Question

solve 7th and 9th please​

Answer 1

$\bf{ Q. \: (vii) \: \: \: {(1 + i)}^{ - 3} } \\ \\ \red{\bf{solution} \implies} \\ \\ \implies \: {(1 + i)}^{ - 3} \\ \\ \implies \: \frac{1}{ {(1 + i)}^{3} } \\ \\ \green{\bf{using \: identity \: {(a + b)}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b)}} \\ \\ \implies \: \frac{1}{1 + {i}^{3} + 3i(1 + i)} \\ \because \: {i}^{3} = - i \: \: {i}^{2} = - 1 \\ \\ \implies \: \frac{1}{1 - i + 3i + 3 {i}^{2} } \\ \\ \implies \frac{1}{1 + 2i - 3} \\ \\ \implies \: \boxed{ \frac{1}{2(i - 1)} }$

$\bf{Q. \: (ix) \: ( - \sqrt{5} + 2 \sqrt{ - 4)} + (1 - \sqrt{ - 9)} + (2 + 3i)(2 - 3i)} \\ \\ \red{ \bf{solution}\implies} \\ \implies \bf{ - \sqrt{5} + 4i + 1 - 3i + (4 - 9 {i}^{2} )} \\ \\ \implies \: 1 - \sqrt{5} + i + 4 + 9 \\ \\ \implies \: \boxed{14 - \sqrt{5} + i} \because{i}^{2}=-1$