====== Question 3, Exercise 1.2 ======
Solutions of Question 3 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. 
====Question 3(i)====
Prove that for $z \in \mathbb{C}$. $z$ is real iff $z=\bar{z}$.

**Solution.** 
Let $$z=a+ib\quad \text{where}\quad a,b\in \mathbb{R}\, ... (1)$$
First suppose that $z$ is real, then we shall prove $\overline{z}=z$.

Since $z$ is real, imaginary part of $z$ is zero. i.e. $b=0$.

Then 
\begin{align}
&z=a \\
\implies &\bar{z}=a \end{align}
This gives $z=\bar{z}$.

Now conversly suppose that $\overline{z}=z$, then we $z$ is real.

As
\begin{align}& z=\bar{z}\\
\Rightarrow \quad & a+ib=a-ib\\
\Rightarrow \quad & 2ib=0\\
\Rightarrow \quad & b=0\quad \because \quad 2i\neq 0\end{align}
Then (1) becomes
$$z=a+i(0)=a$$
This gives $z$ is real.
====Question 3(ii)====
Prove that for $z \in \mathbb{C}$. $\dfrac{z-\bar{z}}{z+\bar{z}}=i\left(\dfrac{I m z}{R e z}\right)$.

**Solution.** 

Suppose $z=x+iy$, then $\overline{z}=x-iy$.

Now
\begin{align}&\quad\dfrac{z-\bar{z}}{z+\bar{z}}\\\
&=\dfrac{x+iy-(x-iy)}{x+iy+x-iy}\\
&=\dfrac{2iy}{2x}\\
&=i\left(\dfrac{I m z}{R e z}\right)\end{align}

====Question 3(iii)====
Prove that for $z \in \mathbb{C}$. $z$ is either real or pure imaginary iff $(\overline{z})^{2}=z^{2}$.

**Solution.** 

For $z=x+iy$, first suppose that $z$ is real or pure imaginary, then
$$z=x \quad \text{ or } \quad z=iy.$$
This gives
$$\bar{z}=x \quad \text{ or } \quad \bar{z}=-iy,$$
implies
$$(\bar{z})^2=x^2 \quad \text{ or } \quad (\bar{z})^2=-y^2. ... (i)$$
Also, we have
$$z^2=x^2 \quad \text{ or } \quad z^2=-y^2. ...(ii)$$
From (i) and (ii), we have
$$(\overline{z})^{2}=z^{2}$$
Conversly, suppose that \begin{align}&(\overline{z})^{2}=z^{2}\\
&(x-iy)^2=(x+iy)^2\\
&x^2-y^2-2ixy=x^2-y^2+2ixy \cdots \cdots (1)\\
&4ixy=0\end{align}
This gives either $x=0$ or $y=0$.

As $z=x+iy$, so either $z=iy$ or $z=x$.

This gives $z$ is real or pure imaginary.
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