Unit 1: Complex Numbers – Exercise 1.3

(i) \( 4 + 3\iota \)
\[ \begin{aligned} |4+3\iota| &= \sqrt{(4)^{2}+(3)^{2}} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5 \end{aligned} \]

(ii) \( -5 - 4\iota \)
\[ \begin{aligned} |-5-4\iota| &= \sqrt{(-5)^{2}+(-4)^{2}} \\ &= \sqrt{25+16} \\ &= \sqrt{41} \end{aligned} \]

(iii) \( \frac{3}{5} - \frac{4}{5}\iota \)
\[ \begin{aligned} \left|\frac{3}{5}-\frac{4}{5}\iota\right| &= \sqrt{\left(\frac{3}{5}\right)^{2}+\left(-\frac{4}{5}\right)^{2}} \\ &= \sqrt{2frac{9}{25}+\frac{16}{25}} \\ &= \sqrt{\frac{9+16}{25}} \\ &= \sqrt{\frac{25}{25}} \\ &= \sqrt{1} \\ &= 1 \end{aligned} \]

(iv) \( -\sqrt{2} - \sqrt{3}\iota \)
\[ \begin{aligned} |-\sqrt{2}-\sqrt{3}\iota| &= \sqrt{(-\sqrt{2})^{2}+(-\sqrt{3})^{2}} \\ &= \sqrt{2+3} \\ &= \sqrt{5} \end{aligned} \]

(i) \( \overline{z_{1}+z_{2}} = \overline{z_{1}}+\overline{z_{2}} \)

L.H.S
\[ \begin{aligned} z_{1}+z_{2} &= (2+7\iota)+(4-3\iota) \\ &= 2+7\iota+4-3\iota \\ &= 2+4+7\iota-3\iota \\ &= 6+4\iota \end{aligned} \] \[ \text{L.H.S} = \overline{z_{1}+z_{2}} = \overline{6+4\iota} = 6-4\iota \]
R.H.S
\( z_{1} = 2+7\iota \implies \overline{z_{1}} = 2-7\iota \)
\( z_{2} = 4-3\iota \implies \overline{z_{2}} = 4+3\iota \)

\[ \begin{aligned} \text{R.H.S} = \overline{z_{1}}+\overline{z_{2}} &= (2-7\iota)+(4+3\iota) \\ &= 2-7\iota+4+3\iota \\ &= 2+4-7\iota+3\iota \\ &= 6-4\iota \end{aligned} \]
Hence, \( \text{L.H.S} = \text{R.H.S} \)

(ii) \( \overline{z_{1}z_{2}} = \overline{z_{1}}\overline{z_{2}} \)

L.H.S
\[ \begin{aligned} z_{1}z_{2} &= (2+7\iota)(4-3\iota) \\ &= 2(4-3\iota)+7\iota(4-3\iota) \\ &= 8-6\iota+28\iota-21\iota^{2} \\ &= 8+22\iota-21(-1) \\ &= 8+22\iota+21 \\ &= 29+22\iota \end{aligned} \] \[ \text{L.H.S} = \overline{z_{1}z_{2}} = \overline{29+22\iota} = 29-22\iota \]
R.H.S
\( z_{1} = 2+7\iota \implies \overline{z_{1}} = 2-7\iota \)
\( z_{2} = 4-3\iota \implies \overline{z_{2}} = 4+3\iota \)

\[ \begin{aligned} \text{R.H.S} = \overline{z_{1}}\overline{z_{2}} &= (2-7\iota)(4+3\iota) \\ &= 2(4+3\iota)-7\iota(4+3\iota) \\ &= 8+6\iota-28\iota-21\iota^{2} \\ &= 8-22\iota-21(-1) \\ &= 8-22\iota+21 \\ &= 29-22\iota \end{aligned} \]
Hence, \( \text{L.H.S} = \text{R.H.S} \)

(iii) \( \overline{\left(\frac{z_{1}}{z_{2}}\right)} = \frac{\overline{z_{1}}}{\overline{z_{2}}} \)

L.H.S
\[ \begin{aligned} \frac{z_{1}}{z_{2}} &= \frac{2+7\iota}{4-3\iota} \\ &= \frac{2+7\iota}{4-3\iota} \times \frac{4+3\iota}{4+3\iota} \\ &= \frac{2(4+3\iota)+7\iota(4+3\iota)}{(4)^{2}-(3\iota)^{2}} \\ &= \frac{8+6\iota+28\iota+21\iota^{2}}{16-9\iota^{2}} \\ &= \frac{8+34\iota+21(-1)}{16-9(-1)} \\ &= \frac{8+34\iota-21}{16+9} \\ &= \frac{-13+34\iota}{25} \\ &= -\frac{13}{25}+\frac{34}{25}\iota \end{aligned} \] \[ \text{L.H.S} = \overline{\left(\frac{z_{1}}{z_{2}}\right)} = \overline{-\frac{13}{25}+\frac{34}{25}\iota} = -\frac{13}{25}-\frac{34}{25}\iota \]
R.H.S
\( z_{1} = 2+7\iota \implies \overline{z_{1}} = 2-7\iota \)
\( z_{2} = 4-3\iota \implies \overline{z_{2}} = 4+3\iota \)

\[ \begin{aligned} \text{R.H.S} = \frac{\overline{z_{1}}}{\overline{z_{2}}} &= \frac{2-7\iota}{4+3\iota} \\ &= \frac{2-7\iota}{4+3\iota} \times \frac{4-3\iota}{4-3\iota} \\ &= \frac{2(4-3\iota)-7\iota(4-3\iota)}{(4)^{2}-(3\iota)^{2}} \\ &= \frac{8-6\iota-28\iota+21\iota^{2}}{16-9\iota^{2}} \\ &= \frac{8-34\iota+21(-1)}{16-9(-1)} \\ &= \frac{8-34\iota-21}{16+9} \\ &= \frac{-13-34\iota}{25} \\ &= -\frac{13}{25}-\frac{34}{25}\iota \end{aligned} \]
Hence, \( \text{L.H.S} = \text{R.H.S} \)

(i) \( \overline{\overline{z}} = z \)
\[ \begin{aligned} z &= 5-2\iota \\ \overline{z} &= 5+2\iota \\ \overline{\overline{z}} &= 5-2\iota \end{aligned} \] Since \( 5-2\iota = z \), hence proved \( \overline{\overline{z}} = z \).

(ii) \( |z| = |\overline{z}| \)
\[ \begin{aligned} z = 5-2\iota \implies |z| &= \sqrt{(5)^{2}+(-2)^{2}} \\ &= \sqrt{25+4} \\ &= \sqrt{29} \end{aligned} \] \[ \begin{aligned} \overline{z} = 5+2\iota \implies |\overline{z}| &= \sqrt{(5)^{2}+(2)^{2}} \\ &= \sqrt{25+4} \\ &= \sqrt{29} \end{aligned} \] Hence proved \( |z| = |\overline{z}| \).

(iii) \( |z| = |-\overline{z}| \)
\[ \begin{aligned} z = 5-2\iota \implies |z| &= \sqrt{(5)^{2}+(-2)^{2}} \\ &= \sqrt{25+4} \\ &= \sqrt{29} \end{aligned} \] \[ \begin{aligned} \overline{z} = 5+2\iota \implies -\overline{z} &= -5-2\iota \\ |-\overline{z}| &= \sqrt{(-5)^{2}+(-2)^{2}} \\ &= \sqrt{25+4} \\ &= \sqrt{29} \end{aligned} \] Hence proved \( |z| = |-\overline{z}| \).

(iv) \( z\overline{z} = |z|^{2} \)
L.H.S
\[ \begin{aligned} z\overline{z} &= (5-2\iota)(5+2\iota) \\ &= (5)^{2}-(2\iota)^{2} \\ &= 25-4\iota^{2} \\ &= 25-4(-1) \\ &= 25+4 \\ &= 29 \end{aligned} \] R.H.S
\[ \begin{aligned} |z| &= \sqrt{(5)^{2}+(-2)^{2}} \\ &= \sqrt{25+4} \\ &= \sqrt{29} \end{aligned} \] \[ |z|^{2} = (\sqrt{29})^{2} = 29 \] Hence proved \( z\overline{z} = |z|^{2} \).

\[ \begin{aligned} z = 4-3\iota \implies |z| &= \sqrt{(4)^{2}+(-3)^{2}} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5 \quad \dots \text{(i)} \end{aligned} \]
\[ \begin{aligned} -z = -4+3\iota \implies |-z| &= \sqrt{(-4)^{2}+(3)^{2}} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5 \quad \dots \text{(ii)} \end{aligned} \]
\[ \begin{aligned} \overline{z} = 4+3\iota \implies \overline{\overline{z}} &= 4-3\iota \\ |\overline{\overline{z}}| &= \sqrt{(4)^{2}+(-3)^{2}} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5 \quad \dots \text{(iii)} \end{aligned} \]
\[ \begin{aligned} -\overline{z} = -4-3\iota \implies |-\overline{z}| &= \sqrt{(-4)^{2}+(-3)^{2}} \\ &= \sqrt{16+9} \\ &= \sqrt{25} \\ &= 5 \quad \dots \text{(iv)} \end{aligned} \]
From equation (i), (ii), (iii) and (iv):
\[ |z|=|-z|=|\overline{\overline{z}}|=|-\overline{z}| \]

First, find \( z_{1}z_{2} \):
\[ \begin{aligned} z_{1}z_{2} &= (2+3\iota)(-1+\iota) \\ &= 2(-1+\iota)+3\iota(-1+\iota) \\ &= -2+2\iota-3\iota+3\iota^{2} \\ &= -2-\iota+3(-1) \\ &= -2-\iota-3 \\ &= -5-\iota \end{aligned} \]
Now, evaluate parts (i) and (ii):
(i) \( \text{Re}(z_{1}z_{2}) = -5 \)
(ii) \( \text{Im}(z_{1}z_{2}) = -1 \)