(UFMS) trigonometria-polinomios-numeros complexos

Passe 3° Etapa 2017-2019

Questão 27 - A forma trigonométrica do número complexo Z = (1 + i) (2 + 2i) (4 + 4i) é equivalente a:

a)  [latex] 8(\cos \frac{3\pi }{2} + i\sin \frac{3\pi }{2}) [/latex]

b)  [latex] 16(\cos \frac{3\pi }{4} + i\sin \frac{3\pi }{4}) [/latex]

c)  [latex] 16 \sqrt{2}(\cos \frac{3\pi }{4} + i\sin \frac{3\pi }{4}) [/latex]

d)  [latex] 8\sqrt{2}(\cos \frac{3\pi }{4} + i\sin \frac{3\pi }{4}) [/latex]

e) [latex] 4(\cos \frac{\pi }{2} + i\sin \frac{\pi }{2}) [/latex]

z = (1 + i) (2 + 2i) (4 + 4i)
z = 2(1 + i) (1 + i) (4 + 4i)
z = 4*2(1 + i) (1 + i) (1 + i)
z = 8(1 + i)³
z= 8(-2+2i)
z=-16+16i
|z|= 16√2
arg(z)=3pi/4

z=16√2[cos(3pi/4)+isen(3pi/4)]

Boa tarde!

Só pra fazer de uma forma diferente
[latex]z_1=1+i\Rightarrow\begin{cases}\left|z_1\right|=\sqrt{1^2+1^2}=\sqrt{2}\\\tan\theta=\frac{1}{1}\Rightarrow\theta=\frac{\pi}{4}\end{cases}\Rightarrow z_1=\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)[latex]
[latex]z_2=2+2i\Rightarrow\begin{cases}\left|z_2\right|=\sqrt{2^2+2^2}=\sqrt{8}\\\tan\theta=\frac{2}{2}\Rightarrow\theta=\frac{\pi}{4}\end{cases}\Rightarrow z_2=\sqrt{8}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)[latex]
[latex]z_3=4+4i\Rightarrow\begin{cases}\left|z_3\right|=\sqrt{4^2+4^2}=\sqrt{32}\\\tan\theta=\frac{4}{4}\Rightarrow\theta=\frac{\pi}{4}\end{cases}\Rightarrow z_3=\sqrt{32}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)[latex]

Produto:
[latex]\\z_1\cdot z_2\cdot z_3=\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\cdot\sqrt{8}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\cdot\sqrt{32}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\\ z_1\cdot z_2\cdot z_3=\sqrt{2}\cdot\sqrt{8}\cdot\sqrt{32}\left[ \cos\left(\frac{\pi}{4}+\frac{\pi}{4}+\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}+\frac{\pi}{4}+\frac{\pi}{4}\right)\right]\\ z_1\cdot z_2\cdot z_3=16\sqrt{2}\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)=16\sqrt{2}\left(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)=-16+16i[/latex]

Somente para fazer diferente, mesmo

Espero ter ajudado!