Question

Dados os numeros complexos z=4+i e w=-3+5i determine a)z×w b)z/w

Answer 1

Lembrando que i² = - 1, vamos lá!
Na divisão iremos utilizar o conjugado de w o qual é w¯ =  -3-5i.

a)
z × w =
(4 + i) × ((-3) + 5i) =
-12 + 20i + (-3i) + 5i² = 
- 12 + 20i - 3i + 5 × (-1) = 
- 12 + 20i - 3i + (-5) =
- 12 + 20i - 3i - 5 =
17i - 17

b) 
$\frac{z}{w}$ =
$\frac{z}{w}$ × $\frac{w(conjugado)}{w(conjugado)}$ =

$\frac{4+i}{(-3)+5i}$ × $\frac{(-3)-5i}{(-3)-5i}$ =

$\frac{(4+i)((-3)-5i)}{((-3)+5i)((-3)-5i)}$ =

$\frac{-12-20i-3i-5i^{2}}{9+15i-15i-25i^{2}}$ =

$\frac{-12-23i-5i^{2}}{9-25i^{2}}$ =

$\frac{-12-23i-5(-1)}{9-25(-1)}$ =

$\frac{-12-23i+5}{9+25}$ =

$\frac{-23i-7}{34}$

 

Answer 2

a)
$z\cdot w=(4+i)\cdot(-3+5i)=-12+20i-3i+5i^2=-17+17i$

b)
$\displaystyle \frac{z}{w}=\frac{z\cdot\bar{w}}{w\cdot\bar{w}}=\frac{4+i}{-3+5i}\cdot\frac{-3-5i}{-3-5i}=\frac{-12-20i-3i-5i^2}{9-25i^2}=\frac{-7-23i}{34}\\\\\boxed{\frac{z}{w}=-\frac{7}{34}-\frac{23}{34}i}$

onde:
$\bar{w}=~conjugado~de~w$