Question

Dados os números complexos z1 e z2, calcule as operações indicadas nos itens abaixo:

Answer 1

Explicação passo-a-passo:

a)

$\sf z_1+z_2=1+3i-2+i$

$\sf z_1+z_2=1-2+3i+i$

$\sf \red{z_1+z_2=-1+4i}$

b)

$\sf z_1\cdot z_2=(1+3i)\cdot(-2+i)$

$\sf z_1\cdot z_2=-2+i-6i+3i^2$

$\sf z_1\cdot z_2=-2+i-6i+3\cdot(-1)$

$\sf z_1\cdot z_2=-2+i-6i-3$

$\sf z_1\cdot z_2=-2-3+i-6i$

$\sf \red{z_1\cdot z_2=-5-5i}$

c)

$\sf (z_1)^2=(1+3i)^2$

$\sf (z_1)^2=(1+3i)\cdot(1+3i)$

$\sf (z_1)^2=1+3i+3i+9i^2$

$\sf (z_1)^2=1+3i+3i+9\cdot(-1)$

$\sf (z_1)^2=1+3i+3i-9$

$\sf (z_1)^2=1-9+3i+3i$

$\sf \red{(z_1)^2=-8+6i}$

d)

$\sf z_1+(z_2)^2=1+3i+(-2+i)^2$

$\sf z_1+(z_2)^2=1+3i+(-2+i)\cdot(-2+i)$

$\sf z_1+(z_2)^2=1+3i+4-2i-2i+i^2$

$\sf z_1+(z_2)^2=1+3i+4-2i-2i-1$

$\sf z_1+(z_2)^2=1+4-1+3i-2i-2i$

$\sf z_1+(z_2)^2=5-1+i-2i$

$\sf \red{z_1+(z_2)^2=4-i}$