Question

Consider the given expression ${{x}_{r}}=\cos \left( \dfrac{\pi }{{{3}^{r}}} \right)-i\sin \left( \dfrac{\pi }{{{3}^{r}}} \right)$ , (where $i=\sqrt{-1}$ ), then the value of the given expression ${{x}_{1}}.{{x}_{2}}..........\infty $ , is
(a) 1
(b) -1
(c) -i
(d) i

Answer 1

Hint: Use definition of i and Euler’s formula ${{e}^{ix}}=\cos x+i\sin x$. It is easier to convert the complex number in euler form as by using exponent property we can simplify the given expression.

Complete step-by-step solution -
Definition of i:
i is an imaginary number which is solution of an equation:
${{x}^{2}}=-1\Rightarrow {{x}^{2}}+1=0$
Use Euler’s formula: ${{e}^{ix}}=\cos x+i\sin x$
The left-hand side can be written as $\text{cis}x$
So, $\text{cis}x=\cos x+i\sin x$
Let $x=\dfrac{\pi }{2}$
By substituting above $x$ value into expression, we get:
$\begin{align}
  & \text{cis}\dfrac{\pi }{2}=\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \\
 & \text{cis}\dfrac{\pi }{2}=i \\
\end{align}$
Required expression:
$\begin{align}
  & {{z}_{1}},{{z}_{2}}......{{z}_{\infty }} \\
 & {{z}_{1}}=\cos \left( \dfrac{\pi }{3} \right)+i\sin \left( \dfrac{\pi }{3} \right)=cis\dfrac{\pi }{3} \\
\end{align}$
We know $\text{cis}x={{e}^{ix}}$
${{z}_{1}},{{z}_{2}},...........{{z}_{\infty }}$ can be assumed ass
${{z}_{1}}=a,{{z}_{2}}=b,{{z}_{3}}=c$
a can be written as
$a={{e}^{i\dfrac{\pi }{3}}}$
b can be written as
$b={{e}^{i\dfrac{\pi }{{{3}^{2}}}}}$
Like that nth term can be written as
${{t}_{n}}={{e}^{i\dfrac{\pi }{{{3}^{n}}}}}$
a.b.c….. is nothing but multiplication of all above terms.
Required equation turns into
${{e}^{i\left( \dfrac{\pi }{3}+\dfrac{\pi }{{{3}^{2}}}+...........+\infty \right)}}=\cos \left( \dfrac{\pi }{3}+\dfrac{\pi }{{{3}^{2}}}+...........+\infty \right)+i\sin \left( \dfrac{\pi }{3}+\dfrac{\pi }{{{3}^{2}}}+...........+\infty \right)$
You can observe an infinite geometric progression with the first term as $\dfrac{\pi }{3}$ and common ratio as $\dfrac{1}{3}$.
If we need infinite sum of geometric progression with first term a and common ratio r, the sum s can be written as
$s=\dfrac{a}{1-r}$
Case 1: Solving real part
$\cos \left( \dfrac{\pi }{3}+\dfrac{\pi }{{{3}^{2}}}+......... \right)$
By applying geometric progression here, we get:
\[\begin{align}
  & \cos \left( \dfrac{\dfrac{\pi }{3}}{1-\dfrac{1}{3}} \right)=\cos \left( \dfrac{\pi }{2} \right) \\
 & =0
\end{align}\]
Case 2: Solving imaginary part
\[=i\sin \left( \dfrac{\pi }{3}+\dfrac{\pi }{{{3}^{2}}}+......... \right)\]
By applying geometric progression here, we get:
$\begin{align}
  & i\sin \left( \dfrac{\dfrac{\pi }{3}}{1-\dfrac{1}{3}} \right)=i\sin \dfrac{\pi }{2} \\
 & =i
\end{align}$
By adding them both, we get i
${{z}_{1}},{{z}_{2}},{{z}_{3}}.......{{z}_{\infty }}=i$
These i is value of required expression
Option (d) is correct.

Note: While using Euler’s formula be careful what to substitute in Cis and always keep $\sin x$ as imaginary if you keep $\cos x$ you may lead to wrong answer.
Idea of using Cis and again using back ${{e}^{ix}}$ is just for convenience you may directly use ${{e}^{ix}}$.