QJEEAdv15Ma8082: JEE Advanced 2015 Mathematics - Class 11 JEEAdv15Ma8082 Solution: For any integer k, let , where . The value of the expression is

Given, ${a_k} = \cos \left( {{{k\pi } \over 7}} \right) + i\sin \left( {{{k\pi } \over 7}} \right) = {e^{{{k\pi } \over 7}i}}$

We have to find ${{\sum\limits_{k = 1}^{12} {\left| {{a_{k + 1}} - {a_k}} \right|} } \over {\sum\limits_{k = 1}^3 {\left| {{a_{4k - 1}} - {a_{4k - 2}}} \right|} }}$

${a_{k + 1}} = \cos \left( {{{k + 1} \over 7}} \right)\pi + i\sin \left( {{{k + 1} \over 7}} \right)\pi = {e^{i\left( {{{k + 1} \over 7}} \right)\pi }}$

$\therefore$ ${a_{k + 1}} - {a_k} = {e^{\left( {{{k + 1} \over 7}} \right)\pi i}} - {e^{{{k\pi } \over 7}i}}$

$= {e^{{{k\pi } \over 7}i}}\,.\,\,{e^{{\pi \over 7}i}} - {e^{{{k\pi } \over 7}i}}$

$= {e^{{{k\pi } \over 7}i}}\left( {{e^{{\pi \over 7}i}} - 1} \right)$

$\therefore$ $\left| {{a_{k + 1}} - {a_k}} \right| = \left| {{e^{{{k\pi } \over 7}i}}\left( {{e^{{\pi \over 7}i}} - 1} \right)} \right|$

$= \left| {{e^{{{k\pi } \over 7}i}}} \right|\left| {{e^{{\pi \over 7}i}} - 1} \right|$

$= \left| {{e^{{\pi \over 7}i}} - 1} \right|$

If $z = {e^{i\theta }} = \cos \theta + i\sin \theta$

then $\left| z \right| = \sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta } = 1$

that is why $\left| {{e^{{{k\pi } \over 7}i}}} \right| = 1$

Now, ${a_{4k - 1}} = {e^{\left( {{{4k - 1} \over 7}} \right)\pi i}}$

${a_{4k - 2}} = {e^{\left( {{{4k - 2} \over 7}} \right)\pi i}}$

${a_{4k - 1}} - {a_{4k - 2}} = {e^{\left( {{{4k - 1} \over 7}} \right)\pi i}} - {e^{\left( {{{4k - 2} \over 7}} \right)\pi i}}$

$= {e^{{{4k\pi } \over 7}i}}\,.\,{e^{ - {\pi \over 7}i}} - {e^{{{4k\pi } \over 7}}}\,.\,{e^{ - {{2\pi } \over 7}i}}$

$= {e^{{{4k\pi } \over 7}i}}\left( {{e^{ - {\pi \over 7}i}} - {e^{ - {{2\pi } \over 7}i}}} \right)$

$\therefore$ $\left| {{a_{4k - 1}} - {a_{4k - 2}}} \right| = \left| {{e^{{{4k\pi } \over 7}i}}} \right|\left| {{e^{ - {\pi \over 7}i}} - {e^{ - {{2\pi } \over 7}i}}} \right|$

$= \left| {{e^{ - {\pi \over 7}i}} - {e^{ - {{2\pi } \over 7}i}}} \right|$

Now, ${{\sum\limits_{k = 1}^{12} {\left| {{a_{k + 1}} - {a_k}} \right|} } \over {\sum\limits_{k = 1}^3 {\left| {{a_{4k - 1}} - {a_{4k - 2}}} \right|} }}$

$= {{12\left| {{e^{{\pi \over 7}i}} - 1} \right|} \over {3\left| {{e^{ - {\pi \over 7}i}} - {e^{ - {{2\pi } \over 7}i}}} \right|}}$

$= 4\,.\,{{\left| {{e^{{\pi \over 7}i}} - 1} \right|} \over {\left| {\left( {{e^{ - {\pi \over 7}i}}1 - {e^{ - {\pi \over 7}i}}} \right)} \right|}}$

$= 4\,.\,{{\left| {{e^{{\pi \over 7}i}} - 1} \right|} \over {\left| {1 - {e^{ - {\pi \over 7}i}}} \right|}}$

$= 4\,.\,{{\left| {{e^{{\pi \over 7}i}}\left( {1 - {e^{ - {\pi \over 7}i}}} \right)} \right|} \over {\left| {1 - {e^{ - {\pi \over 7}i}}} \right|}}$

$= 4\,.\,\left| {{e^{{\pi \over 7}i}}} \right|$

= 4 as $\left| {{e^{{\pi \over 7}i}}} \right| = 1$