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

Hint: 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,
\[\begin{align}
  & \Rightarrow {{z}^{7}}-1=0 \\
 & \Rightarrow {{z}^{7}}=1 \\
 & \Rightarrow z={{\left( 1 \right)}^{\dfrac{1}{7}}} \\
 & \Rightarrow z={{\left( \cos 0+i\sin 0 \right)}^{\dfrac{1}{7}}} \\
\end{align}\]
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 \[z\].