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Equations Reducible to a Pair of Linear Equations in Two Variables

Express the f...

Question

Express the following complex numbers in the standard form a + i b:
(i) $(1 + i) (1 + 2 i)$
(ii) $\frac{3 + 2 i}{- 2 + i}$
(iii) $\frac{1}{(2 + i)^{2}}$
(iv) $\frac{1 - i}{1 + i}$
(v) $\frac{(2 + i)^{3}}{2 + 3 i}$
(vi) $\frac{(1 + i) (1 + \sqrt{3} i)}{1 - i}$
(vii) $\frac{2 + 3 i}{4 + 5 i}$
(viii) $\frac{(1 - i)^{3}}{1 - i^{3}}$
(ix) $(1 + 2 i)^{- 3}$
(x) $\frac{3 - 4 i}{(4 - 2 i) (1 + i)}$
(xi) $(\frac{1}{1 - 4 i} - \frac{2}{1 + i}) (\frac{1 - 4 i}{5 + i})$
(xii) $\frac{5 + \sqrt{2} i}{1 - 2 \sqrt{i}}$

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Solution

$(i) (1 + i) (1 + 2 i) = 1 + 2 i + i + 2 i^{2} = 1 + 3 i - 2 (∵ i^{2} = - 1) = - 1 + 3 i (ii) \frac{3 + 2 i}{- 2 + i} = \frac{3 + 2 i}{- 2 + i} \times \frac{- 2 - i}{- 2 - i} = \frac{- 6 - 3 i - 4 i - 2 i^{2}}{4 - i^{2}} (∵ i^{2} = - 1) = \frac{- 6 - 7 i + 2}{4 + 1} = \frac{- 4 - 7 i}{5} = \frac{- 4}{5} - \frac{7}{5} i (iii) \frac{1}{{(2 + i)}^{2}} = \frac{1}{4 + i^{2} + 4 i} (∵ i^{2} = - 1) = \frac{1}{3 + 4 i} = \frac{1}{3 + 4 i} \times \frac{3 - 4 i}{3 - 4 i} = \frac{3 - 4 i}{9 - 16 i^{2}} = \frac{3 - 4 i}{9 + 16} = \frac{3}{25} - \frac{4}{25} i$

$(iv) \frac{1 - i}{1 + i} = \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{1 + i^{2} - 2 i}{1 - i^{2}} (∵ i^{2} = - 1) = \frac{- 2 i}{2} = - i (v) \frac{{(2 + i)}^{3}}{2 + 3 i} = \frac{(4 + i^{2} + 4 i) (2 + i)}{2 + 3 i} (∵ i^{2} = - 1) = \frac{8 + 2 i^{2} + 8 i + 4 i + i^{3} + 4 i^{2}}{2 + 3 i} = \frac{2 + 11 i}{2 + 3 i} \times \frac{2 - 3 i}{2 - 3 i} = \frac{4 - 6 i + 22 i - 33 i^{2}}{4 - 9 i^{2}} = \frac{37 + 16 i}{4 + 9} = \frac{37}{13} + \frac{16}{13} i (vi) \frac{(1 + i) (1 + \sqrt{3 i})}{1 - i} = \frac{(1 + i) (1 + \sqrt{3 i})}{1 - i} \times \frac{1 + i}{1 + i} = \frac{(1 + \sqrt{3} i) (1 + i^{2} + 2 i)}{1 - i^{2}} (∵ i^{2} = - 1) = \frac{(1 + \sqrt{3} i) 2 i}{2} = i (1 + \sqrt{3} i) = i + \sqrt{3} i^{2} = - \sqrt{3} + i (vii) \frac{2 + 3 i}{4 + 5 i} = \frac{2 + 3 i}{4 + 5 i} \times \frac{4 - 5 i}{4 - 5 i} = \frac{8 - 10 i + 12 i - 15 i^{2}}{16 - 25 i^{2}} (∵ i^{2} = - 1) = \frac{23 + 2 i}{16 + 25} = \frac{23}{41} + \frac{2}{41} i$

$(viii) \frac{{(1 - i)}^{3}}{1 - i^{3}} = \frac{(1 + i^{2} - 2 i) (1 - i)}{1 - i^{3}} (∵ i^{2} = - 1) = \frac{- 2 i (1 - i)}{1 - i^{3}} \times \frac{1 + i^{3}}{1 + i^{3}} = \frac{- 2 i (1 + i^{3} - i - i^{4})}{1 - i^{6}} = \frac{- 2 i (1 - i - i - 1)}{1 - i^{2}} = \frac{- 2 i (- 2 i)}{2} = - 2 + 0 i$

$(ix) {(1 + 2 i)}^{- 3} = \frac{1}{{(1 + 2 i)}^{3}} = \frac{1}{1 + 8 i^{3} + 6 i + 12 i^{2}} = \frac{1}{1 - 8 i + 6 i - 12} (∵ i^{2} = - 1 & i^{3} = - i) = \frac{1}{- 2 i - 11} = \frac{1}{- 2 i - 11} \times \frac{- 2 i + 11}{- 2 i + 11} = \frac{- 2 i + 11}{4 i^{2} - 121} = \frac{- 2 i + 11}{- 4 - 121} = \frac{- 2 i + 11}{- 125} = - \frac{11}{125} + \frac{2 i}{125}$

$(x) \frac{3 - 4 i}{(4 - 2 i) (1 + i)} = \frac{3 - 4 i}{4 + 2 i - 2 i^{2}} (∵ i^{2} = - 1) = \frac{3 - 4 i}{6 + 2 i} = \frac{3 - 4 i}{6 + 2 i} \times \frac{6 - 2 i}{6 - 2 i} = \frac{18 - 6 i - 24 i + 8 i^{2}}{36 - 4 i^{2}} = \frac{18 - 30 i - 8}{36 + 4} = \frac{10 - 30 i}{40} = \frac{1}{4} - \frac{3}{4} i$

$(xi) (\frac{1}{1 - 4 i} - \frac{2}{1 + i}) (\frac{3 - 4 i}{5 + i}) = (\frac{1 + i - 2 + 8 i}{(1 - 4 i) (1 + i)}) (\frac{3 - 4 i}{5 + i}) = (\frac{- 1 + 9 i}{1 - 3 i + 4 i^{2}}) (\frac{3 - 4 i}{5 + i}) = (\frac{- 1 + 9 i}{5 - 3 i}) (\frac{3 - 4 i}{5 + i}) (∵ i^{2} = - 1) = \frac{- 3 + 4 i + 27 i - 36 i^{2}}{25 + 5 i - 15 i - 3 i^{2}} = \frac{33 + 31 i}{28 - 10 i} = \frac{33 + 31 i}{28 - 10 i} \times \frac{28 + 10 i}{28 + 10 i} = \frac{924 + 330 i + 868 i + 310 i^{2}}{784 - 100 i^{2}} = \frac{614 + 1198 i}{884} = \frac{307}{442} + \frac{599}{442} i$

$(xii) \frac{5 + \sqrt{2} i}{1 - \sqrt{2} i} = \frac{5 + \sqrt{2} i}{1 - \sqrt{2} i} \times \frac{1 + \sqrt{2} i}{1 + \sqrt{2} i} = \frac{5 + 5 \sqrt{2} i + \sqrt{2} i + 2 i^{2}}{1 - 2 i^{2}} = \frac{5 + 6 \sqrt{2} i - 2}{1 + 2} (∵ i^{2} = - 1) = \frac{3 + 6 \sqrt{2} i}{3} = 1 + 2 \sqrt{2} i$

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