Question

Find the value of ${\left[ {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right]^2}$ where, $i = \sqrt { - 1}$.
(a) Cannot be determined
(b) $- 4$
(c) $i$
(d) $4$

Answer 1

We will be going to use the most curious concept of complex relations that to be recognised by an imaginary unit ‘$i$’ Mathematically, the imaginary identity is noted as $i = \sqrt { - 1}$. As a result, substituting the respective complex identity in the expression, the desired value is obtained.

Complete step by step solution:
The condition is related to the complex number as there exists the parameter ‘i’ where the instance ‘i’ is an ‘imaginary unit’ $i = \sqrt { - 1}$ respectively.
Here, we have given the expression as ${\left[ {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right]^2}$
Since, using the rules of indices, adjusting the powers so that we get the respective output as per the terminology,
${\left[ {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right]^2} = {\left[ {{i^3}{i^{16}} + \dfrac{1}{{{i^{24}}i}}} \right]^2}$
(Calculations of complex identity ‘i’):
Now, since further expanding the term $i = \sqrt { - 1}$ to achieve the desire value, we get
$\because$We know that, $i = \sqrt { - 1}$

$$\therefore {i^2} = {\left( {\sqrt { - 1} } \right)^2} = - 1 \\\ {i^3} = {i^2}i = - i \\\ {i^4} = {i^3}i = ( - i)(i) = - {i^2} = 1 \\\ {i^5} = {i^4}i = (1)(i) = i \\\ {i^6} = {i^5}i = (i)(i) = {i^2} = - 1 \\\ {i^7} = {i^6}i = ( - 1)(i) = - i \\\ {i^8} = {i^7}i = ( - i)(i) = - {i^2} = 1 \\\ {i^9} = {i^8}i = (1)(i) = i \\\ {i^{10}} = {i^9}i = (i)(i) = {i^2} = - 1 \\\ {i^{11}} = {i^{10}}i = ( - 1)(i) = - i \\\ {i^{12}} = {i^{11}}i = ( - i)(i) = {i^2} = - 1 \\\ {i^{13}} = {i^{12}}i = ( - 1)(i) = - i \\\ {i^{14}} = {i^{13}}i = ( - i)(i) = - {i^2} = 1 \\\ {i^{15}} = {i^{14}}i = (1)(i) = i \\\ {i^{16}} = {i^{15}}i = (i)(i) = {i^2} = - 1 \\\ {i^{17}} = {i^{16}}i = ( - 1)(i) = - i \\\ {i^{18}} = {i^{17}}i = ( - i)(i) = - {i^2} = 1 \\\ {i^{19}} = {i^{18}}i = (1)(i) = i \\\ {i^{20}} = {i^{19}}i = (i)(i) = {i^2} = - 1 \\\ {i^{21}} = {i^{20}}i = ( - 1)(i) = - i \\\ {i^{22}} = {i^{21}}i = ( - i)(i) = - {i^2} = 1 \\\ {i^{23}} = {i^{22}}i = (1)i = i \\\ {i^{24}} = {i^{23}}i = (i)i = {i^2} = - 1 \\\ {i^{25}} = {i^{24}}i = ( - 1)i = - i \\\$$

From the above assumptions, we get
${\left[ {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right]^2} = {\left( { - i \times - 1 + \dfrac{1}{{ - i}}} \right)^2}$
Hence, the equation becomes,
${\left[ {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right]^2} = {\left( {i - \dfrac{1}{i}} \right)^2}$
Simplifying the above equation as per the algebraic expansion formula ${(a - b)^2} = {a^2} - 2ab - {b^2}$, we get
${\left[ {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right]^2} = {i^2} - 2i\dfrac{1}{i} + \dfrac{1}{{{i^2}}}$
Substituting the respective calculation that is ${i^2} = - 1$, we get
${\left[ {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right]^2} = - 1 - 2 - 1$
Hence, the required value for the expression is,
${\left[ {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right]^2} = - 4$
$\therefore$The correct option is option (b).

Note:
One must know the complex identities for such kinds of complications in the problem such as ${i^2}$, ${i^3}$, ${i^4}$, etc… where, $i = \sqrt { - 1}$. Since, we also know that, $1 + \omega + {\omega ^2} = 0$ and ${\omega ^3} = 1$. As a result, know the mug-up the algebraic identities also, known as algebraic expression such as ${(a + b)^2} = {a^2} + 2ab + {b^2}$, ${(a - b)^2} = {a^2} - 2ab + {b^2}$, etc… As a result, formulate the solutions so as to get the absolute value.