Question

Sendo Z1=4(cosπ+i senπ), Z2=3(cos 7π/4+i sen 7π/4) e Z3=2(cos 5π/6+i sen 5π/6), determine:
a)Z1. Z2
b)Z1 .Z3

Answer 1

Fórmula de Euler:

$\boxed{\boxed{e^{i\theta}=cos\,\theta+i\,sen\,\theta}}~~\Rightarrow~~\boxed{\boxed{xe^{i\theta}=x\,(cos\,\theta+i\,sen\,\theta)}}$
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Escrevendo $z_{1},~z_{2},~z_{3}$ na fórmula de Euler:

$z_{1}=4(cos\,\pi+i\,sen\,\pi)=4e^{i\pi}\\\\z_{2}=3(cos\,\frac{7\pi}{4}+i\,sen\,\frac{7\pi}{4})=3e^{i(7\pi/4)}\\\\z_{3}=2(cos\,\frac{5\pi}{6}+i\,sen\,\frac{5\pi}{6})=2e^{i(5\pi/6)}$

Portanto:

$z_{1}\cdot z_{2}=4e^{i\pi}\cdot3e^{i(7\pi/4)}=12e^{i\pi+i(7\pi/4)}=12e^{i[\pi+(7\pi/4)]}=12e^{i(11\pi/4)}$

Reescrevendo o número (usando a fórmula), temos

$\boxed{\boxed{z_{1}\cdot z_{2}=12\bigg(cos\,\frac{11\pi}{4}+i\,sen\,\frac{11\pi}{4}\bigg)}}$

_________

$z_{1}\cdot z_{3}=4e^{i\pi}\cdot2e^{i(5\pi/6)}=8e^{i\pi+i(5\pi/6)}=8e^{i[\pi+(5\pi/6)]}=8e^{i(11\pi/6)}$

Portanto:

$\boxed{\boxed{z_{1}\cdot z_{3}=8\bigg(cos\,\frac{11\pi}{6}+i\,sen\,\frac{11\pi}{6}\bigg)}}$