Question

Let $z = \cos \theta + i\sin \theta$. Then the value of $\sum\limits_{m = 1}^{15} {Im\left( {{z^{2m - 1}}} \right)}$ at $\theta {\text{ }} = {\text{ }}{2^0}$ is
${\text{A}}{\text{. }}\dfrac{1}{{\sin {2^0}}} \\\ {\text{B}}{\text{. }}\dfrac{1}{{3\sin {2^0}}} \\\ {\text{C}}{\text{. }}\dfrac{1}{{2\sin {2^0}}} \\\ {\text{D}}{\text{. }}\dfrac{1}{{4\sin {2^0}}} \\\$

Answer 1

Write the complex number in exponential form and substitute it in the given summation and it will be seen that the term inside the summation will form a geometric progression so we will solve the series with the help of suitable formulas and hence we will solve the summation and find its final value.

Complete step-by-step answer :
We can represent a complex number as:
$z = \cos \theta + i\sin \theta = {e^{i\theta }}$.
We need to find out the value of $\sum\limits_{m = 1}^{15} {Im\left( {{z^{2m - 1}}} \right)}$. We can substitute the value of z as ${e^{i\theta }}$.
$\sum\limits_{m = 1}^{15} {Im\left( {{z^{2m - 1}}} \right)} = \sum\limits_{m = 1}^{15} {Im\left( {{{\left( {{e^{i\theta }}} \right)}^{2m - 1}}} \right)} \\\ \sum\limits_{m = 1}^{15} {Im\left( {{z^{2m - 1}}} \right)} = \sum\limits_{m = 1}^{15} {Im\left( {{{\left( e \right)}^{i(2m - 1)\theta }}} \right)} \\\ \sum\limits_{m = 1}^{15} {Im\left( {{z^{2m - 1}}} \right)} = Im\sum\limits_{m = 1}^{15} {\left( {{{\left( e \right)}^{i(2m - 1)\theta }}} \right)} \\\$
The series formed inside the summation is in G.P, whose first term can be found by substituting the initial value of m i.e. 1 and therefore,
$a = {e^{i\theta }}$
Now, to find the common ratio of this G.P, we have to take the ratio of second term and first term.
$r = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{e^{i3\theta }}}}{{{e^{i\theta }}}} = {e^{i2\theta }}$
Now, to calculate the sum of a geometric progression we can use the formula of sum of G.P.
The formula to find sum of a geometric progression is represented as:
$S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$
Now, we will substitute this equation in the above equation that has been obtained and by substituting, we get,
$\operatorname{Im} (\dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}})$ and here, we substitute the value of n as 15.
= $\operatorname{Im} \left( {\dfrac{{a\left( {{r^{15}} - 1} \right)}}{{r - 1}}} \right)$, where the values of a and r are,
$a = {e^{i\theta }}$
$r = {e^{i2\theta }}$
Now, in the above equation we substitute the values of a and r,
$= \operatorname{Im} \left( {\dfrac{{{e^{i\theta }}\left( {{e^{i2\theta .15}} - 1} \right)}}{{{e^{i2\theta }} - 1}}} \right) \\\ = \operatorname{Im} \left( {\dfrac{{{e^{i\theta }}\left( {{e^{i(15)\theta }}\left( {{e^{i(15)\theta }} - {e^{ - i(15)\theta }}} \right)} \right)}}{{{e^{i\theta }}\left( {{e^{i\theta }} - {e^{ - i\theta }}} \right)}}} \right) \\\$,
Here a formula has to be used:
$\sin \theta = \dfrac{{{e^{i\theta }} - {e^{ - i\theta }}}}{{2i}} \\\ 2i.\sin \theta = {e^{i\theta }} - {e^{ - i\theta }} \\\ \sin \left( {n\theta } \right) = \dfrac{{{e^{in\theta }} - {e^{ - in\theta }}}}{{2i}} \\\ 2i.\sin \left( {n\theta } \right) = {e^{in\theta }} - {e^{ - in\theta }} \\\$
So, by substituting the values of these formula in the above equation, we get,
$= \operatorname{Im} \left( {\dfrac{{{e^{15i\theta }}\left( {2i.\sin \left( {15\theta } \right)} \right)}}{{2i.\sin \left( \theta \right)}}} \right) \\\ = \operatorname{Im} \left( {\dfrac{{{e^{15i\theta }}\left( {\sin \left( {15\theta } \right)} \right)}}{{\sin \left( \theta \right)}}} \right) \\\ = \operatorname{Im} \left( {\dfrac{{\left( {\cos \left( {15\theta } \right) + i\left( {\sin \left( {15\theta } \right)} \right)} \right)\left( {\sin \left( {15\theta } \right)} \right)}}{{\sin \left( \theta \right)}}} \right) \\\ = \dfrac{{\sin \left( {15\theta } \right)}}{{\sin \left( \theta \right)}}\operatorname{Im} \left( {\cos \left( {15\theta } \right) + i\left( {\sin \left( {15\theta } \right)} \right)} \right) \\\ = \dfrac{{\sin \left( {15\theta } \right)}}{{\sin \left( \theta \right)}}.\sin \left( {15\theta } \right) \\\ = \dfrac{{{{\sin }^2}\left( {15\theta } \right)}}{{\sin \left( \theta \right)}} \\\$,

Now, it is given in the question that the value of theta is ${2^0}$.
$= \dfrac{{{{\sin }^2}\left( {{{30}^0}} \right)}}{{\sin \left( {{2^0}} \right)}} \\\ (\because {\text{sin3}}{{\text{0}}^0} = \dfrac{1}{2}) \\\$
Hence on putting value we get,
$\dfrac{1}{{4\sin {2^0}}}$
Therefore, the final answer is $\dfrac{1}{{4\sin {2^0}}}$ and the correct option is: D.

Note : Whenever we get this type of question the key concept of solving is we have to remember all the formulae of complex number and GP like $S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$,
$z = \cos \theta + i\sin \theta = {e^{i\theta }}$ These formulae help in solving these types of questions easily.