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 ${{z}^{4}}+{{z}^{2}}+1=0$.

$$z^4+z^2+1=0$$ $$z^4+2z^2+1-z^2=0$$ $$( z^2+1 )^2-z^2=0$$ $$( z^2+1+z)( z^2+1-z )=0$$ $$( z^2+z+1 )( z^2-z+1 )=0$$ $$(z^2+z+1 )=0$$ By using quadratic formula, we have $$z=\dfrac{-1\pm \sqrt{1-4}}{2}$$ $$z=\dfrac{-1\pm \sqrt{3}i}{2}$$ $$(z^2-z+1 )=0$$ By using quadratic formula, we have $$z=\dfrac{1\pm \sqrt{1-4}}{2}$$ $$z=\dfrac{1\pm \sqrt{3}i}{2}$$ The value of $z$ from both equations, we have $$z=\pm \dfrac{1}{2}\pm \dfrac{\sqrt{3}}{2}i$$

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

Given: $$z^3=-8.$$ This gives \begin{align} & z^3+2^3=0\\ \implies &(z+2)\left(z^2-2z+4 \right)=0 \\ & \quad \because a^3+b^3=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\\ \implies & z+2=0 \quad \text{or} \quad z^2-2z+4=0.\end{align} Now $$z^2-2z+4=0$$ According to the quadratic formula, we have $a=1$, $b=-2$ and $c=4$ Thus, we have \begin{align} z&=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\ &=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( 4 \right)}}{2\left( 1 \right)}\\ &=\dfrac{2\pm \sqrt{4-16}}{2}\\ &=\dfrac{2\pm \sqrt{-12}}{2}\\ &=\dfrac{2\pm 2\sqrt{3}i}{2}\\ &=1\pm \sqrt{3}i\end{align} Thus $-2$, $1\pm \sqrt{3}i$ are the solutions of the required equations.

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

Given: $$(z-1)^3=-1.$$ Since we have $$(a-b)^3=a^3-b^3-3ab(a-b).$$ Therefore, we have \begin{align} &z^3-1-3z(z-1)=-1\\ \implies &{{z}^{3}}-1-3{{z}^{2}}+3z+1=0\\ \implies &{{z}^{3}}-3{{z}^{2}}+3z=0\\ \implies &z\left(z^2-3z+3 \right)=0.\end{align} This gives $$z=0 \quad \text{or} \quad z^2-3z+3=0.$$ According to the quadratic formula, we have $a=1$, $b=-3$ and $c=3$. So, we have \begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ &=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\left( 1 \right)\left( 3 \right)}}{2\left( 1 \right)}\\ &=\dfrac{3\pm \sqrt{9-12}}{2}\\ &=\dfrac{3\pm \sqrt{-3}}{2}\\ &=\dfrac{3\pm \sqrt{3}i}{2}.\end{align} Hence solutions of the given equation are $0$, $\dfrac{3}{2}\pm \dfrac{\sqrt{3}i}{2}$.

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

Given $$z^3=1.$$ Thus, we have \begin{align}&z^3-1^3=0\\ \implies &(z-1)\left(z^2+z+1\right)=0.\end{align} This gives $z-1=0$ or $z^2+z+1=0$.
This gives $z=1$ and we solve $z^2+z+1=0$. According to the quadratic formula, we have $a=1$, $b=1$ and $c=1$. Thus, we have \begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ &=\dfrac{-\left( 1 \right)\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 1 \right)\left( 1 \right)}}{2\left( 1 \right)}\\ &=\dfrac{-1\pm \sqrt{1-4}}{2}\\ &=\dfrac{-1\pm \sqrt{-3}}{2}\\ &=\dfrac{-1\pm \sqrt{3}i}{2}\\ &=-\dfrac{1}{2}\pm \dfrac{\sqrt{3}i}{2}\end{align} Thus, the solutions of the given equations are $1$, $-\dfrac{1}{2}\pm \dfrac{\sqrt{3}i}{2}$.

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