Question

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

Answer 1

Convert the complex number from given trigonometric form to Euler’s form in order to make it easier for calculation as $\cos \theta + i\sin \theta = {e^{i\theta }}$. In order to calculate the modulus of the function $\left| {\cos \theta + i\sin \theta } \right| = \left| {{e^{i\theta }}} \right|$ it will always be one. Hence, it is quiet clear that the modulus of complex number will always be 1 as by using the concept of $z = \sqrt {{x^2} + {y^2}}$. Hence, through the mentioned concept put the values in the above equation and do the summation of the above equation and calculate to obtain the final answer.

Complete step by step solution: As the given equation are ${\alpha _k} = \cos \dfrac{{k\pi }}{7} + i\sin \dfrac{{k\pi }}{7}$where $i = \sqrt { - 1}$. The expression is $\dfrac{{\sum\limits_{k = 1}^{12} {\left| {{\alpha _{k + 1}} - {\alpha _k}} \right|} }}{{\sum\limits_{k = 1}^3 {\left| {{\alpha _{4k - 1}} - {\alpha _{4k - 2}}} \right|} }}$

Hence, firstly convert all the complex number from given trigonometric form to Euler’s form, which is $\cos \theta + i\sin \theta = {e^{i\theta }}$

Hence, ${\alpha _k} = \cos \dfrac{{k\pi }}{7} + i\sin \dfrac{{k\pi }}{7}$can also be given as ${e^{i\dfrac{{k\pi }}{7}}}$and so calculating the value of all the expressions as

${\alpha _k} = {e^{i\dfrac{{k\pi }}{7}}}$

Similarly, ${\alpha _{k + 1}} = {e^{i\dfrac{{\left( {k + 1} \right)\pi }}{7}}}$ also calculate the terms for denominators values as ${\alpha _{4k - 1}} = {e^{i\dfrac{{\left( {4k - 1} \right)\pi }}{7}}}$ and for the another term will
be ${\alpha _{4k - 2}} = {e^{i\dfrac{{\left( {4k - 2} \right)\pi }}{7}}}$.

Hence, substitute the value in above equation of $\dfrac{{\sum\limits_{k = 1}^{12} {\left| {{\alpha _{k + 1}} - {\alpha _k}} \right|} }}{{\sum\limits_{k = 1}^3 {\left| {{\alpha _{4k - 1}} - {\alpha _{4k - 2}}} \right|} }}$ and simplify it

\Rightarrow $$$$\dfrac{{\sum\limits_{k = 1}^{12} {\left| {{e^{i\dfrac{{\left( {k + 1} \right)\pi }}{7}}} - {e^{i\dfrac{{\left( k \right)\pi }}{7}}}} \right|} }}{{\sum\limits_{k = 1}^3 {\left| {{e^{i\dfrac{{\left( {4k - 1} \right)\pi }}{7}}} - {e^{i\dfrac{{\left( {4k - 2} \right)\pi }}{7}}}} \right|} }}

Take the terms common from both numerator and denominator as

$\Rightarrow \dfrac{{\sum\limits_{k = 1}^{12} {\left| {{e^{i\dfrac{{\left( k \right)\pi }}{7}}}} \right|\left| {{e^{i\dfrac{\pi }{7}}} - 1} \right|} }}{{\sum\limits_{k = 1}^3 {\left| {{e^{i\dfrac{{\left( {4k - 2} \right)\pi }}{7}}}} \right|\left| {{e^{i\dfrac{\pi }{7}}} - 1} \right|} }}$

Cancel out the common terms from both numerator and denominator as

$\Rightarrow \dfrac{{\sum\limits_{k = 1}^{12} {\left| {{e^{i\dfrac{{\left( k \right)\pi }}{7}}}} \right|} }}{{\sum\limits_{k = 1}^3 {\left| {{e^{i\dfrac{{\left( {4k - 2} \right)\pi }}{7}}}} \right|} }}$

Hence, as discussed that for the concept of modulus of$\left| {\cos \theta + i\sin \theta } \right| = \left| {{e^{i\theta }}} \right|$ such term will always be $1$.

As $\left| {{e^{i\theta }}} \right| = \sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta } = \sqrt 1 = 1$

Hence, $= \dfrac{{\sum\limits_{k = 1}^{12} 1 }}{{\sum\limits_{k = 1}^3 1 }}$

As, we know that $\sum\limits_{i = 1}^n 1 = n$

So, applying the concept of summation in the above equation it can be simplified to

$= \dfrac{{12}}{3}$

Hence, on dividing the above values our answer will be obtained as

$= 4$

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

Note: A complex number is a number that can be expressed in the form $a{\text{ }} + {\text{ }}bi$, where a and b are real numbers, and i represents the imaginary unit, satisfying the equation ${i^2} = - 1$. Because no real number satisfies this equation, i is called an imaginary number.