Question

Se z = 1 + i, então w = z^16 na forma trigonométrica e na forma algébrica é, respectivamente:

( ) w = 256(cos⁡2 π + isen2π) e w = 256.
( ) w = 2^16(cos⁡2 π + isen2π) e w = 2^16.
( ) w = 256(cos⁡4 π + isen4π) e w = 2^16.
( ) w = 2^16(cos⁡4 π + isen4π) e w = 256.

Answer 1

Resposta:

$\textsf{letra A}$

Explicação passo a passo:

$\mathsf{z = 1 + i}$

$\mathsf{\rho = \sqrt{a^2 + b^2}}$

$\mathsf{\rho = \sqrt{1^2 + 1^2}}$

$\mathsf{\rho = \sqrt{1 + 1}}$

$\mathsf{\rho = \sqrt{2}}$

$\mathsf{sen\:\Theta = \dfrac{b}{\rho} = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}}$

$\mathsf{cos\:\Theta = \dfrac{a}{\rho} = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}}$

$\mathsf{z = \sqrt{2}\left(cos\:\dfrac{\pi }{4} + i\:sen\:\dfrac{\pi }{4}\right)}$

$\mathsf{w = z^{16}}$

$\mathsf{w = \left(\sqrt{2}\left(cos\:\dfrac{\pi }{4} + i\:sen\:\dfrac{\pi }{4}\right)\right)^{16}}$

$\mathsf{w = (\sqrt{2})^{16}\left(cos\:\dfrac{16\pi }{4} + i\:sen\:\dfrac{16\pi }{4}\right)\right)}$

$\boxed{\boxed{\mathsf{w = 256\:(cos\:2 \pi + i\:sen\:2\pi)}}}$

$\mathsf{w = 256\:(1 + 0)}$

$\boxed{\boxed{\mathsf{w = 256}}}$