====== Question 8, Exercise 1.2 ======
Solutions of Question 8 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 8(i)=====
Show that  $z+\overline{z}=2\operatorname{Re}\left( z \right)$.
====Solution==== 
Assume $z=a+ib$, then $\overline{z}=a-ib$.
\begin{align}z+\overline{z}&=\left( a+ib \right)+\left( a-ib \right)\\
&=a+ib+a-ib\\
&=2a\\
z+\overline{z}&=2\operatorname{Re}\left( z \right)\end{align}

=====Question 8(ii)=====
Show that  $z-\overline{z}=2i\operatorname{Im}\left( z \right)$.
====Solution==== 
Assume that $z=a+ib$, then $\overline{z}=a-ib$.
\begin{align}z-\overline{z}&=\left( a+ib \right)-\left( a-ib \right)\\
&=a+ib-a+ib\\
z-\overline{z}&=2bi\\
z-\overline{z}&=2i\operatorname{Im}(z)\end{align}

=====Question 8(iii)=====
Show that  $z\overline{z}={{\left[ \operatorname{Re}\left( z \right) \right]}^{2}}+{{\left[ \operatorname{Im}\left( z \right) \right]}^{2}}$.
====Solution==== 
Suppose $z=a+ib$, then $\overline{z}=a-ib$. Then
\begin{align}z\overline{z}&=\left( a+ib \right)\cdot \left( a-ib \right)\\ 
&={{a}^{2}}-{{bi}^{2}}\\
&={{a}^{2}}-b^2 (-1)\\
\implies z\overline{z}&={{a}^{2}}+b^2. \ldots (1) 
\end{align}
Now
\begin{align}
{{\left[\operatorname{Re}\left( z \right) \right]}^{2}}+{{\left[ \operatorname{Im}\left( z \right) \right]}^{2}}&={{a}^{2}}+{{b}^{2}}. \ldots (2)
\end{align}
Using (1) and (2), we get 
$$z\overline{z}={{\left[\operatorname{Re}\left( z \right) \right]}^{2}}+{{\left[ \operatorname{Im}\left( z \right) \right]}^{2}}.$$
This is required.


=====Question 8(iv)=====
Show that  $z=\overline{z}\Rightarrow z$ is real. 
====Solution==== 
Suppose $z=a+bi$ ... (1) 

Then $\overline{z}=a-bi$.

We have given \begin{align}&z=\overline{z} \\
\implies &a+bi=a-bi \\
\implies &bi=bi \\
\implies &2bi=0 \\
\implies &b=0 \end{align}
Using it in (1), we get $z=a+0i=a$, that is, $z$ is real.





=====Question 8(v)=====
Show that  $\overline{z}=-z$ if and only if $z$ is pure imaginary.
====Solution==== 
Suppose that $z=a+bi$ ... (1)

Then $\overline{z}=a-bi.$

suppose that
\begin{align} &\overline{z}=-{z}\\
\implies &a-bi=-(a+bi)\\
\implies &a-bi=-a-bi\\
\implies &a+a=0\\
\implies &2a=0\\
\implies &a=0 \end{align}
Using it in (1), we get
$$z=0+bi=bi,$$
that is, $z$ is pure imaginary.

Conversly, suppose that $z$ is pure imaginary, then its real part will be zero, that is, $a=0$.

Using it in (1), we get
$$z=bi \ldots (2)$$
Then
\begin{align}&\overline{z}=-bi \\
\implies &\overline{z}=-z \quad \text{by using (2).}
\end{align}
This was required.

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