Question

If ${x_r} = \cos \left( {\dfrac{\pi }{{{3^r}}}} \right) + i\sin \left( {\dfrac{\pi }{{{3^r}}}} \right)$ then ${x_1} \cdot {x_2} \cdot {x_3}......\infty$ equals
A.$- i$
B.$i$
C.$- 1$
D.$1$

Answer 1

Here, we have to find the product of the values. First, we will convert the complex number into Euler form and then simplify it using the exponential formula. Then we will substitute different values of $r$ and find the product of these values by using the complex form.

Formula used:
We will use the following formulas:
1.Euler form: $\cos \theta + i\sin \theta = {e^{i\theta }}$
2.Exponential formula: ${({e^m})^n} = {e^{mn}}$ ; ${a^m} \cdot {a^n} \cdot {a^o} \cdot ........ = {a^{m + n + o + .......}}$
3.Trigonometric Ratio: $\cos \dfrac{\pi }{2} = 0$ and $\sin \dfrac{\pi }{2} = 1$
4.The formula for the sum of an infinite geometric series is given by ${S_n} = \dfrac{a}{{1 - r}}$ , where $a$ is the first term and $r$ is the common ratio.

Complete step-by-step answer:
We are given a value of $x$ in the form of a complex number.
${x_r} = cos\left( {\dfrac{\pi }{{{3^r}}}} \right) + isin\left( {\dfrac{\pi }{{{3^r}}}} \right)$
Now, we will convert a complex number into a Euler form $\cos \theta + i\sin \theta = {e^{i\theta }}$. Therefore, we get
$\Rightarrow {x_r} = {e^{\dfrac{{i\pi }}{{{3^r}}}}}$
Now, by using the exponential formula ${({e^m})^n} = {e^{mn}}$, we get
$\Rightarrow {x_r} = {\left( {{e^{i\pi }}} \right)^{\dfrac{1}{{{3^r}}}}}$
Now Let us assume $y = {e^{i\pi }}$. So, we get
$\Rightarrow {x_r} = {y^{\dfrac{1}{{{3^r}}}}}$
$\Rightarrow {x_1} = \dfrac{1}{3}$; ${x_2} = \dfrac{1}{{{3^2}}} = \dfrac{1}{9}$;…………….
Exponential formula:
Now, by using the exponential formula ${a^m} \cdot {a^n} \cdot {a^o} \cdot ........ = {a^{m + n + o + .......}}$, we have
$\Rightarrow {x_1}{x_2}{x_3}....\infty = {y^{\left( {\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + \dfrac{1}{{81}} + ..... + \infty } \right)}}$
Since the power is the sum of an infinite geometric series, we have to use the formula for the sum of an infinite geometric series is given by ${S_n} = \dfrac{a}{{1 - r}}$, we have
$\Rightarrow {x_1}{x_2}{x_3}......\infty = {y^{\left( {\dfrac{{\dfrac{1}{3}}}{{1 - \dfrac{1}{3}}}} \right)}}$

By cross multiplication, we have
$\Rightarrow {x_1}{x_2}{x_3}......\infty = {y^{\left( {\dfrac{{\dfrac{1}{3}}}{{\dfrac{3}{3} - \dfrac{1}{3}}}} \right)}}$
Subtracting the like terms, we have
$\Rightarrow {x_1}{x_2}{x_3}......\infty = {y^{\left( {\dfrac{{\dfrac{1}{3}}}{{\dfrac{2}{3}}}} \right)}}$
Cancelling the denominators, we have
$\Rightarrow {x_1}{x_2}{x_3}......\infty = {y^{\left( {\dfrac{1}{2}} \right)}}$
Again substituting $y$ as ${e^{i\pi }}$, we have
$\Rightarrow {x_1}{x_2}{x_3}......\infty = {\left( {{e^{i\pi }}} \right)^{\dfrac{1}{2}}}$
Now, by using the exponential formula ${({e^m})^n} = {e^{mn}}$, we have
$\Rightarrow {x_1}{x_2}{x_3}......\infty = \left( {{e^{\dfrac{{i\pi }}{2}}}} \right)$
Now, again converting Euler form into a complex number, we have
$\Rightarrow {x_1}{x_2}{x_3}......\infty = \cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}$
Now, by using the trigonometric ratio, we have
$\Rightarrow {x_1}{x_2}{x_3}......\infty = 0 + i(1)$
$\Rightarrow {x_1}{x_2}{x_3}......\infty = i$

Note: The polar form of a complex number is a different way to represent a complex number apart from rectangular form. Usually, we represent the complex numbers, in the form of $z = x + iy$ where $i$ the imaginary number. But in polar form, the complex numbers are represented as the combination of modulus and argument. The modulus of a complex number is also called absolute value. This polar form is represented with the help of polar coordinates of real and imaginary numbers in the coordinate system.