Resposta :

rikardoa
a) [tex]Z_1 . Z_2 = (\frac{1}{2} + 3i)(2 - 5i)[/tex]

[tex]Z_1 . Z_2 = \frac{1}{2} . 2 - \frac{1}{2} . 5i + 2 . 3i - 3i . 5i[/tex]

[tex]Z_1 . Z_2 = 1 - \frac{5i}{2} + 6i - 15i^2 = 1 + 15 + \frac{2 . 6i - 5i}{2}[/tex]

[tex]Z_1 . Z_2 = 16 + \frac{12i - 5i}{2} = 16 + \frac{7}{2}i[/tex]

Devemos lembrar que [tex] i^{2} = (\sqrt{-1})^2 = -1 [/tex]

b) [tex]{Z_2}^2 = (2 - 5i)^2 = 2^2 - 2 . 2 . 5i + (5i)^2[/tex]

[tex]{Z_2}^2 = 4 - 20i + 25i^2 = 4 - 20i + 25(-1)[/tex]

[tex]{Z_2}^2 = 4 - 20i - 25 = -21 - 20i = -(21 + 20i)[/tex]
3478elc

a) (1/2 + 3i).(2-5i)= 1 - 5i/2 +6i - 15i^2 ==> 1 +15 (- 5i +12i)/2 ==>16 + 7i/2

b) (2-5i)2 =  4 - 20i + 25i^2 ==> 4 - 25 - 20i ==> - 21 - 20i

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