Unit 1: Complex Numbers – Exercise 1.4

(i) \((8 - 3\iota)^2\) \[ \begin{aligned} z &= (8 - 3\iota)^2 \\ &= 8^2 + (3\iota)^2 - 2(8)(3\iota) \\ &= 64 + 9\iota^2 - 48\iota \\ &= 64 + 9(-1) - 48\iota \\ &= 64 - 9 - 48\iota \\ &= 55 - 48\iota \end{aligned} \] Hence \(\operatorname{Re}(z)=55,\quad \operatorname{Im}(z)=-48\).

(ii) \((5 + 3\iota)^{-1}\) \[ \begin{aligned} z &= (5 + 3\iota)^{-1} = \frac{1}{5 + 3\iota} \\ &= \frac{1}{5 + 3\iota} \times \frac{5 - 3\iota}{5 - 3\iota} \\ &= \frac{5 - 3\iota}{(5)^2 - (3\iota)^2} \\ &= \frac{5 - 3\iota}{25 - 9\iota^2} = \frac{5 - 3\iota}{25 - 9(-1)} \\ &= \frac{5 - 3\iota}{25 + 9} = \frac{5 - 3\iota}{34} \\ &= \frac{5}{34} - \frac{3}{34}\iota \end{aligned} \] Hence \(\operatorname{Re}(z)=\dfrac{5}{34},\quad \operatorname{Im}(z)=-\dfrac{3}{34}\).

(iii) \((4 - 5\iota)^{-1}\) \[ \begin{aligned} z &= \frac{1}{4 - 5\iota} = \frac{1}{4 - 5\iota} \times \frac{4 + 5\iota}{4 + 5\iota} \\ &= \frac{4 + 5\iota}{4^2 - (5\iota)^2} = \frac{4 + 5\iota}{16 - 25\iota^2} \\ &= \frac{4 + 5\iota}{16 - 25(-1)} = \frac{4 + 5\iota}{16 + 25} \\ &= \frac{4 + 5\iota}{41} = \frac{4}{41} + \frac{5}{41}\iota \end{aligned} \] Hence \(\operatorname{Re}(z)=\dfrac{4}{41},\quad \operatorname{Im}(z)=\dfrac{5}{41}\).

(iv) \((4 - 3\iota)^{-2}\) \[ \begin{aligned} z &= \frac{1}{(4 - 3\iota)^2} = \frac{1}{16 + 9\iota^2 - 24\iota} \\ &= \frac{1}{16 + 9(-1) - 24\iota} = \frac{1}{7 - 24\iota} \\ &= \frac{1}{7 - 24\iota} \times \frac{7 + 24\iota}{7 + 24\iota} = \frac{7 + 24\iota}{49 - (24\iota)^2} \\ &= \frac{7 + 24\iota}{49 - 576\iota^2} = \frac{7 + 24\iota}{49 - 576(-1)} \\ &= \frac{7 + 24\iota}{49 + 576} = \frac{7 + 24\iota}{625} = \frac{7}{625} + \frac{24}{625}\iota \end{aligned} \] Hence \(\operatorname{Re}(z)=\dfrac{7}{625},\quad \operatorname{Im}(z)=\dfrac{24}{625}\).

(v) \(\dfrac{3 + 2\iota}{4 + 3\iota}\) \[ \begin{aligned} z &= \frac{3 + 2\iota}{4 + 3\iota} \times \frac{4 - 3\iota}{4 - 3\iota} \\ &= \frac{(3)(4) + 3(-3\iota) + 2\iota(4) + 2\iota(-3\iota)}{(4)^2 - (3\iota)^2} \\ &= \frac{12 - 9\iota + 8\iota - 6\iota^2}{16 - 9\iota^2} \\ &= \frac{12 - \iota -6(-1)}{16 - 9(-1)} = \frac{12 - \iota + 6}{16 + 9} \\ &= \frac{18 - \iota}{25} = \frac{18}{25} - \frac{1}{25}\iota \end{aligned} \] Hence \(\operatorname{Re}(z)=\dfrac{18}{25},\quad \operatorname{Im}(z)=-\dfrac{1}{25}\).

(vi) \(\left( \dfrac{2-\iota}{2+\iota} \right)^{-2}\) \[ \begin{aligned} z &= \left( \frac{2+\iota}{2-\iota} \right)^{2} = \frac{(2+\iota)^2}{(2-\iota)^2} \\ &= \frac{4 + \iota^2 + 4\iota}{4 + \iota^2 - 4\iota} = \frac{4 - 1 + 4\iota}{4 - 1 - 4\iota} \\ &= \frac{3 + 4\iota}{3 - 4\iota} \times \frac{3 + 4\iota}{3 + 4\iota} = \frac{9 + 12\iota + 12\iota + 16\iota^2}{9 - 16\iota^2} \\ &= \frac{9 + 24\iota + 16(-1)}{9 - 16(-1)} = \frac{9 + 24\iota - 16}{9 + 16} \\ &= \frac{-7 + 24\iota}{25} = -\frac{7}{25} + \frac{24}{25}\iota \end{aligned} \] Hence \(\operatorname{Re}(z)=-\dfrac{7}{25},\quad \operatorname{Im}(z)=\dfrac{24}{25}\).

(vii) \(\left( \dfrac{1-2\iota}{1+\iota} \right)^{2}\) \[ \begin{aligned} z &= \frac{(1-2\iota)^2}{(1+\iota)^2} = \frac{1 + 4\iota^2 - 4\iota}{1 + \iota^2 + 2\iota} \\ &= \frac{1 + 4(-1) - 4\iota}{1 + (-1) + 2\iota} = \frac{1 - 4 - 4\iota}{2\iota} \\ &= \frac{-3 - 4\iota}{2\iota} \times \frac{2\iota}{2\iota} = \frac{-6\iota - 8\iota^2}{4\iota^2} \\ &= \frac{-6\iota - 8(-1)}{4(-1)} = \frac{-6\iota + 8}{-4} \\ &= \frac{8 - 6\iota}{-4} = -2 + \frac{3}{2}\iota \end{aligned} \] Hence \(\operatorname{Re}(z)=-2,\quad \operatorname{Im}(z)=\dfrac{3}{2}\).

(i) \[ \begin{cases} 3z + (2+\iota)w = 11 - \iota \\ (2-\iota)z - w = -1 + \iota \end{cases} \] \[ \begin{aligned} &\text{From }(2-\iota)z - w = -1+\iota \implies w = (2-\iota)z + 1 - \iota. \\ &\text{Substitute into first equation:}\\ 3z + (2+\iota)\big[(2-\iota)z + 1 - \iota\big] &= 11 - \iota \\ 3z + (2+\iota)(2-\iota)z + (2+\iota)(1 - \iota) &= 11 - \iota \\ 3z + (4 - \iota^2)z + \big[2 - 2\iota + \iota - \iota^2\big] &= 11 - \iota \\ 3z + (4 + 1)z + \big[2 - \iota - (-1)\big] &= 11 - \iota \\ 3z + 5z + (2 - \iota + 1) &= 11 - \iota \\ 8z + (3 - \iota) &= 11 - \iota \\ 8z &= 8 \implies z = 1. \end{aligned} \] Then \(w = (2-\iota)(1) + 1 - \iota = 2 - \iota + 1 - \iota = 3 - 2\iota\).
\(\boxed{z=1,\; w=3-2\iota}\).

(ii) \[ \begin{cases} 2z + (3+\iota)w = 9 - \iota \\ -\iota z - \iota w = -1 + \iota \end{cases} \] \[ \begin{aligned} &\text{From } -\iota(z+w) = -1+\iota \;\Rightarrow\; z+w = \frac{-1+\iota}{-\iota} = \frac{-1+\iota}{-\iota}\cdot \frac{\iota}{\iota} \\ &= \frac{-\iota + \iota^2}{-\iota^2} = \frac{-\iota -1}{1} = -1 - \iota \\ &\Rightarrow z = -w -1 - \iota. \end{aligned} \] Substitute into first equation: \[ 2(-w -1 - \iota) + (3+\iota)w = 9 - \iota \\ -2w -2 -2\iota + 3w + \iota w = 9 - \iota \\ w + \iota w -2 -2\iota = 9 - \iota \\ w(1+\iota) = 9 - \iota + 2 + 2\iota = 11 + \iota. \] \[ w = \frac{11+\iota}{1+\iota} = \frac{11+\iota}{1+\iota}\cdot\frac{1-\iota}{1-\iota} = \frac{11-11\iota + \iota -\iota^2}{1 - \iota^2} = \frac{11 -10\iota +1}{1+1} = \frac{12 -10\iota}{2} = 6 -5\iota. \] Then \(z = -(6-5\iota) -1 - \iota = -6+5\iota -1 -\iota = -7 + 4\iota\).
\(\boxed{z = -7+4\iota,\; w = 6-5\iota}\).

(iii) \[ \begin{cases} z - 4w = 3\iota \\ 2z + 3w = 11 - 5\iota \end{cases} \] From first: \(z = 3\iota + 4w\). Substitute into second: \[ 2(3\iota + 4w) + 3w = 11 - 5\iota \implies 6\iota + 8w + 3w = 11 - 5\iota \implies 11w + 6\iota = 11 - 5\iota. \] \[ 11w = 11 - 5\iota - 6\iota = 11 - 11\iota \implies w = 1 - \iota. \] Then \(z = 3\iota + 4(1 - \iota) = 3\iota + 4 - 4\iota = 4 - \iota\).
\(\boxed{z = 4 - \iota,\; w = 1 - \iota}\).

(iv) \[ \begin{cases} z + w = 3\iota \\ 2z + 3w = 2 \end{cases} \] From first: \(z = 3\iota - w\). Insert into second: \[ 2(3\iota - w) + 3w = 2 \implies 6\iota - 2w + 3w = 2 \implies 6\iota + w = 2 \implies w = 2 - 6\iota. \] Then \(z = 3\iota - (2 - 6\iota) = 3\iota - 2 + 6\iota = -2 + 9\iota\).
\(\boxed{z = -2 + 9\iota,\; w = 2 - 6\iota}\).

(v) \[ \begin{cases} 2z + (3+\iota)w = 1 \\ -z - (1-\iota)w = 2 \end{cases} \] From second: \(-z = 2 + (1-\iota)w \;\Rightarrow\; z = -2 - (1-\iota)w = -2 - w + \iota w\). Substitute into first: \[ 2\big[-2 - w + \iota w\big] + (3+\iota)w = 1 \\ -4 - 2w + 2\iota w + 3w + \iota w = 1 \\ -4 + w + 3\iota w = 1 \implies w + 3\iota w = 5 \\ w(1+3\iota) = 5 \implies w = \frac{5}{1+3\iota}. \] \[ w = \frac{5}{1+3\iota} \cdot \frac{1-3\iota}{1-3\iota} = \frac{5 - 15\iota}{1 - 9\iota^2} = \frac{5 - 15\iota}{1 + 9} = \frac{5 - 15\iota}{10} = \frac{1}{2} - \frac{3}{2}\iota. \] Then \(z = -2 - \left(\frac12 - \frac32\iota\right) + \iota\left(\frac12 - \frac32\iota\right)\): \[ z = -2 - \frac12 + \frac32\iota + \frac12\iota - \frac32\iota^2 = -\frac52 + 2\iota - \frac32(-1) = -\frac52 + 2\iota + \frac32 = -1 + 2\iota. \] \(\boxed{z = -1 + 2\iota,\; w = \dfrac{1}{2} - \dfrac{3}{2}\iota}\).