Explicație pas cu pas:
[tex] \: \: \: \: \: \: \: \: z_1=-2 +2 \sqrt{3}i \\z_2=4+4i \\ z_3= \sqrt{3} -i[/tex]
a)
[tex]4 \sqrt{3} \times z_3 + 2 \times z_1-z_2 = 4 \sqrt{3}( \sqrt{3} - i) + 2( - 2 + 2 \sqrt{3}i) - (4 + 4i) = 12 - 4 \sqrt{3}i - 4 + 4 \sqrt{3} i - 4 - 4i = 4 - 4i = 4(1 - i) [/tex]
b)
[tex] |z_1| = | - 2 + 2 \sqrt{3}i| = \sqrt{ {( - 2)}^{2} + {(2 \sqrt{3} )}^{2} } = \sqrt{4 + 12} = \sqrt{16} = 4 [/tex]
c)
[tex](z_2)^{2} = {(4 + 4i})^{2} =16 + 32i + 16 {i}^{2} = 16 + 32i - 16 = 32i[/tex]
d)
[tex]\frac{z_3}{z_2} = \frac{-2 +2 \sqrt{3}i}{4 + 4i} = \frac{2( - 1 + \sqrt{3}i)}{4(1 + i)}= \frac{ - 1 + \sqrt{3}i}{2(1 + i)} = \frac{( - 1 + \sqrt{3}i)(1 - i)}{2(1 + i)(1 - i)} = \frac{(-1 \times 1 + \sqrt{3} \times 1) + ( \sqrt{3} \times 1 - ( - 1) \times 1)i}{2(1^{2} + {1}^{2})} = \frac{( - 1 + \sqrt{3}) + (\sqrt{3} + 1)i}{2 \times 2} = \frac{ \sqrt{3} - 1}{4} + \frac{ \sqrt{3} + 1}{4}i [/tex]
