Question

Dados os números z_1= 2-i z_2= 1/3 +i z_3= -2 - (1/4)i Calcule: a) z_1 + z_2 *

b) z_3 - z_2 = *

c) z_1 . z_2=

d) z_3^( 2)= *

e) z_1 / z_2 = *

Answer 1

Explicação passo-a-passo:

a)

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

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

$\sf z_1+z_2=2+\dfrac{1}{3}\not{-i}+\not{i}$

$\sf z_1+z_2=2+\dfrac{1}{3}$

$\sf z_1+z_2=\dfrac{6}{3}+\dfrac{1}{3}$

$\sf z_1+z_2=\dfrac{6+1}{3}$

$\sf \red{z_1+z_2=\dfrac{7}{3}}$

b)

$\sf z_3-z_2=-2-\dfrac{i}{4}-\left(\dfrac{1}{3}+i\right)$

$\sf z_3-z_2=-2-\dfrac{i}{4}-\dfrac{1}{3}-i$

$\sf z_3-z_2=-2-\dfrac{1}{3}-\dfrac{i}{4}-i$

$\sf z_3-z_2=\dfrac{-6}{3}-\dfrac{1}{3}-\dfrac{i}{4}-\dfrac{4i}{4}$

$\sf z_3-z_2=\dfrac{-6-1}{3}-\dfrac{1+4i}{4}$

$\sf \red{z_3-z_2=\dfrac{-7}{3}-\dfrac{5i}{4}}$

c)

$\sf z_1\cdot z_2=(2-i)\cdot\left(\dfrac{1}{3}+i\right)$

$\sf z_1\cdot z_2=\dfrac{2}{3}+2i-\dfrac{i}{3}-i^2$

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

$\sf z_1\cdot z_2=\dfrac{2}{3}+2i-\dfrac{i}{3}+1$

$\sf z_1\cdot z_2=\dfrac{2}{3}+1+2i-\dfrac{i}{3}$

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

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

$\sf \red{z_1\cdot z_2=\dfrac{5}{3}+\dfrac{5i}{3}}$

d)

$\sf (z_3)^2=\left(-2-\dfrac{i}{4}\right)^2$

$\sf (z_3)^2=(-2)^2-2\cdot(-2)\cdot\cdot\dfrac{i}{4}+\left(\dfrac{i}{4}\right)^2$

$\sf (z_3)^2=4+\dfrac{4i}{4}+\dfrac{i^2}{16}$

$\sf (z_3)^2=4+i-\dfrac{1}{16}$

$\sf (z_3)^2=4-\dfrac{1}{16}+i$

$\sf (z_3)^2=\dfrac{64}{16}-\dfrac{1}{16}+i$

$\sf (z_3)^2=\dfrac{64-1}{16}+i$

$\sf \red{(z_3)^2=\dfrac{63}{16}+i}$

e)

$\sf \dfrac{z_1}{z_2}=\dfrac{2-i}{\frac{1}{3}+i}$

$\sf \dfrac{z_1}{z_2}=\dfrac{2-i}{\frac{1}{3}+i}\cdot\dfrac{\frac{1}{3}-i}{\frac{1}{3}-i}$

$\sf \dfrac{z_1}{z_2}=\dfrac{\frac{2}{3}-\frac{i}{3}-2i+i^2}{\left(\frac{1}{3}\right)^2-i^2}$

$\sf \dfrac{z_1}{z_2}=\dfrac{\frac{2}{3}-\frac{i}{3}-2i-1}{\frac{1}{9}-(-1)}$

$\sf \dfrac{z_1}{z_2}=\dfrac{\frac{2}{3}-\frac{i}{3}-2i-1}{\frac{1}{9}+1}$

$\sf \dfrac{z_1}{z_2}=\dfrac{\frac{2}{3}-1-\frac{i}{3}-2i}{\frac{1}{9}+\frac{9}{9}}$

$\sf \dfrac{z_1}{z_2}=\dfrac{\frac{2}{3}-\frac{3}{3}-\frac{i}{3}-\frac{6i}{3}}{\frac{1}{9}+\frac{9}{9}}$

$\sf \dfrac{z_1}{z_2}=\dfrac{\frac{2-3}{3}-\frac{i+6i}{3}}{\frac{1+9}{9}}$

$\sf \dfrac{z_1}{z_2}=\dfrac{\frac{-1-7i}{3}}{\frac{10}{9}}$

$\sf \dfrac{z_1}{z_2}=\dfrac{-1-7i}{3}\cdot\dfrac{9}{10}$

$\sf \dfrac{z_1}{z_2}=\dfrac{-9-63i}{30}$

$\sf \dfrac{z_1}{z_2}=\dfrac{-3-21i}{10}$

$\sf \red{\dfrac{z_1}{z_2}=\dfrac{-3}{10}-\dfrac{21i}{10}}$