Question

Sendo z1 = 5 –2i e z2 = 1 + 6i e z3 = –4i , calcule: a)$\frac{Z1}{Z2}$ = b)$\frac{Z3}{Z1}$ = c)$\frac{Z2}{Z1}$ =

Answer 1

Explicação passo-a-passo:

a)

$\sf \dfrac{z_1}{z_2}=\dfrac{5-2i}{1+6i}$

$\sf \dfrac{z_1}{z_2}=\dfrac{5-2i}{1+6i}\cdot\dfrac{1-6i}{1-6i}$

$\sf \dfrac{z_1}{z_2}=\dfrac{5-2i-30i+12i^2}{1^2-(6i)^2}$

$\sf \dfrac{z_1}{z_2}=\dfrac{5-2i-30i+12\cdot(-1)}{1-36i^2}$

$\sf \dfrac{z_1}{z_2}=\dfrac{5-2i-30i-12}{1-36\cdot(-1)}$

$\sf \dfrac{z_1}{z_2}=\dfrac{5-12-2i-30i}{1+36}$

$\sf \dfrac{z_1}{z_2}=\dfrac{-7-32i}{37}$

$\sf \red{\dfrac{z_1}{z_2}=-\dfrac{7}{37}-\dfrac{32i}{37}}$

b)

$\sf \dfrac{z_3}{z_1}=\dfrac{-4i}{5-2i}$

$\sf \dfrac{z_3}{z_1}=\dfrac{-4i}{5-2i}\cdot\dfrac{5+2i}{5+2i}$

$\sf \dfrac{z_3}{z_1}=\dfrac{-20i-8i^2}{5^2-(2i)^2}$

$\sf \dfrac{z_3}{z_1}=\dfrac{-20i-8\cdot(-1)}{25-4i^2}$

$\sf \dfrac{z_3}{z_1}=\dfrac{-20+8}{25-4\cdot(-1)}$

$\sf \dfrac{z_3}{z_1}=\dfrac{8-20i}{25+4}$

$\sf \dfrac{z_3}{z_1}=\dfrac{8-20i}{29}$

$\sf \red{\dfrac{z_3}{z_1}=\dfrac{8}{29}-\dfrac{20i}{29}}$

c)

$\sf \dfrac{z_2}{z_1}=\dfrac{1+6i}{5-2i}$

$\sf \dfrac{z_2}{z_1}=\dfrac{1+6i}{5-2i}\cdot\dfrac{5+2i}{5+2i}$

$\sf \dfrac{z_2}{z_1}=\dfrac{5+30i+2i+12i^2}{5^2-(2i)^2}$

$\sf \dfrac{z_2}{z_1}=\dfrac{5+30i+2i+12\cdot(-1)}{25-4i^2}$

$\sf \dfrac{z_2}{z_1}=\dfrac{5+30i+2i-12}{25-4\cdot(-1)}$

$\sf \dfrac{z_2}{z_1}=\dfrac{5-12+30i+2i}{25+4}$

$\sf \dfrac{z_2}{z_1}=\dfrac{-7+32i}{29}$

$\sf \red{\dfrac{z_2}{z_1}=-\dfrac{7}{29}+\dfrac{32i}{29}}$