Question 6, Exercise 1.3

Solutions of Question 6 of Exercise 1.3 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.

Find the solutions of the equation ${{z}^{4}}+{{z}^{2}}+1=0$

\begin{align}{{z}^{4}}+{{z}^{2}}+1&=0\\ {{z}^{4}}+2\left( \dfrac{1}{2} \right){{z}^{2}}+\dfrac{1}{4}-\dfrac{1}{4}+1&=0\\ {{\left( {{z}^{2}}+\dfrac{1}{2} \right)}^{2}}+\dfrac{4-1}{4}&=0\\ {{\left( {{z}^{2}}+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}=0\\ {{\left( {{z}^{2}}+\dfrac{1}{2} \right)}^{2}}&=-\dfrac{3}{4}\end{align}
Take square root on both sides.
\begin{align}\left( {{z}^{2}}+\dfrac{1}{2} \right)&=\pm \sqrt{-\dfrac{3}{4}}\\ \left( {{z}^{2}}+\dfrac{1}{2} \right)&=\pm \sqrt{\dfrac{3}{4}}i\\ {{z}^{2}}&=-\dfrac{1}{2}\pm \dfrac{\sqrt{3}}{2}i\\ z&={{\left( -\dfrac{1}{2}\pm \dfrac{\sqrt{3}}{2}i \right)}^{\dfrac{1}{2}}}\end{align}

Find the solutions of the equation ${{z}^{3}}=-8$

\begin{align}{{z}^{3}}&=-8\\ {{z}^{3}}+{{2}^{3}}&=0\\ \left( z+2 \right)\left( {{z}^{2}}-2z+4 \right)&=0\quad\quad \because \left( {{a}^{3}}+{{b}^{3}} \right)=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\\ z+2&=0\\ z&=-2\end{align} Now
$\left( {{z}^{2}}-2z+4 \right)=0$
According to the quadratic formula, we have
$a=1,\,\,\,b=-2$ and $c=4$
Quadratic formula is
\begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ z&=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( 4 \right)}}{2\left( 1 \right)}\\ z&=\dfrac{2\pm \sqrt{4-16}}{2}\\ z&=\dfrac{2\pm \sqrt{-12}}{2}\\ z&=\dfrac{2\pm 2\sqrt{3}i}{2}\\ z&=1\pm \sqrt{3}i\end{align} Then
$$z=-2,1\pm \sqrt{3}i$$

Find the solutions of the equation ${{\left( z-1 \right)}^{3}}=-1$.

\begin{align}{{\left( z-1 \right)}^{3}}&=-1\\ {{z}^{3}}-1-3z\left( z-1 \right)&=-1\quad\quad \because {{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)\\ {{z}^{3}}-1-3{{z}^{2}}+3z+1&=0\\ {{z}^{3}}-3{{z}^{2}}+3z&=0\\ z\left( {{z}^{2}}-3z+3 \right)&=0\end{align} $z=0$ , ${{z}^{2}}-3z+3=0$
${{z}^{2}}-3z+3=0$
According to the quadratic formula, we have
$a=1,\,\,\,b=-3$ and $c=3$
Quadratic formula is
\begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ z&=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\left( 1 \right)\left( 3 \right)}}{2\left( 1 \right)}\\ z&=\dfrac{3\pm \sqrt{6-12}}{2}\\ z&=\dfrac{3\pm \sqrt{-6}}{2}\\ z&=\dfrac{3\pm \sqrt{6}i}{2}\end{align} $$\Rightarrow z=0,\dfrac{3}{2}\pm \dfrac{\sqrt{6}i}{2}$$

Find the solutions of the equation ${{z}^{3}}=1$

\begin{align}{{z}^{3}}&=1\\ {{z}^{3}}-{{1}^{3}}&=0\\ \left( z-1 \right)\left( {{z}^{2}}+z+1 \right)&=0\end{align}
$\left( z-1 \right)=0$, $\left( {{z}^{2}}+z+1 \right)=0$
\begin{align}\left( z-1 \right)&=0\\ z&=1\\ {{z}^{2}}+z+1&=0\end{align} According to the quadratic formula, we have
$a=1,\quad\quad b=1$ and $c=1$
Quadratic formula is
\begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ z&=\dfrac{-\left( 1 \right)\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 1 \right)\left( 1 \right)}}{2\left( 1 \right)}\\ z&=\dfrac{-1\pm \sqrt{1-4}}{2}\\ z&=\dfrac{-1\pm \sqrt{-3}}{2}\\ z&=\dfrac{-1\pm \sqrt{3}i}{2}\\ z&=1,-\dfrac{1}{2}\pm \dfrac{\sqrt{3}i}{2}\end{align}

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