Question

URGENTE!
Dado o número complexo z = 1 + i / i , qual é a forma trigonométrica de z?

Answer 1

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

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

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

$\boxed{\begin{array}{l}\sf z=\dfrac{(1+i)}{i}\cdot\dfrac{-i}{-i}\\\sf z=\dfrac{-1-i}{-i^2}\\\sf z=\dfrac{-1-i}{1}\\\sf z=-1-i\end{array}}$

$\boxed{\begin{array}{l}\sf \rho=\sqrt{(-1)^2+(-1)^2}\\\sf\rho=\sqrt{1+1}\\\sf\rho=\sqrt{2}\\\begin{cases}\sf cos(\theta)=\dfrac{-1}{\sqrt{2}}=-\dfrac{\sqrt{2}}{2}\\\sf sen(\theta)=\dfrac{-1}{\sqrt{2}}=-\dfrac{\sqrt{2}}{2}\end{cases}\longrightarrow \theta=\dfrac{5\pi}{4}\\\sf z=\rho[cos(\theta)+i~sen(\theta)]\\\sf z=\sqrt{2}\bigg[cos\bigg(\dfrac{5\pi}{4}\bigg)+i~sen\bigg(\dfrac{5\pi}{4}\bigg)\bigg]\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~c}}}}\end{array}}$

Answer 2

Resposta:

 

n. d. a.

Explicação passo a passo:

 

n. d. a.