } \cdot \dfrac{\sqrt{2}}{\sqrt{2}} }\\\\\\\mathsf{cos\theta = \dfrac{\sqrt{2}}{2} }$




Calculando o senθ


$\mathsf{sen\theta = \dfrac{b}{\rho} }\\\\\\\mathsf{sen\theta~=~\dfrac{\diagup\!\!\!\!2}{\diagup\!\!\!\!2\sqrt{2}}}\\\\\\\mathsf{sen\theta= \dfrac{1}{ \sqrt{2} } \cdot \dfrac{\sqrt{2}}{\sqrt{2}} }\\\\\\\mathsf{sen\theta = \dfrac{\sqrt{2}}{2} }$



O angulo √2/2 de seno e cosseno é 45º.

45º equivale a π/4


Então


θ = π/4



E já que a forma trigonométrica de um número complexo é 

z = ρ.( cosθ + isenθ )



∴



$\displaystyle \boxed{\mathsf{z= 2\sqrt{2}\cdot \left[cos \left( \frac{\pi}{4} \right)~+~isen \left( \frac{\pi}{4} \right) \right] }}$

Answer 2

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$\boxed{\sf{\underline{M\acute{o}dulo~de~um~n\acute{u}mero~complexo~z=a+bi}}}\\\huge\boxed{\boxed{\boxed{\boxed{\boxed{\sf{\rho=\sqrt{a^2+b^2}}}}}}}$

$\boxed{\sf{\underline{Argumento~de~um~n\acute{u}mero~complexo}}}\\\huge\begin{array}\boxed{\boxed{\boxed{\boxed{\boxed{\sf{cos(\theta)=\dfrac{a}{\rho}~~~sen(\theta)=\dfrac{b}{\rho}}}}}}\end{array}}$

$\boxed{\sf{\underline{Forma~trigonom\acute{e}trica~de~um~n\acute{u}mero~complexo}}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf{Z=\rho\left[cos(\theta)+i~sen(\theta)\right]}}}}}$

$\sf z=2+2i\\\sf \rho=\sqrt{2^2+2^2}=\sqrt{2\cdot2^2}=2\sqrt{2}\\\begin{cases}\sf cos(\theta)=\dfrac{\diagdown\!\!\!2}{\diagdown\!\!\!2\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\\sf sen(\theta)=\dfrac{\diagdown\!\!\!2}{\diagdown\!\!\!2\sqrt{2}}=\dfrac{\sqrt{2}}{2}\end{cases}\implies\tt \theta=\dfrac{\pi}{4}\\\huge\boxed{\boxed{\boxed{\boxed{\sf z=2\sqrt{2}\bigg[cos\bigg(\dfrac{\pi}{4}\bigg)+i~sen\bigg(\dfrac{\pi}{4}\bigg)\bigg]}}}}$