Question

Find the value of \[{\left( {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^2}\].
A. 4
B. \[ - 4\]
C. 5
D. \[ - 5\]

Answer 1

Hint In the given question, one expression is given which contains the imaginary number \[i\]. We will rewrite the \[{i^{19}}\] such that \[{i^{16}} \cdot {i^3}\] and plug the value of \[{i^{16}}\] and \[{i^3}\]. In the second term, first we will apply the indices property that is \[{\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}\]. Then we simplify the denominator only because \[{1^n} = 1\] for any natural number \[n\]. Again rewrite \[{i^{25}}\] as \[{i^{24}} \cdot i\] and put \[{i^{24}} = 1\]. Then rewrite 1 as \[{i^4}\]. Then cancel out the common term and simplify the expression \[{\left( { - i + {i^3}} \right)^2}\]. We know that \[{i^3} = - i\]. So the expression becomes \[{\left( { - 2i} \right)^2}\]. Now we can easily derive the value \[{\left( { - 2i} \right)^2}\].

Formula used:
The values of various powers of imaginary number \[i = \sqrt { - 1} \] are:
\[i = \sqrt { - 1} \]
\[{i^2} = - 1\]
\[{i^3} = - i\]
\[{i^4} = 1\]

Complete step by step solution:
The given expression is \[{\left( {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^2}\].
Let’s simplify the given expression.
\[{\left( {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^2} = {\left( {{i^{16}} \times {i^3} + \left( {\dfrac{{{1^{25}}}}{{{i^{24}} \times i}}} \right)} \right)^2}\]
\[ \Rightarrow \]\[{\left( {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^2} = {\left( {\left( 1 \right) \cdot \left( { - i} \right) + \left( {\dfrac{1}{{\left( 1 \right) \cdot i}}} \right)} \right)^2}\] [Since, \[{i^4} = 1\] and \[{i^3} = - i\]]
\[ \Rightarrow \]\[{\left( {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^2} = {\left( { - i + \dfrac{1}{i}} \right)^2}\]
\[ \Rightarrow \]\[{\left( {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^2} = {\left( { - i + \dfrac{{{i^4}}}{i}} \right)^2}\] [Since, \[{i^4} = 1\]]
\[ \Rightarrow \]\[{\left( {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^2} = {\left( { - i + {i^3}} \right)^2}\]
\[ \Rightarrow \]\[{\left( {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^2} = {\left( { - i - i} \right)^2}\] [Since, \[{i^3} = - i\]]
\[ \Rightarrow \]\[{\left( {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^2} = {\left( { - 2i} \right)^2}\]
\[ \Rightarrow \]\[{\left( {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^2} = 4{i^2}\]
\[ \Rightarrow \]\[{\left( {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^2} = 4\left( { - 1} \right)\] [Since, \[{i^2} = - 1\]]
\[ \Rightarrow \]\[{\left( {{i^{19}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right)^2} = - 4\]

Hence the correct option is option B.

Note: Students are often confused with the values of the powers of the imaginary number \[i = \sqrt { - 1} \]. To find the values of powers correctly, multiply the previous value by \[i\]. If the power of \[i\] is a multiple of 4 then value of term will be 1.