====== 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.

=====Question 5(i)=====
Let ${{z}_{1}}=2+4i$and ${{z}_{2}}=1-3i$. Verify that $\overline{{{z}_{1}}+{{z}_{2}}}=\overline{{{z}_{1}}}+\overline{{{z}_{2}}}$.
====Solution==== 
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.
=====Question 5(ii)=====
Let ${{z}_{1}}=2+3i$ and ${{z}_{2}}=2-3i$. Verify that $\overline{{{z}_{1}}{{z}_{2}}}=\overline{{{z}_{1}}}\overline{{{z}_{2}}}$.
====Solution==== 
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
$$\overline{z_1 z_2}=\overline{z_1}\overline{z_2}.$$
=====Question 5(iii)=====
Let ${{z}_{1}}=-a-3bi$ and ${{z}_{2}}=2a-3bi$. Verify that $\overline{\left(\dfrac{{{z}_{1}}}{{{z}_{2}}}\right)}=\dfrac{\overline{{{z}_{1}}}}{\overline{{{z}_{2}}}}$.
====Solution==== 
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
$$\overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac{\overline{{{z}_{1}}}}{\overline{{{z}_{2}}}}.$$
====Go to====
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