Question

If $x = {9^{\dfrac{1}{3}}}{9^{\dfrac{1}{9}}}{9^{\dfrac{1}{{27}}}}....\infty ,y = {4^{\dfrac{1}{3}}}{4^{\dfrac{{ - 1}}{9}}}{4^{\dfrac{1}{{27}}}}{4^{\dfrac{{ - 1}}{{81}}}}.....\infty ,{\text{ and z}} = \sum\limits_{r = 1}^\infty {{{\left( {1 + i} \right)}^{ - r}}} $, then arg (x + yz) is equal to
$\left( a \right)$ 0
$\left( b \right)\pi - {\tan ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{3}} \right)$
$\left( c \right) - {\tan ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{3}} \right)$
$\left( d \right) - {\tan ^{ - 1}}\left( {\dfrac{2}{{\sqrt 3 }}} \right)$

Answer 1

Hint: In this particular type of question use the concept that that if the multiplication of two or more numbers having same base is written as, ${a^p}{a^q} = {a^{p + q}}$so apply this in all the given equation, later on use the concept of infinite term G.P series formula which is given as, ${S_\infty } = \dfrac{a}{{1 - r}},r < 1$, and use the concept that arg (x + iy) = ${\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given equation
$x = {9^{\dfrac{1}{3}}}{9^{\dfrac{1}{9}}}{9^{\dfrac{1}{{27}}}}....\infty $.......................... (1)
$y = {4^{\dfrac{1}{3}}}{4^{\dfrac{{ - 1}}{9}}}{4^{\dfrac{1}{{27}}}}{4^{\dfrac{{ - 1}}{{81}}}}.....\infty $....................... (2)
And ${\text{z}} = \sum\limits_{r = 1}^\infty {{{\left( {1 + i} \right)}^{ - r}}} $.................... (3)
Now first solve equation (1) we have,
$x = {9^{\dfrac{1}{3}}}{9^{\dfrac{1}{9}}}{9^{\dfrac{1}{{27}}}}....\infty $
Now as we know that ${a^p}{a^q} = {a^{p + q}}$ so use this property we have,
$ \Rightarrow x = {9^{\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ......\infty }}$.................. (4)
Now as we see that $\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + .....\infty $ forms an infinite series G.P,
Where first term a = $\dfrac{1}{3}$, common ratio (r) = $\dfrac{{\dfrac{1}{9}}}{{\dfrac{1}{3}}} = \dfrac{1}{3}$, the sum of the infinite series G.P is given as,
$ \Rightarrow {S_\infty } = \dfrac{a}{{1 - r}},r < 1$
Now substitute the values we have,
$ \Rightarrow \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + .....\infty = \dfrac{{\dfrac{1}{3}}}{{1 - \dfrac{1}{3}}} = \dfrac{1}{2}$
Now from equation (4) we have,
$ \Rightarrow x = {9^{\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ......\infty }} = {9^{\dfrac{1}{2}}} = {\left( {{3^2}} \right)^{\dfrac{1}{2}}} = 3$
Now solve equation (2) we have,
$y = {4^{\dfrac{1}{3}}}{4^{\dfrac{{ - 1}}{9}}}{4^{\dfrac{1}{{27}}}}{4^{\dfrac{{ - 1}}{{81}}}}.....\infty $
Now as we know that ${a^p}{a^q} = {a^{p + q}}$ so use this property we have,
$ \Rightarrow y = {4^{\dfrac{1}{3} + \dfrac{{ - 1}}{9} + \dfrac{1}{{27}} + \dfrac{{ - 1}}{{81}}......\infty }}$.................. (5)
Now as we see that $\dfrac{1}{3} + \dfrac{{ - 1}}{9} + \dfrac{1}{{27}} + \dfrac{{ - 1}}{{81}} + ....\infty $ forms an infinite series G.P,
Where first term a = $\dfrac{1}{3}$, common ratio (r) = $\dfrac{{ - \dfrac{1}{9}}}{{\dfrac{1}{3}}} = - \dfrac{1}{3}$, the sum of the infinite series G.P is given as,
$ \Rightarrow {S_\infty } = \dfrac{a}{{1 - r}},r < 1$
Now substitute the values we have,
$ \Rightarrow \dfrac{1}{3} + \dfrac{{ - 1}}{9} + \dfrac{1}{{27}} + \dfrac{{ - 1}}{{81}} + ....\infty = \dfrac{{\dfrac{1}{3}}}{{1 - \dfrac{{ - 1}}{3}}} = \dfrac{1}{4}$
Now from equation (5) we have,
$ \Rightarrow y = {4^{\dfrac{1}{3} + \dfrac{{ - 1}}{9} + \dfrac{1}{{27}} + \dfrac{{ - 1}}{{81}}......\infty }} = {4^{\dfrac{1}{4}}} = {\left( {{2^2}} \right)^{\dfrac{1}{4}}} = {2^{\dfrac{1}{2}}} = \sqrt 2 $
Now solve equation (3) we have,
${\text{z}} = \sum\limits_{r = 1}^\infty {{{\left( {1 + i} \right)}^{ - r}}} $
Now expand this summation we have,
$ \Rightarrow z = \dfrac{1}{{1 + i}} + \dfrac{1}{{{{\left( {1 + i} \right)}^2}}} + \dfrac{1}{{{{\left( {1 + i} \right)}^3}}} + ..... + \infty $
Now as we see that $\dfrac{1}{{1 + i}} + \dfrac{1}{{{{\left( {1 + i} \right)}^2}}} + \dfrac{1}{{{{\left( {1 + i} \right)}^3}}} + ..... + \infty $ forms an infinite series G.P,
Where first term a = $\dfrac{1}{{1 + i}}$, common ratio (r) = $\dfrac{{\dfrac{1}{{{{\left( {1 + i} \right)}^2}}}}}{{\dfrac{1}{{\left( {1 + i} \right)}}}} = \dfrac{1}{{\left( {1 + i} \right)}}$, the sum of the infinite series G.P is given as,
$ \Rightarrow {S_\infty } = \dfrac{a}{{1 - r}},r < 1$
Now substitute the values we have,
\[ \Rightarrow z = \dfrac{1}{{1 + i}} + \dfrac{1}{{{{\left( {1 + i} \right)}^2}}} + \dfrac{1}{{{{\left( {1 + i} \right)}^3}}} + ..... + \infty = \dfrac{{\dfrac{1}{{1 + i}}}}{{1 - \dfrac{1}{{\left( {1 + i} \right)}}}}\]
Now simplify we have,
$ \Rightarrow z = \dfrac{{\dfrac{1}{{1 + i}}}}{{1 - \dfrac{1}{{\left( {1 + i} \right)}}}} = \dfrac{1}{i} = \dfrac{i}{{{i^2}}} = - i$, $\left[ {\because i = \sqrt { - 1} \Rightarrow {i^2} = - 1} \right]$
Now we have to find the argument of x + yz.
$ \Rightarrow x + yz = 3 + \sqrt 2 \left( { - i} \right) = 3 - i\sqrt 2 $
Now as we know that the arg (x + iy) = ${\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$
$ \Rightarrow \arg \left( {x + iy} \right) = \arg \left( {3 - i\sqrt 2 } \right) = {\tan ^{ - 1}}\left( {\dfrac{{ - \sqrt 2 }}{3}} \right)$
Now as we know that tan (-x) = tan (x) so we have,
$ \Rightarrow \arg \left( {x + iy} \right) = \arg \left( {3 - i\sqrt 2 } \right) = {\tan ^{ - 1}}\left( {\dfrac{{ - \sqrt 2 }}{3}} \right) = - {\tan ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{3}} \right)$
So this is the required answer.
Hence option (d) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the formula of infinite series of G.P which is stated above, then use this formula to simplify all the given equations as above, then calculate the value of x + yz as above and then take the argument of x + yz, we will get the required answer.