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.

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.

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

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.

< Question 2 Question 4 >