Question

Sa se calculeze :
urgent !!!!
Vezi in poza :

Answer 1

Explicație pas cu pas:

a)

$(1 + i) {(1 - i)}^{ - 1} = \dfrac{1 + i}{1 - i} = \dfrac{ {(1 + i)}^{2} }{(1 - i)(1 + i)} = \\ = \dfrac{1 + 2i - 1}{1 + 1} = \dfrac{2i}{2} = \bf i$

b)

$- 6i {(1 + 3i)}^{ - 1} = \dfrac{ - 6i}{1 + 3i} = \dfrac{ - 6i(1 - 3i)}{(1 + 3i)(1 - 3i)} = \\ = \dfrac{ - 6i - 18}{1 + 9} = \dfrac{ - 6(i + 3)}{10} = - \dfrac{3(3 + i)}{5} = - \dfrac{9}{5} - \dfrac{3}{5}i$

c)

$\dfrac{1 - i \sqrt{3} }{1 + i \sqrt{3}} = \dfrac{ {(1 - i \sqrt{3})}^{2} }{(1 + i \sqrt{3})(1 - i \sqrt{3})} = \dfrac{1 - 2i \sqrt{3} - 3}{1 + 3} = \\ = \dfrac{ - 2 - 2i \sqrt{3}}{4} = - \dfrac{2(1 + i \sqrt{3})}{4} = - \dfrac{1 + i \sqrt{3}}{2} = - \dfrac{1}{2} - \dfrac{ \sqrt{3} }{2}i$

d)

$\dfrac{ \sqrt{3} + \sqrt{2}i}{\sqrt{3} - \sqrt{2}i} = \dfrac{ {(\sqrt{3} + \sqrt{2}i)}^{2} }{(\sqrt{3} - \sqrt{2}i)(\sqrt{3} + \sqrt{2}i)} = \\ = \dfrac{3 + 2 \sqrt{6}i - 2}{3 + 2} = \dfrac{1 + 2 \sqrt{6}i}{5} = \dfrac{1}{5} + \dfrac{2 \sqrt{6}}{5}i$