Question

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Determine o produto z1z2 e o quociente z1/z2 para:

a) Z1 = 2 (cos 2π/3 + i. sen 2π/3) e Z2 = 3 (cos π/4 + i. sen π/4)
b) Z1 = 4 (cos π/6 + i. sen π/6) e Z2 = 2 (cos 3π/4 + i. sen 3π/4)
c) Z1 = 2 (cos π/3 + i. sen π/3) e Z2 = 5 (cos π/2 + i. sen π/2)
d) Z1 = (cos 3π/4 + i. sen 3π/4) e Z2 = (cos 2π/3 + i. sen 2π/3)

Answer 1

a)

$\mathsf{z_{1}.z_{2} = } \\\mathsf{ 2.3[cos(\dfrac{2\pi}{3}+\dfrac{3\pi}{4})+isen(\dfrac{2\pi}{3}+\dfrac{3\pi}{4})]}\\\mathsf{z_{1}.z_{2}=6[cos(\dfrac{17\pi}{12})+isen(\dfrac{17\pi}{12})]}$

$\mathsf{\dfrac{z_{1}}{z_{2}}=}\\\mathsf{\dfrac{2}{3}[cos(\dfrac{2\pi}{3}-\dfrac{3\pi}{4})+isen(\dfrac{2\pi}{3}-\dfrac{3\pi}{4})]}\\\mathsf{\dfrac{z_{1}}{z_{2}}=\dfrac{2}{3}[cos(-\dfrac{\pi}{12})+isen(-\dfrac{\pi}{12})]}$

b)

$\mathsf{z_{1}.z_{2} = } \\\mathsf{ 4.2[cos(\dfrac{\pi}{6}+\dfrac{3\pi}{4})+isen(\dfrac{\pi}{6}+\dfrac{3\pi}{4})]}\\\mathsf{z_{1}.z_{2}=8[cos(\dfrac{11\pi}{12})+isen(\dfrac{11\pi}{12})]}$

$\mathsf{\dfrac{z_{1}}{z_{2}}=}\\\mathsf{\dfrac{4}{2}[cos(\dfrac{\pi}{6}-\dfrac{3\pi}{4})+isen(\dfrac{\pi}{6}-\dfrac{3\pi}{4})]}\\\mathsf{\dfrac{z_{1}}{z_{2}}=2[cos(-\dfrac{7\pi}{12})+isen(-\dfrac{7\pi}{12})]}$

c)

$\mathsf{z_{1}.z_{2} = } \\\mathsf{ 2.5[cos(\dfrac{\pi}{3}+\dfrac{\pi}{2})+isen(\dfrac{\pi}{3}+\dfrac{\pi}{2})]}\\\mathsf{z_{1}.z_{2}=10[cos(\dfrac{5\pi}{6})+isen(\dfrac{5\pi}{6})]}$

$\mathsf{\dfrac{z_{1}}{z_{2}}=}\\\mathsf{\dfrac{2}{5}[cos(\dfrac{\pi}{3}-\dfrac{\pi}{2})+isen(\dfrac{\pi}{3}-\dfrac{\pi}{2})]}\\\mathsf{\dfrac{z_{1}}{z_{2}}=\dfrac{2}{5}[cos(-\dfrac{\pi}{6})+isen(-\dfrac{\pi}{6})]}$

d)

$\mathsf{z_{1}.z_{2} = } \\\mathsf{ 1.1[cos(\dfrac{3\pi}{4}+\dfrac{2\pi}{3})+isen(\dfrac{3\pi}{4}+\dfrac{2\pi}{3})]}\\\mathsf{z_{1}.z_{2}=[cos(\dfrac{17\pi}{12})+isen(\dfrac{17\pi}{12})]}$

$\mathsf{\dfrac{z_{1}}{z_{2}}=}\\\mathsf{\dfrac{1}{1}[cos(\dfrac{3\pi}{4}-\dfrac{2\pi}{3})+isen(\dfrac{3\pi}{4}-\dfrac{2\pi}{3})]}\\\mathsf{\dfrac{z_{1}}{z_{2}}=[cos(\dfrac{\pi}{12})+isen(\dfrac{\pi}{12})]}$