Question

Prove that the general value of $\theta $ which satisfy the equation $\left( \cos \theta +i\sin \theta \right){{\left( \cos \theta +i\sin \theta \right)}^{2}}{{\left( \cos \theta +i\sin \theta \right)}^{3}}...=1$ is $2n\pi $, where ‘n’ is an integer.

Answer 1

Hint: Simplify the given expression using the formula ${{e}^{i\theta }}=\cos \theta +i\sin \theta $. Further use the fact that the product of an infinite sequence of increasing terms can take value 1 if and only if the value of each term is 1. Write equations based on this data and simplify the equation to calculate the value of $\theta $ which satisfies the given equation.

Complete step-by-step solution-
We have to prove that the general value of $\theta $ which satisfy the equation $\left( \cos \theta +i\sin \theta \right){{\left( \cos \theta +i\sin \theta \right)}^{2}}{{\left( \cos \theta +i\sin \theta \right)}^{3}}...=1$ is $2\pi n$, where ‘n’ is an integer.
We know that ${{e}^{i\theta }}=\cos \theta +i\sin \theta $.
Thus, we can rewrite the equation $\left( \cos \theta +i\sin \theta \right){{\left( \cos \theta +i\sin \theta \right)}^{2}}{{\left( \cos \theta +i\sin \theta \right)}^{3}}...=1$ as $\left( {{e}^{i\theta }} \right){{\left( {{e}^{i\theta }} \right)}^{2}}{{\left( {{e}^{i\theta }} \right)}^{3}}...=1$.
We know that $y={{e}^{i\theta }}$ is an increasing function whose range is always greater than or equal to zero.
We also know that the product of an infinite sequence of increasing terms can take value 1 if and only if the value of each term is 1.
Thus, we have ${{e}^{i\theta }}=1,{{\left( {{e}^{i\theta }} \right)}^{2}}=1,{{\left( {{e}^{i\theta }} \right)}^{3}}=1,...$.
So, we must have ${{e}^{i\theta }}=1$.
Thus, we have $\cos \theta +i\sin \theta =1$, which means $\cos \theta =1,\sin \theta =0$.
We will now calculate solutions to the equations $\cos \theta =1$ and $\sin \theta =0$.
We know that the general solutions of the equation $\cos \theta =1$ are $\theta =2n\pi ,n\in I$.
We also know that the general solutions of the equation $\sin \theta =0$ are $\theta =n\pi ,n\in I$.
So, the common solutions of the equations $\cos \theta =1$ and $\sin \theta =0$ is $\theta =2n\pi ,n\in I$.
Hence, we have proved that the general value of $\theta $ which satisfy the equation $\left( \cos \theta +i\sin \theta \right){{\left( \cos \theta +i\sin \theta \right)}^{2}}{{\left( \cos \theta +i\sin \theta \right)}^{3}}...=1$ is $2\pi n$, where ‘n’ is an integer.

Note: We must know how to write general solutions of trigonometric equations; otherwise, we won’t be able to prove the given statement. We must also keep in mind that we must consider the solutions common to both the trigonometric equations.