Question

Dados os números complexos Resolva

Answer 1

Explicação passo-a-passo:

a)

$\sf z+v=3+2i+4-3i$

$\sf z+v=3+4+2i-3i$

$\sf z+v=7-i$

b)

$\sf v+w=4-3i-2+2i$

$\sf v+w=4-2-3i+2i$

$\sf v+w=2-i$

c)

$\sf z-v=3+2i-(4-3i)$

$\sf z-v=3+2i-4+3i$

$\sf z-v=3-4+2i+3i$

$\sf z-v=-1+5i$

d)

$\sf v-w=4-3i-(-2+2i)$

$\sf v-w=4-3i+2-2i$

$\sf v-w=4+2-3i-2i$

$\sf v-w=6-5i$

e)

$\sf z+v+w=3+2i+4-3i-2+2i$

$\sf z+v+w=3+4-2+2i-3i+2i$

$\sf z+v+w=5+i$

f)

$\sf z\cdot v=(3+2i)\cdot(4-3i)$

$\sf z\cdot v=12-9i+8i-6i^2$

$\sf z\cdot v=12-i-6\cdot(-1)$

$\sf z\cdot v=12-i+6$

$\sf z\cdot v=18-i$

g)

$\sf z\cdot w=(3+2i)\cdot(-2+2i)$

$\sf z\cdot w=-6+6i-4i+4i^2$

$\sf z\cdot w=-6+2i+4\cdot(-1)$

$\sf z\cdot w=-6+2i-4$

$\sf z\cdot w=-10+2i$

h)

$\sf z^2=(3+2i)^2$

$\sf z^2=3^2+2\cdot3\cdot2i+(2i)^2$

$\sf z^2=9+12i+4i^2$

$\sf z^2=9+12i+4\cdot(-1)$

$\sf z^2=9+12i-4$

$\sf z^2=5+12i$

i)

$\sf \dfrac{z}{v}=\dfrac{3+2i}{4-3i}$

$\sf \dfrac{z}{v}=\dfrac{3+2i}{4-3i}\cdot\dfrac{4+3i}{4+3i}$

$\sf \dfrac{z}{v}=\dfrac{12+8i+9i+6i^2}{4^2-(3i)^2}$

$\sf \dfrac{z}{v}=\dfrac{12+17i+6\cdot(-1)}{16-9i^2}$

$\sf \dfrac{z}{v}=\dfrac{12+17i-6}{16-9\cdot(-1)}$

$\sf \dfrac{z}{v}=\dfrac{6+17i}{16+9}$

$\sf \dfrac{z}{v}=\dfrac{6+17i}{25}$

$\sf \dfrac{z}{v}=\dfrac{6}{25}+\dfrac{17i}{25}$

j)

$\sf \dfrac{z}{w}=\dfrac{3+2i}{-2+2i}$

$\sf \dfrac{z}{v}=\dfrac{3+2i}{-2+2i}\cdot\dfrac{-2-2i}{-2-2i}$

$\sf \dfrac{z}{w}=\dfrac{-6-4i-6i-4i^2}{(-2)^2-(2i)^2}$

$\sf \dfrac{z}{w}=\dfrac{-6-10i-4\cdot(-1)}{4-4i^2}$

$\sf \dfrac{z}{w}=\dfrac{-6-10i+4}{4-4\cdot(-1)}$

$\sf \dfrac{z}{w}=\dfrac{-2-10i}{4+4}$

$\sf \dfrac{z}{w}=\dfrac{-2-10i}{8}$

$\sf \dfrac{z}{w}=\dfrac{-2}{8}-\dfrac{10i}{8}$

$\sf \dfrac{z}{w}=-\dfrac{1}{4}-\dfrac{5i}{4}$