Question

If ${\text{x}} = {9^{\dfrac{1}{3}}}\,{9^{\dfrac{1}{9}}}\,{9^{\dfrac{1}{{27}}}}.....\infty ,{\text{y}} = \,{4^{\dfrac{1}{3}}}\,{4^{ - \dfrac{1}{9}}}\,{4^{\dfrac{1}{{27}}}}\,{4^{ - \dfrac{1}{{81}}}}.......\infty ,\,{\text{and z}} = \sum\limits_{{\text{r}} = 1}^\infty {{{(1 + i)}^{ - {\text{r}}}}} ,$then ${\text{arg}}(x + yz)$ is equal to

Answer 1

Hint: From the given it is clearly saying that the given series is in geometric progression. By finding the first term and the common ratio of the series we can be able to find the required values.

Formula used: The formula used in the question are,
The geometric series is $\dfrac{{\text{a}}}{{1 - {\text{r}}}}$
The argument of the given term can be calculated by,
 $\arg ({\text{z)}} = \pi - {\tan ^{ - 1}}\left( {\dfrac{{\text{y}}}{x}} \right)$
Where,
 ${\text{z}} = {\text{x + iy}}$
 ${\text{a}}$ is the first term of the series
 ${\text{r}}$ is the common ratio

Complete step-by-step answer:
The ${\text{x}}$ value is given as,
 ${\text{x}} = {9^{\dfrac{1}{3}}}\,{9^{\dfrac{1}{9}}}\,{9^{\dfrac{1}{{27}}}}...\infty $
 ${\text{y}}$ value is given as
  ${\text{y}} = \,{4^{\dfrac{1}{3}}}\,{4^{ - \dfrac{1}{9}}}\,{4^{\dfrac{1}{{27}}}}\,{4^{ - \dfrac{1}{{81}}}}...\infty ,$
 Z value is given as
 \[{\text{z}} = \sum\limits_{{\text{r}} = 1}^\infty {{{(1 + i)}^{ - {\text{r}}}}} \]
First step we have to find the solution of ${\text{x}}$ ,
 ${\text{x}} = {9^{\dfrac{1}{3}}}\,{9^{\dfrac{1}{9}}}\,{9^{\dfrac{1}{{27}}}}...\infty ,$
From the above, the first term is $\dfrac{1}{3}$ and the common ratio is $\dfrac{1}{3}$ and substituting in the formula $\dfrac{{\text{a}}}{{1 - {\text{r}}}}$ we get,
 ${\text{x}}\, = {9^{\left( {\dfrac{{\dfrac{1}{3}}}{{1 - \dfrac{1}{3}}}} \right)}}$,
Finally by solving we get ${\text{x}}$ value as $3$.
Second step we have to find the solution of ${\text{y}}$ ,
 ${\text{y}} = \,{4^{\dfrac{1}{3}}}\,{4^{ - \dfrac{1}{9}}}\,{4^{\dfrac{1}{{27}}}}\,{4^{ - \dfrac{1}{{81}}}}...\infty $
From the above the first term is $\dfrac{1}{3}$ and the common ratio is $ - \dfrac{1}{3}$ and substituting in the formula $\dfrac{{\text{a}}}{{1 - {\text{r}}}}$ ,
 ${\text{y}} = {4^{\left( {\dfrac{{\dfrac{1}{3}}}{{1 + \dfrac{1}{3}}}} \right)}}$
And we get the ${\text{y}}$ value as $\sqrt 2 $
Now the value ${\text{z}}$ is taken as per the derivation,
 \[{\text{z}} = \sum\limits_{{\text{r}} = 1}^\infty {{{(1 + i)}^{ - {\text{r}}}}} \]
If we use the limits in ${\text{z}}$ that ${\text{r}} = 1$ up to infinity this above equation becomes,
 \[{\text{z}} = {(1 + {\text{i}})^{ - 1}} + {(1 + {\text{i}})^{ - 2}} + {(1 + {\text{i}})^{ - 3}}...\infty \]
Since it is a G.P with a common ratio of $\dfrac{1}{{1 + {\text{i}}}}$ and the first term is $\dfrac{1}{{1 + {\text{i}}}}$ ,
Finally, by solving in the formula we get,
 ${\text{z}} = \dfrac{{\dfrac{1}{{1 + {\text{i}}}}}}{{1 - \dfrac{1}{{1 + {\text{i}}}}}}$
By simplifying the above we get,
 ${\text{z}} = - {\text{i}}$
Thus the value of ${\text{z}}$ now becomes $ - {\text{i}}$ .
Now substitute all the values in the ${\text{x}} + {\text{yz}}$ we get,
 $ \Rightarrow 3 - \sqrt 2 i$
Then the ${\text{arg}}({\text{x}} + {\text{yz}})$ will be,
 ${\text{arg}}({\text{x}} + {\text{yz}}) = \pi - {\tan ^{ - 1}}\left( {\dfrac{{ - \sqrt 2 }}{3}} \right)$
Hence, the value of ${\text{arg}}({\text{x}} + {\text{yz}})$ is $\pi - {\tan ^{ - 1}}\left( {\dfrac{{ - \sqrt 2 }}{3}} \right)$

Note: In the argument formula we have to compare the values of z to get the x and y values. Since it is in a geometric series we are using a common ratio. For the arithmetic series, we will be using the common difference.