Question

For any integer $k,$ let ${{\rm{\alpha }}_k} = \cos \dfrac{{k\pi }}{7} + i\sin \dfrac{{k\pi }}{7},$where $i = \sqrt { - 1}$. The value of expression $\dfrac{{\sum\nolimits_{k = 1}^{12} {\left| {{\alpha _{k + 1}} - {\alpha _k}} \right|} }}{{\sum\nolimits_{k = 1}^3 {\left| {{\alpha _{4k - 1}} - {\alpha _{4k - 2}}} \right|} }}$ is

Answer 1

Though this expression looks quite frightening, complex and difficult, it is really easy to solve. We just have to make the substitution of the sine and cosine defined for the ${\alpha _k}$and then just substitute the required values and we are going to get the answer.

Formula Used:
We are going to use the Euler’s formula for making the required substitutions, which is:
${e^{ix}} = \cos x + i\sin x$ …(i)

Complete step by step solution:
This question seems quite lengthy and complex as it includes summation combined with sine and cosine and to make it all worse, the Euler’s constant. And it is if we directly start putting in the values. But, if we closely observe the value of ${\alpha _k}$, we see that it can be replaced by a single function if we know the correct formula and know how to mold it according to the given parameters. And when the substitution is done, it becomes simple as a bit.
In this question, ${{\rm{\alpha }}_k} = \cos \dfrac{{k\pi }}{7} + i\sin \dfrac{{k\pi }}{7},$and also,
${e^{ix}} = \cos x + i\sin x$
If we put $x = \dfrac{{k\pi }}{7}$, we get,
${e^{i\dfrac{{k\pi }}{7}}} = \cos \dfrac{{k\pi }}{7} + i\sin \dfrac{{k\pi }}{7}$
And now, we can substitute this ${e^{i\dfrac{{k\pi }}{7}}}$ into ${\alpha _k}$ , and we have:
${\alpha _k} = {e^{i\dfrac{{k\pi }}{7}}}$
So, ${\alpha _{k + 1}} = {e^{i\left( {\dfrac{{k + 1}}{7}} \right)\pi }}$
and similarly, ${\alpha _{4k - 1}} = {e^{i\left( {\dfrac{{4k - 1}}{7}} \right)\pi }}$
and finally, ${\alpha _{4k - 2}} = {e^{i\left( {\dfrac{{4k - 2}}{7}} \right)\pi }}$

Now, let’s substitute the alphas into the problem equation, and we get:
$\dfrac{{\sum\nolimits_{k = 1}^{12} {\left| {{\alpha _{k + 1}} - {\alpha _k}} \right|} }}{{\sum\nolimits_{k = 1}^3 {\left| {{\alpha _{4k - 1}} - {\alpha _{4k - 2}}} \right|} }} = \dfrac{{\sum\nolimits_{k = 1}^{12} {\left| {{e^{i\left( {\dfrac{{k + 1}}{7}} \right)\pi }} - {e^{i\left( {\dfrac{k}{7}} \right)\pi }}} \right|} }}{{\sum\nolimits_{k = 1}^3 {\left| {{e^{i\left( {\dfrac{{4k - 1}}{7}} \right)\pi }} - {e^{i\left( {\dfrac{{4k - 2}}{7}} \right)\pi }}} \right|} }}$
Taking ${{\rm{e}}^{{\rm{i}}\left( {\dfrac{{{\rm{k + 1}}}}{{\rm{7}}}} \right){\rm{\pi }}}}$common from the numerator and ${e^{i\left( {\dfrac{{4k - 2}}{7}} \right)\pi }}$common from the denominator, we get:
$\dfrac{{\sum\nolimits_{k = 1}^{12} {\left| {{\alpha _{k + 1}} - {\alpha _k}} \right|} }}{{\sum\nolimits_{k = 1}^3 {\left| {{\alpha _{4k - 1}} - {\alpha _{4k - 2}}} \right|} }} = \dfrac{{\sum\nolimits_{k = 1}^{12} {\left| {{e^{i\left( {\dfrac{k}{7}} \right)\pi }}} \right|\left| {{e^{\dfrac{{i\pi }}{7}}} - 1} \right|} }}{{\sum\nolimits_{k = 1}^3 {\left| {{e^{i\left( {\dfrac{{4k - 2}}{7}} \right)\pi }}} \right|\left| {{e^{\dfrac{{i\pi }}{7}}} - 1} \right|} }} = \dfrac{{\sum\nolimits_{k = 1}^{12} {\left| {{e^{i\left( {\dfrac{k}{7}} \right)\pi }}} \right|} }}{{\sum\nolimits_{k = 1}^3 {\left| {{e^{i\left( {\dfrac{{4k - 2}}{7}} \right)\pi }}} \right|} }}$ …(ii)

Now, let’s go back to the Euler’s formula:
${e^{ix}} = \cos x + i\sin x$
If we take modulus on both the sides of the equation, we get:
$\left| {{e^{ix}}} \right| = \left| {\cos x + i\sin x} \right|$
$\left| {{e^{ix}}} \right| = \sqrt {{{\cos }^2}x + {{\sin }^2}x} = 1$
Hence, $\left| {{e^{ix}}} \right| = 1$
So, if we substitute this value into (ii), we have:
$\dfrac{{\sum\nolimits_{k = 1}^{12} {\left| {{\alpha _{k + 1}} - {\alpha _k}} \right|} }}{{\sum\nolimits_{k = 1}^3 {\left| {{\alpha _{4k - 1}} - {\alpha _{4k - 2}}} \right|} }} = \dfrac{{\sum\nolimits_{k = 1}^{12} 1 }}{{\sum\nolimits_{k = 1}^3 1 }}$
or, $\dfrac{{\sum\nolimits_{k = 1}^{12} {\left| {{\alpha _{k + 1}} - {\alpha _k}} \right|} }}{{\sum\nolimits_{k = 1}^3 {\left| {{\alpha _{4k - 1}} - {\alpha _{4k - 2}}} \right|} }} = \dfrac{{12}}{3} = 4$ (applying the definition of summation)

Hence, $\dfrac{{\sum\nolimits_{k = 1}^{12} {\left| {{\alpha _{k + 1}} - {\alpha _k}} \right|} }}{{\sum\nolimits_{k = 1}^3 {\left| {{\alpha _{4k - 1}} - {\alpha _{4k - 2}}} \right|} }} = 4$

Note:
This question is one out of infinite examples in mathematics which seem really difficult, complex at the beginning, but when used the right pathway to solve them, they are nothing as difficult as such. The only thing that is needed for working like this is that we must know the formula to replace a complex expression and its related information. And then, we just apply the same and see the magic happen as we witnessed it here.