Question

Solutions of the equation ${{z}^{7}}-1=0$ are given by
(a) $z=-1,z=\cos \dfrac{2k\pi }{7}+i\sin \dfrac{2k\pi }{7},k=0,1,2,3,4,5$
(b) $z=1,z=\cos \dfrac{2k\pi }{7}+i\sin \dfrac{2k\pi }{7},k=1,2,3,4,5,6$
(c) $z=-1,z=\cos \dfrac{k\pi }{7}+i\sin \dfrac{k\pi }{7},k=0,1,2,3,4,5$
(d) $z=1,z=\cos \dfrac{k\pi }{7}+i\sin \dfrac{k\pi }{7},k=0,1,2,3,4,5$

Answer 1

In this type of question we have to use the concept of complex numbers. We know that a complex number $z=x+iy$ can be expressed in polar form as $z=r\left( \cos \theta +i\sin \theta \right)$ where $r=\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$ and $\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$. Here, we should know how to solve the equation and then use these values to make new functions. In this case we have to use De-Moivre’s theorem which states that, for any real number $x$ we have ${{\left( \cos x+i\sin x \right)}^{n}}=\left( \cos nx+i\sin nx \right),n=0,1,2,3,\cdots \cdots \cdots n-1$

Complete step by step answer:
Now, we have to solve the equation ${{z}^{7}}-1=0$
Let us consider,

& \Rightarrow {{z}^{7}}-1=0 \\\ & \Rightarrow {{z}^{7}}=1 \\\ & \Rightarrow z={{\left( 1 \right)}^{\dfrac{1}{7}}} \\\ & \Rightarrow z={{\left( \cos 0+i\sin 0 \right)}^{\dfrac{1}{7}}} \\\ \end{aligned}$$ As we know that, $$\cos \theta =\cos \left( 2n\pi +\theta \right),\sin \theta =\sin \left( 2n\pi +\theta \right)$$ we can write the above equation as $$\Rightarrow z={{\left( \cos 2k\pi +i\sin 2k\pi \right)}^{\dfrac{1}{7}}}$$ Now, we have to use De-Moivre’s theorem which states that, for any real number $$x$$ we have $${{\left( \cos x+i\sin x \right)}^{n}}=\left( \cos nx+i\sin nx \right),n=0,1,2,3,\cdots \cdots \cdots n-1$$ and hence we get, $$\Rightarrow z=\left( \cos \dfrac{2k\pi }{7}+i\sin \dfrac{2k\pi }{7} \right),k=0,1,2,3,4,5,6$$ Now if we substitute the value for $$k=0$$ we will get the value of $$z$$ as $$\Rightarrow z=\cos 0+i\sin 0$$ But we know that, $$\cos 0=1$$ and hence we can write $$\Rightarrow z=1$$ Thus the possible values of $$z$$ that means the solutions of the equation $${{z}^{7}}-1=0$$ are $$\Rightarrow z=1,z=\left( \cos \dfrac{2k\pi }{7}+i\sin \dfrac{2k\pi }{7} \right),k=1,2,3,4,5,6$$ **So, the correct answer is “Option b”.** **Note:** To solve such types of questions students must be well familiar with the De-Moivre’s rule. Also students have to take care when they convert a complex number into its polar form for that they must know the certain values of trigonometric functions such as $$\cos 0=1,\sin 0=0,\cos \pi =-1,\sin \left( \dfrac{\pi }{2} \right)=1$$ etc. Also they have to remember that in De-Moivre’s theorem the value of $$k$$ varies from 0 to $$n-1$$ and by substituting $$k=0$$ they can obtain one particular value of $$z$$.