Question 5, Exercise 1.2

Solutions of Question 5 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Let ${{z}_{1}}=2+4i$and ${{z}_{2}}=1-3i$. Verify that $\overline{{{z}_{1}}+{{z}_{2}}}=\overline{{{z}_{1}}}+\overline{{{z}_{2}}}$.

Given ${{z}_{1}}=2+4i$ and ${{z}_{2}}=1-3i$.

Thus $\overline{{{z}_{1}}}=2-4i$ and $\overline{{{z}_{2}}}=1+3i$.

Now $$\begin{align}z_1+z_2&=2+4i+1-3i\\ &=3+i \end{align}$$ Now $$\begin{align}\overline{z_1+z_2}=3-i \ldots (1)\end{align}$$ and $$\begin{align} \overline{z_1}+\overline{z_2}&=2-4i+1+3i\\ &=3-i \ldots (2)\end{align}$$ From (1) and (2), we have $$\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}$$ as required.

Let ${{z}_{1}}=2+3i$ and ${{z}_{2}}=2-3i$. Verify that $\overline{{{z}_{1}}{{z}_{2}}}=\overline{{{z}_{1}}}\overline{{{z}_{2}}}$.

Given ${{z}_{1}}=2+3i$ and ${{z}_{2}}=2-3i$
Thus $\overline{{{z}_{1}}}=2-3i$ and $\overline{{{z}_{2}}}=2+3i$.

$$\begin{align}z_1 z_2 &=(2+3i)(2-3i)\\ &=2^2-(3i)^2\\ &=13\end{align}$$ Now $$\begin{align}\overline{z_1 z_2}=13 \ldots (1)\end{align}$$ Now $$\begin{align}\overline{z_1}\overline{z_2} &=(2-3i)(2+3i)\\ &=2^2-(3i)^2\\ &=13 \ldots (2)\end{align}$$ From (1) and (2), we have $$\overline{z_1 z_2}=\overline{z_1}\overline{z_2}.$$

Let ${{z}_{1}}=-a-3bi$ and ${{z}_{2}}=2a-3bi$. Verify that $\overline{\left(\dfrac{{{z}_{1}}}{{{z}_{2}}}\right)}=\dfrac{\overline{{{z}_{1}}}}{\overline{{{z}_{2}}}}$.

Given ${{z}_{1}}=-a-3bi$ and ${{z}_{2}}=2a-3bi$.
Thus $\overline{z_1}=-a+3bi$ and $\overline{z_2}=2a+3bi$.

Take $$\begin{align}\dfrac{z_1}{z_2}&=\dfrac{-a-3bi}{2a-3bi}\\ &=\dfrac{-a-3bi}{2a-3bi}\times \dfrac{2a+3bi}{2a+3bi} \quad (\text{by rationalizing})\\ &=\dfrac{-2{{a}^{2}}+9{{b}^{2}}-9abi}{2{{a}^{2}}+9{{b}^{2}}}.\end{align}$$ This gives $$\begin{align} \overline{\left(\dfrac{z_1}{z_2}\right)}&=\dfrac{-2{{a}^{2}}+9{{b}^{2}}+9abi}{2{{a}^{2}}+9{{b}^{2}}} \ldots (1) \end{align}$$ Now $$\begin{align} \dfrac{\overline{{{z}_{1}}}}{\overline{{{z}_{2}}}}&=\dfrac{-a+3bi}{2a+3bi}\\ &=\dfrac{-a+3bi}{2a+3bi}\times \dfrac{2a-3bi}{2a-3bi}\\ &=\dfrac{-2{{a}^{2}}+9{{b}^{2}}+6abi+3abi}{2{{a}^{2}}+9{{b}^{2}}}\\ &=\dfrac{-2{{a}^{2}}+9{{b}^{2}}+9abi}{2{{a}^{2}}+9{{b}^{2}}} \ldots (2) \end{align}$$ From (1) and (2), we have $$\overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac{\overline{{{z}_{1}}}}{\overline{{{z}_{2}}}}.$$

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