# Question 10, Exercise 1.2

Solutions of Question 10 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

For $z_{1}=-3+2 i$, verify: $$\left|z_{1}\right|=\left|-z_{1}\right|=\left|\overline{z_{!}}\right|=\left|-\overline{z_{!}}\right|.$$

Solution. \begin{align} |z_1| &= \sqrt{(-3)^2 + (2)^2} \\ &= \sqrt{9 + 4} = \sqrt{13} \,\, -- (1) \end{align} Now

\begin{align} -z_1 &= -(-3 + 2i) = 3 - 2i\\ \implies |-z_1| &= \sqrt{(3)^2 + (-2)^2} \\ &= \sqrt{9 + 4} = \sqrt{13} \,\, -- (2) \end{align}

Also \begin{align} \overline{z_1} &= -3 - 2i \\ \implies |\overline{z_1}| &= \sqrt{(-3)^2 + (-2)^2} \\ & = \sqrt{9 + 4} = \sqrt{13} \,\, -- (3) \end{align}

Now \begin{align} -\overline{z_1} &= -(-3 - 2i) = 3 + 2i \\ \implies |-\overline{z_1}| &= \sqrt{(3)^2 + (2)^2} \\ & = \sqrt{9 + 4} = \sqrt{13} \,\, -- (4) \end{align} From (1), (2), (3) and (4), we have: $$\left| z_1 \right| = \left| -z_1 \right| = \left| \overline{z_1} \right| = \left| -\overline{z_1} \right| = \sqrt{13}.$$ As required.

For $z_{1}=-3+2 i$ and $z_{2}=1-3 i$ verify: $\overline{\left(\frac{z_{1}}{z_{2}}\right)}=\frac{\overline{z_{1}}}{\overline{z_{2}}}$.

Solution.

Given $z_1 = -3 + 2i, \quad z_2 = 1 - 3i$ Then \begin{align} \frac{z_1}{z_2} &= \frac{-3 + 2i}{1 - 3i}\\ &= \frac{(-3 + 2i)(1 + 3i)}{(1 - 3i)(1 + 3i)} \\ &= \frac{-3 + 2i + -9i + 6i^2}{1^2 - (3i)^2} \\ &= \frac{-9 - 7i}{1 + 9} \\ &= -\frac{9}{10} - \frac{7}{10}i. \end{align} This give \begin{align} \overline{\left( \frac{z_1}{z_2} \right)} = -\frac{9}{10} + \frac{7}{10}i. \,\, -- (i) \end{align} Now, we have \begin{align} \overline{z_1} = -3 - 2i, \quad \overline{z_2} = 1 + 3i. \end{align} Then \begin{align} \frac{\overline{z_1}}{\overline{z_2}} & = \frac{-3 - 2i}{1 + 3i} \\ & = \frac{(-3 - 2i)(1 - 3i)}{(1 + 3i)(1 - 3i)} \\ &= \frac{-3 + 9i - 2i + 6i^2}{1^2 - (3i)^2}\\ &= \frac{-3 + 7i - 6}{1 - 9(-1)} \\ &= \frac{-9 + 7i}{10} \end{align} $$\implies \frac{\overline{z_1}}{\overline{z_2}} = -\frac{9}{10} + \frac{7}{10}i.\,\, -- (ii)$$ From (i) and (ii), we have $\overline{\left( \frac{z_1}{z_2} \right)} = \frac{\overline{z_1}}{\overline{z_2}}.$

For $z_{1}=-3+2 i$ and $z_{2}=1-3 i$ verify: $\overline{z_{1} z_{2}}=\overline{z_{1}}\,\, \overline{z_{2}}$.

Solution.

Given $z_1 = -3 + 2i, \quad z_2 = 1 - 3i.$ First we calculate \begin{align} z_1 z_2 &= (-3 + 2i)(1 - 3i) \\ &= -3 + 9i + 2i - 6i^2 \\ &= -3 + 11i + 6\\ &= 3 + 11i. \end{align} Then $\overline{z_1 z_2}= 3 - 11i. -- (i)$ Now $\overline{z_1} = -3 - 2i, \quad \overline{z_2} = 1 + 3i.$ Then \begin{align} \overline{z_1}\,\, \overline{z_2} &= (-3 - 2i)(1 + 3i)\\ & = -3 - 9i - 2i - 6i^2 \\ &= -3 - 11i + 6 \\ \implies \overline{z_1}\,\, \overline{z_2} & = 3 - 11i. -- (ii) \end{align}

From (i) and (ii), we have $\overline{z_1 z_2} = \overline{z_1} \overline{z_2},$ as required.

For $z_{1}=-3+2 i$ and $z_{2}=1-3 i$ verify: $\overline{z_{1}+z_{2}}=\overline{z_{1}}+\overline{z_{2}}$.

Solution.

Given $z_1 = -3 + 2i, \quad z_2 = 1 - 3i.$ First we calculate \begin{align} z_1 + z_2 & = (-3 + 2i) + (1 - 3i) \\ &= -2 - i. \end{align} Then $\overline{z_1 + z_2} = -2 + i. \,\, -- (i)$ Now $\overline{z_1} = -3 - 2i, \quad \overline{z_2} = 1 + 3i,$ then \begin{align} \overline{z_1} + \overline{z_2} &= (-3 - 2i) + (1 + 3i) \\ &= -2 + i \,\, -- (ii) \end{align} From (i) and (ii), we have $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}.$

For $z_{1}=-3+2 i$ and $z_{2}=1-3 i$ verify: $\left|z_{1} z_{2}\right|=\left|z_{1}\right|\left|z_{2}\right|$.

Solution.

Given: \begin{align} z_1 = -3 + 2i, \quad z_2 = 1 - 3i \end{align} First we calculate \begin{align} z_1 z_2 &= (-3 + 2i)(1 - 3i) \\ &= -3 - 9i + 2i + 6i^2 \\ &= -3 - 7i - 6\\ &= -9 - 7i, \end{align} then \begin{align} \left|z_1 z_2\right| &= \left|-9 - 7i\right|\\ &= \sqrt{(-9)^2 + (-7)^2} \\ \implies \left|z_1 z_2\right| &= \sqrt{130} \,\, -- (i) \end{align} Now \begin{align}\left|z_1\right| &= \sqrt{(-3)^2 + (2)^2}\\ &= \sqrt{13} \end{align} \begin{align} \left|z_2\right| &= \sqrt{(1)^2 + (-3)^2} \\ &= \sqrt{10}. \end{align} Then \begin{align} \left|z_1\right| \left|z_2\right| &= \sqrt{13} \cdot \sqrt{10} = \sqrt{130} \,\, --(ii) \end{align} From (i) and (ii), we have $\left|z_1 z_2\right| = \left|z_1\right| \left|z_2\right|,$ as required.

For $z_{1}=-3+2 i$ and $z_{2}=1-3 i$ verify: $\left|z_{1}+z_{2}\right| \leq\left|z_{1}\right|+\left|z_{2}\right|$

Solution.

Given \begin{align} z_1= -3 + 2i, \quad z_2 = 1 - 3i \end{align} First, we calculate \begin{align} z_1 + z_2 &= (-3 + 2i) + (1 - 3i) \\ &= -2 - i, \end{align} then \begin{align} \left|z_1 + z_2\right| = \sqrt{(-2)^2 + (-1)^2} = \sqrt{5}. \,\, -- (i) \end{align} Now \begin{align}\left|z_1\right| = \sqrt{(-3)^2 + (2)^2} = \sqrt{13}. \end{align} \begin{align}\left|z_2\right| = \sqrt{(1)^2 + (-3)^2} = \sqrt{10}. \end{align} This gives \begin{align} \left|z_1\right| + \left|z_2\right| &= \sqrt{13} + \sqrt{10} \,\, -- (ii) \end{align} Since $\sqrt{5} \leq \sqrt{13} + \sqrt{10},$ therefore, from (i) and (ii), we have $\left|z_1 + z_2\right| \leq \left|z_1\right| + \left|z_2\right|.$