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.
Question 6(i)
Find the solutions of the equation z4+z2+1=0.
Solution
z4+z2+1=0 z4+2z2+1−z2=0 (z2+1)2−z2=0 (z2+1+z)(z2+1−z)=0 (z2+z+1)(z2−z+1)=0 (z2+z+1)=0 By using quadratic formula, we have z=−1±√1−42 z=−1±√3i2 (z2−z+1)=0 By using quadratic formula, we have z=1±√1−42 z=1±√3i2 The value of z from both equations, we have z=±12±√32i
Question 6(ii)
Find the solutions of the equation z3=−8.
Solution
Given: z3=−8. This gives z3+23=0⟹(z+2)(z2−2z+4)=0∵ 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.
Question 6(iii)
Find the solutions of the equation {{\left( z-1 \right)}^{3}}=-1.
Solution
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}.
Question 6(iv)
Find the solutions of the equation {{z}^{3}}=1.
Solution
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}.
Go To