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.

Show that $z+\overline{z}=2\operatorname{Re}\left( z \right)$.

Assume $z=a+ib$, then $\overline{z}=a-ib$. $$\begin{align}z+\overline{z}&=\left( a+ib \right)+\left( a-ib \right)\\ &=a+ib+a-ib\\ &=2a\\ z+\overline{z}&=2\operatorname{Re}\left( z \right)\end{align}$$

Show that $z-\overline{z}=2i\operatorname{Im}\left( z \right)$.

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

Show that $z\overline{z}={{\left[ \operatorname{Re}\left( z \right) \right]}^{2}}+{{\left[ \operatorname{Im}\left( z \right) \right]}^{2}}$.

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.

Show that $z=\overline{z}\Rightarrow z$ is real.

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.

Show that $\overline{z}=-z$ if and only if $z$ is pure imaginary.

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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