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

Hint: We can use the Euler’s formula ${{e}^{i\theta }}=\cos \theta +i\sin \theta $ to find value of ${{z}_{1}},{{z}_{2}},..{{z}_{n}}$. Then put the values and use formula of infinite G.P.- S=$\dfrac{a}{1-r}$ where ‘a’ is the first term and r is the common ratio and $\left| \text{r} \right|<1$

Complete step-by-step answer:

Given, ${{z}_{r}}=\cos \left( \dfrac{\pi }{{{3}^{r}}} \right)$ $+i\sin \left( \dfrac{\pi }{{{3}^{r}}} \right)$-- (i)
Where r=$1,2,3,...$
Then we have to find the value of ${{z}_{1}}.{{z}_{2}}.{{z}_{3}}...\infty $
We know that Euler’s formula is-
$\Rightarrow {{e}^{i\theta }}=\cos \theta +i\sin \theta $
On putting the value of $\theta =\dfrac{\pi }{{{3}^{r}}}$ , we get
$\Rightarrow {{e}^{i\dfrac{\pi }{{{3}^{\text{r}}}}}}=\cos \dfrac{\pi }{{{3}^{r}}}+i\sin \dfrac{\pi }{{{3}^{r}}}$ --- (ii)
From eq. (i) and (ii), we get-
$\Rightarrow {{z}_{\text{r}}}={{e}^{i\dfrac{\pi }{{{3}^{\text{r}}}}}}$-- (ii)
Now on putting the value of r=$1,2,3,...$, we get-
$\Rightarrow {{z}_{1}}={{e}^{i\dfrac{\pi }{{{3}^{1}}}}}$
$\Rightarrow {{z}_{2}}={{e}^{i\dfrac{\pi }{{{3}^{2}}}}}$
- - - - - -
$\Rightarrow {{z}_{n}}={{e}^{i\dfrac{\pi }{{{3}^{n}}}}}$
Then we can write
${{z}_{1}}.{{z}_{2}}.{{z}_{3}}...\infty =$${{e}^{i\dfrac{\pi }{{{3}^{1}}}}}.{{e}^{i\dfrac{\pi }{{{3}^{2}}}}}.{{e}^{i\dfrac{\pi }{{{3}^{3}}}}}...\infty $
Here the base of the multiplication is identical or same so the raised powers are added.
So the eq. becomes-
$\Rightarrow {{z}_{1}}.{{z}_{2}}.{{z}_{3}}...\infty ={{e}^{i\dfrac{\pi }{3}+i\dfrac{\pi }{{{3}^{2}}}+i\dfrac{\pi }{{{3}^{3}}}+...\infty }}$
On taking iota common we get,
$\Rightarrow {{z}_{1}}.{{z}_{2}}.{{z}_{3}}...\infty ={{e}^{i\left( \dfrac{\pi }{3}+\dfrac{\pi }{{{3}^{2}}}+\dfrac{\pi }{{{3}^{3}}}+...\infty \right)}}$
Here the raised power of e is in infinite G.P. so we can use the following formula to the sum of the terms-
$\Rightarrow $ S=$\dfrac{a}{1-r}$ where ‘a’ is the first term and r is the common ratio and $\left| \text{r} \right|<1$
Here the first term a=$\dfrac{\pi }{3}$ and common ratio is=$\dfrac{1}{3}$
On putting these values in the formula we get,
$\Rightarrow S=\dfrac{\dfrac{\pi }{3}}{1-\dfrac{1}{3}}$
On solving we get,
$\Rightarrow S=\dfrac{\dfrac{\pi }{3}}{\dfrac{3-1}{3}}=\dfrac{\pi }{2}$
On putting this value in the equation we get,
$\Rightarrow {{z}_{1}}.{{z}_{2}}.{{z}_{3}}...\infty ={{e}^{i\dfrac{\pi }{2}}}$ -- (iii)
Now we know that
$\Rightarrow {{e}^{i\theta }}=\cos \theta +i\sin \theta $
On using this formula we get,
$\Rightarrow {{e}^{i\dfrac{\pi }{2}}}=\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2}$
We know that$\cos \dfrac{\pi }{2}=0$ and $\sin \dfrac{\pi }{2}=1$ . On putting these values we get,
$\Rightarrow {{e}^{i\dfrac{\pi }{2}}}=0+i\times 1$ $=i$
On putting this value in Eq. (iii) we get
$\Rightarrow {{z}_{1}}.{{z}_{2}}.{{z}_{3}}...\infty =i$
Hence the correct answer is A.

Note: Don’t confuse the formula of infinite G.P. with finite G.P. The formula of finite G.P. is-
$\Rightarrow {{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$
Where r is the common ratio, ‘a’ is the first term and n is the number of terms.
In Infinite G.P. series-
1) The series is said to be convergent when $-1< r < 1$ which means it has a sum.
2) The series is said to be divergent when $r>1$ or $r<-1$ which means it has no sum.
3) In the series, if $r\ge 1$ then the sum of the infinite G.P. series tends to infinity.