Question

If ${z_r} = \cos \left( {\dfrac{\pi }{{{3^r}}}} \right) + i\sin \left( {\dfrac{\pi }{{{3^r}}}} \right)$ , $r = 1,2,3,...,$ then ${z_1}.{z_2}.{z_3}...\infty$ =
A) i
B) -i
C) 1
D) -1

Answer 1

We are given equation ${z_r} = \cos \left( {\dfrac{\pi }{{{3^r}}}} \right) + i\sin \left( {\dfrac{\pi }{{{3^r}}}} \right)$ . Now we write it in terms of exponential, then equate it to ${z_r}$ . Then assume $({e^{i\pi }}) = t$ . Put this value in ${z_1}.{z_2}.{z_3}...\infty$ to find the product, then we solve the series using G.P. On simplification we will get our result.

Complete step by step solution:
Given,
$\Rightarrow {z_r} = \cos \left( {\dfrac{\pi }{{{3^r}}}} \right) + i\sin \left( {\dfrac{\pi }{{{3^r}}}} \right)$ ….(1)
We know that,
$\Rightarrow {e^{i\theta }} = \cos \theta + i\sin \theta$
Therefore, we can write
$\Rightarrow \cos \left( {\dfrac{\pi }{{{3^r}}}} \right) + i\sin \left( {\dfrac{\pi }{{{3^r}}}} \right) = {e^{\dfrac{{i\pi }}{{{3^r}}}}}$ ….(2)
From equation (1) and (2)
$\Rightarrow {z_r} = {e^{\dfrac{{i\pi }}{{{3^r}}}}}$
On simplification
$\Rightarrow {z_r} = {({e^{i\pi }})^{\dfrac{1}{{{3^r}}}}}$ ….(3)
Consider $({e^{i\pi }}) = t$ ….(4)
Using equation (4) in (3)
$\Rightarrow {z_r} = {\left( t \right)^{\dfrac{1}{{{3^r}}}}}$ ….(5)
We have to find the value of ${z_1}.{z_2}.{z_3}...\infty$ …(6)
Therefore, from equations (5) and (6)
$\Rightarrow {z_1}.{z_2}.{z_3}...\infty = {\left( t \right)^{\dfrac{1}{{{3^r}}}}}$
Taking r as 1, 2, 3, … $\infty$
$\Rightarrow {z_1}.{z_2}.{z_3}...\infty = {t^{\left( {\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ... + \infty } \right)}}$ ….(5)
Since, $\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ... + \infty$ is an infinite G.P.
Whose first term
$\Rightarrow a = \dfrac{1}{3}$
And common ratio
$\Rightarrow r = \dfrac{1}{3}$
$\because \left| r \right| < 1$
Therefore,
$\Rightarrow {s_\infty } = \dfrac{a}{{1 - r}}$ ….(6)
Here ${s_\infty }$ denotes the sum of infinite terms.
Therefore, putting the values of a and r in equation (6)
$\Rightarrow {s_\infty } = \left( {\dfrac{{\dfrac{1}{3}}}{{1 - \dfrac{1}{3}}}} \right)$
On simplifying the denominator, we get,
$\Rightarrow {s_\infty } = \left( {\dfrac{{\dfrac{1}{3}}}{{\dfrac{{3 - 1}}{3}}}} \right)$
On simplification we get,
$\Rightarrow {s_\infty } = \dfrac{1}{2}$ …..(7)
Therefore, the value of $\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ... + \infty = \dfrac{1}{2}$ ….(
Now we will put the value of (8) in (5)
$\Rightarrow {z_1}.{z_2}.{z_3}...\infty = {t^{\dfrac{1}{2}}}$ ….(9)
Put the value of (4) in (9)
$\Rightarrow {z_1}.{z_2}.{z_3}...\infty = {({e^{i\pi }})^{\dfrac{1}{2}}}$ ….(10)
We can write
$\Rightarrow {e^{\dfrac{{i\pi }}{2}}} = \cos \left( {\dfrac{\pi }{2}} \right) + i\sin \left( {\dfrac{\pi }{2}} \right)$ ….(11)
From, equation (10) and (11)
$\Rightarrow {z_1}.{z_2}.{z_3}...\infty = \cos \left( {\dfrac{\pi }{2}} \right) + i\sin \left( {\dfrac{\pi }{2}} \right)$ ….(12)
The value of $\cos \left( {\dfrac{\pi }{2}} \right)$ is 0 and the value of $\sin \left( {\dfrac{\pi }{2}} \right)$ is 1. Putting these values in equation (12).
We have,
$\Rightarrow {z_1}.{z_2}.{z_3}...\infty = 0 + i$
$\Rightarrow {z_1}.{z_2}.{z_3}...\infty = i$
Therefore, the value of ${z_1}.{z_2}.{z_3}...\infty = i$

Hence, Option (A) is correct.

Note:
You may get confused in solving equations and may think like which equation is to be put in. You can find difficulty in converting equations containing exponential values to equations containing sine and cosine values.
Euler's formula is a mathematical formula in complex analysis that establishes the fundamental relationship between the trigonometric functions and the complex exponential function. Euler's formula states that for any real number x:
${e^{ix}} = \cos x + i\sin x$
where e is the base of the natural logarithm, i is the imaginary unit, and cos and sin are the trigonometric functions cosine and sine respectively.