Explicație pas cu pas:
1)
numărul z:
[tex]{z}^{2} = 4 + 6i \sqrt{5}[/tex]
substituim z = x + yi, x,y ∈ R
[tex]{(x + yi)}^{2} = 4 + 6 \sqrt{5}i[/tex]
[tex]{x}^{2} + 2xyi - {y}^{2} = 4 + 6 \sqrt{5}i \\ ({x}^{2} - {y}^{2}) + 2xyi = 4 + 6 \sqrt{5}i[/tex]
[tex]\left \{ {{ {x}^{2} - {y}^{2} = 4} \atop {2xy = 6 \sqrt{5} }} \right. \\[/tex]
[tex]x = \frac{6 \sqrt{5} }{2y} = \frac{3 \sqrt{5} }{y} \\ [/tex]
[tex]{( \frac{3 \sqrt{5}}{y} )}^{2} - {y}^{2} = 4 \\ \frac{45}{ {y}^{2} } - {y}^{2} = 4 \\ {y}^{4} + 4 {y}^{2} - 45 = 0 \\ ( {y}^{2} + 9)({y}^{2} - 5) = 0[/tex]
[tex]{y}^{2} - 5 = 0 < = > {y}^{2} = 5 \\ = > y = - \sqrt{5}; \: y = \sqrt{5} [/tex]
[tex]y = - \sqrt{5} = > x = \frac{3 \sqrt{5} }{ - \sqrt{5} } = > x = - 3 \\ [/tex]
[tex]= > z = - 3 - \sqrt{5}i[/tex]
[tex]|z| = \sqrt{ {( - 3)}^{2} + {( - \sqrt{5} )}^{2} } = \sqrt{9 + 5} = \sqrt{14} \\[/tex]
[tex]y = \sqrt{5} = > x = \frac{3 \sqrt{5} }{\sqrt{5} } = > x = 3 \\ [/tex]
[tex] = > z = 3 + \sqrt{5}i[/tex]
[tex]|z| = \sqrt{ {3}^{2} + {( \sqrt{5} )}^{2} } = \sqrt{9 + 5} = \sqrt{14} \\[/tex]
2)
[tex]z_{1} = \frac{ \sqrt{2} + i}{ \sqrt{2} - i} = \frac{ {(\sqrt{2} + i)}^{2} }{(\sqrt{2} - i)(\sqrt{2} + i)} \\ = \frac{1 + 2 \sqrt{2} i}{3} = \frac{1}{3} + \frac{2 \sqrt{2} }{3}i[/tex]
[tex]z_{2} = \frac{i}{ - 2 \sqrt{2} + 2i} = \frac{i(- 2 \sqrt{2} - 2i)}{(- 2 \sqrt{2} - 2i)(- 2 \sqrt{2} + 2i)} \\ = \frac{2 - 2 \sqrt{2}i}{12} = \frac{1 - \sqrt{2}i }{6} = \frac{1}{6} - \frac{ \sqrt{2} }{6}i [/tex]
[tex]z_{1} + z_{2} = \frac{1 + 2 \sqrt{2} i}{3} + \frac{1 - \sqrt{2}i }{6} = \frac{2 + 4 \sqrt{2}i + 1 - \sqrt{2}i}{6} \\ = \frac{3 + 3 \sqrt{2}i}{6} = \frac{1 + \sqrt{2}}{2}i = \frac{1}{2} + \frac{\sqrt{2}}{2}i[/tex]
[tex]z_{1}z_{2} = (\frac{1 + 2 \sqrt{2} i}{3})(\frac{1- \sqrt{2}i }{6}) \\ = \frac{1- \sqrt{2}i + 2 \sqrt{2}i + 4}{18} = \frac{5 + \sqrt{2}i}{18} \\ = \frac{5}{18} + \frac{ \sqrt{2}}{18}i[/tex]
[tex]z_{1}^{2} + z_{2}^{2} = {(z_{1} + z_{2})}^{2} - 2z_{1}z_{2} \\ = \left( \frac{1 + \sqrt{2}i}{2}\right) ^{2} - \frac{2\left(5 + \sqrt{2}i \right)}{18} = \frac{1 + 2 \sqrt{2}i - 2 }{4} - \frac{5 + \sqrt{2}i}{9} \\ = \frac{9( - 1 + 2 \sqrt{2}i) - 4(5 + \sqrt{2}i)}{36} = \frac{ - 9 + 18 \sqrt{2}i - 20 - 4 \sqrt{2}i}{36} \\ = \frac{ - 29 + 14 \sqrt{2}i}{36} = - \frac{29}{36} + \frac{7 \sqrt{2}}{18}i[/tex]
