Unit 1: Complex Numbers – Exercise 1.2

(i) \((2 + 5\iota) + (3 - z\iota)\)
\[ \begin{aligned} (2 + 5\iota) + (3 - z\iota) &= 2 + 5\iota + 3 - z\iota \\ &= (2+3) + (5\iota - z\iota) \\ &= 5 + (5 - z)\iota \end{aligned} \]

(ii) \((16 - 3\iota) + (9 + 2\iota)\)
\[ \begin{aligned} (16 - 3\iota) + (9 + 2\iota) &= 16 - 3\iota + 9 + 2\iota \\ &= (16+9) + (-3\iota + 2\iota) \\ &= 25 - \iota \end{aligned} \]

(iii) \((9 - 2\iota) - (7 - 3\iota)\)
\[ \begin{aligned} (9 - 2\iota) - (7 - 3\iota) &= 9 - 2\iota - 7 + 3\iota \\ &= (9-7) + (-2\iota + 3\iota) \\ &= 2 + \iota \end{aligned} \]

(iv) \((11 + 9\iota) - (9 - 7\iota)\)
\[ \begin{aligned} (11 + 9\iota) - (9 - 7\iota) &= 11 + 9\iota - 9 + 7\iota \\ &= (11-9) + (9\iota + 7\iota) \\ &= 2 + 16\iota \end{aligned} \]

(v) \((3 + 4\iota)(2 - 3\iota)\)
\[ \begin{aligned} (3 + 4\iota)(2 - 3\iota) &= 3(2 - 3\iota) + 4\iota(2 - 3\iota) \\ &= 6 - 9\iota + 8\iota - 12\iota^2 \\ &= 6 - \iota - 12(-1) \\ &= 6 - \iota + 12 \\ &= 18 - \iota \end{aligned} \]

(vi) \((5 - 2\iota)(3 - 4\iota)\)
\[ \begin{aligned} (5 - 2\iota)(3 - 4\iota) &= 5(3 - 4\iota) - 2\iota(3 - 4\iota) \\ &= 15 - 20\iota - 6\iota + 8\iota^2 \\ &= 15 - 26\iota + 8(-1) \\ &= 15 - 26\iota - 8 \\ &= 7 - 26\iota \end{aligned} \]

(vii) \((3 - 5\iota) \div (2 - 4\iota)\)
\[ \begin{aligned} \frac{3 - 5\iota}{2 - 4\iota} &= \frac{3 - 5\iota}{2 - 4\iota} \times \frac{2 + 4\iota}{2 + 4\iota} \\ &= \frac{3(2+4\iota) - 5\iota(2+4\iota)}{(2)^2 - (4\iota)^2} \\ &= \frac{6 + 12\iota - 10\iota - 20\iota^2}{4 - 16\iota^2} \\ &= \frac{6 + 2\iota - 20(-1)}{4 - 16(-1)} \\ &= \frac{6 + 2\iota + 20}{4 + 16} \\ &= \frac{26 + 2\iota}{20} = \frac{13}{10} + \frac{1}{10}\iota \end{aligned} \]

(viii) \((5 + 2\iota) \div (6 - 3\iota)\)
\[ \begin{aligned} \frac{5 + 2\iota}{6 - 3\iota} &= \frac{5+2\iota}{6-3\iota} \times \frac{6+3\iota}{6+3\iota} \\ &= \frac{(5+2\iota)(6+3\iota)}{36 - 9\iota^2} \\ &= \frac{30 + 15\iota + 12\iota + 6\iota^2}{36 - 9(-1)} \\ &= \frac{30 + 27\iota + 6(-1)}{36 + 9} \\ &= \frac{30 - 6 + 27\iota}{45} = \frac{24 + 27\iota}{45} = \frac{8}{15} + \frac{3}{5}\iota \end{aligned} \]

(i) \(3 + 2\iota\)   →   Additive inverse: \(-3 - 2\iota\)

(ii) \(4 - 3\iota\)   →   Additive inverse: \(-4 + 3\iota\)

(iii) \(5 - 7\iota\)   →   Additive inverse: \(-5 + 7\iota\)

(iv) \(-\frac{2}{3} + \frac{5}{4}\iota\)   →   Additive inverse: \(\frac{2}{3} - \frac{5}{4}\iota\)

(i) \(4 + 5\iota\)
\[ \begin{aligned} z^{-1} &= \frac{1}{4 + 5\iota} \times \frac{4 - 5\iota}{4 - 5\iota} \\ &= \frac{4 - 5\iota}{16 - 25\iota^2} = \frac{4 - 5\iota}{16 - 25(-1)} = \frac{4 - 5\iota}{41} = \frac{4}{41} - \frac{5}{41}\iota \end{aligned} \]

(ii) \(6 + 2\iota\)
\[ \begin{aligned} z^{-1} &= \frac{1}{6 + 2\iota} \times \frac{6 - 2\iota}{6 - 2\iota} \\ &= \frac{6 - 2\iota}{36 - 4\iota^2} = \frac{6 - 2\iota}{36 - 4(-1)} = \frac{6 - 2\iota}{40} = \frac{3}{20} - \frac{1}{20}\iota \end{aligned} \]

(iii) \(7 - 3\iota\)
\[ \begin{aligned} z^{-1} &= \frac{1}{7 - 3\iota} \times \frac{7 + 3\iota}{7 + 3\iota} \\ &= \frac{7 + 3\iota}{49 - 9\iota^2} = \frac{7 + 3\iota}{49 - 9(-1)} = \frac{7 + 3\iota}{58} = \frac{7}{58} + \frac{3}{58}\iota \end{aligned} \]

(iv) \(\sqrt{5} - 4\iota\)
\[ \begin{aligned} z^{-1} &= \frac{1}{\sqrt{5} - 4\iota} \times \frac{\sqrt{5} + 4\iota}{\sqrt{5} + 4\iota} \\ &= \frac{\sqrt{5} + 4\iota}{5 - 16\iota^2} = \frac{\sqrt{5} + 4\iota}{5 - 16(-1)} = \frac{\sqrt{5} + 4\iota}{21} = \frac{\sqrt{5}}{21} + \frac{4}{21}\iota \end{aligned} \]

(i) \(z_1 + z_2 = z_2 + z_1\)
\[ \begin{aligned} \text{L.H.S} &= (2+5\iota) + (1-3\iota) = 3 + 2\iota \\ \text{R.H.S} &= (1-3\iota) + (2+5\iota) = 3 + 2\iota \quad \Rightarrow \text{verified} \end{aligned} \]

(ii) \(z_1 z_2 = z_2 z_1\)
\[ \begin{aligned} \text{L.H.S} &= (2+5\iota)(1-3\iota) = 2 -6\iota +5\iota -15\iota^2 = 2 - \iota +15 = 17 - \iota \\ \text{R.H.S} &= (1-3\iota)(2+5\iota) = 2 +5\iota -6\iota -15\iota^2 = 2 - \iota +15 = 17 - \iota \end{aligned} \]

(iii) \((z_1 + z_2) + z_3 = z_1 + (z_2 + z_3)\)
\[ \begin{aligned} \text{L.H.S} &= [(2+5\iota)+(1-3\iota)] + (2+\iota) = (3+2\iota)+(2+\iota) = 5 + 3\iota \\ \text{R.H.S} &= (2+5\iota) + [(1-3\iota)+(2+\iota)] = (2+5\iota)+(3-2\iota) = 5 + 3\iota \end{aligned} \]

(iv) \((z_1 z_2) z_3 = z_1 (z_2 z_3)\)
\[ \begin{aligned} z_1z_2 &= 17 - \iota, \quad z_2z_3 = (1-3\iota)(2+\iota) = 2+\iota -6\iota -3\iota^2 = 2 -5\iota +3 = 5 -5\iota \\ \text{L.H.S} &= (17 - \iota)(2+\iota) = 34 + 17\iota -2\iota -\iota^2 = 34 + 15\iota +1 = 35 + 15\iota \\ \text{R.H.S} &= (2+5\iota)(5 - 5\iota) = 10 -10\iota +25\iota -25\iota^2 = 10 +15\iota +25 = 35 + 15\iota \end{aligned} \]

(v) \(z_1 + (-z_1) = (-z_1) + z_1 = 0\)
\[ \begin{aligned} z_1 = 2+5\iota,\; -z_1 = -2-5\iota \\ z_1+(-z_1) = (2-2)+(5\iota-5\iota)=0, \quad (-z_1)+z_1 = 0 \end{aligned} \]

\[ \begin{aligned} \frac{(1+\iota)^2}{2-\iota} &= \frac{1 + \iota^2 + 2\iota}{2-\iota} = \frac{1 - 1 + 2\iota}{2-\iota} = \frac{2\iota}{2-\iota} \\ &= \frac{2\iota}{2-\iota} \times \frac{2+\iota}{2+\iota} = \frac{2\iota(2+\iota)}{4 - \iota^2} = \frac{4\iota + 2\iota^2}{4 - (-1)} = \frac{4\iota + 2(-1)}{5} \\ &= \frac{-2 + 4\iota}{5} = -\frac{2}{5} + \frac{4}{5}\iota \end{aligned} \] Equating real and imaginary parts: \(x = -\frac{2}{5},\; y = \frac{4}{5}\).

\[ \begin{aligned} (2x + \iota y)(1 - \iota) &= 4 + 2\iota \\ 2x + \iota y &= \frac{4+2\iota}{1-\iota} = \frac{4+2\iota}{1-\iota} \times \frac{1+\iota}{1+\iota} \\ &= \frac{(4+2\iota)(1+\iota)}{1 - \iota^2} = \frac{4+4\iota+2\iota+2\iota^2}{1 - (-1)} = \frac{4 + 6\iota + 2(-1)}{2} \\ &= \frac{4 - 2 + 6\iota}{2} = \frac{2 + 6\iota}{2} = 1 + 3\iota \end{aligned} \] Therefore \(2x = 1 \Rightarrow x = \frac{1}{2}\), and \(y = 3\).

\[ \begin{aligned} (a + b\iota)(2 - \iota) &= 2a - a\iota + 2b\iota - b\iota^2 = 2a - a\iota + 2b\iota + b \\ &= (2a + b) + (-a + 2b)\iota = 6 + 5\iota \end{aligned} \] Equating real and imaginary parts: \[ \begin{cases} 2a + b = 6 \\ -a + 2b = 5 \end{cases} \] Solving: multiply first by 2: \(4a + 2b = 12\), subtract second: \((4a+2b) - (-a+2b) = 12-5 \Rightarrow 5a = 7 \Rightarrow a = \frac{7}{5}\). Then \(2(\frac{7}{5}) + b = 6 \Rightarrow \frac{14}{5} + b = 6 \Rightarrow b = 6 - \frac{14}{5} = \frac{30-14}{5} = \frac{16}{5}\). Thus \(a = \frac{7}{5},\; b = \frac{16}{5}\).