Question

Find the value of ${\left[ {{i^{17}} + \dfrac{1}{{{i^{315}}}}} \right]^9}$.
A) $32i$
B) $ - 512$
C) $512$
D) $512i$

Answer 1

Hint: In order to find the value of the given equation, we should know what is the word $i$ here. Basically, $i$ stands for iota here, whose value is $\sqrt { - 1} $ and it is used in complex numbers. Since, earlier we knew that we cannot find the root of negative numbers, but actually, we can in terms of iota.

Complete step by step solution:
Since, we know that $\sqrt { - 1} = i$, on squaring both the sides we get:
 $
  \sqrt { - 1} = i \\
    \Rightarrow {\left( {\sqrt { - 1} } \right)^2} = {i^2} \\
    \Rightarrow {i^2} = - 1 \\
 $
Similarly, multiplying iota to the above term and we get:
$
  {i^2} \times i = \left( { - 1} \right)i \\
    \Rightarrow {i^3} = \left( { - i} \right) \\
 $
Again, we are multiplying iota:
$
  {i^3} \times i = \left( { - i} \right)i \\
    \Rightarrow {i^4} = - \left( {{i^2}} \right) \\
    \Rightarrow {i^4} = 1 \\
 $
But if the power of integrals is greater then $4$, then it becomes difficult to expand the terms.
So, suppose we have ${i^n}$, so dividing $n$ by $4$, we obtain a remainder $r$ and let the quotient be $m$:
Overall, the equation becomes:
$n = 4m + r$
Substituting the value of $n = 4m + r$ in ${i^n}$ and we get:
${i^n} = {i^{4m + r}}$
From the law of radicals, we know that ${p^a}.{p^b} = {p^{a + b}}$. So, applying this above, we can write the equation as:
${i^n} = {i^{4m + r}} = {i^{4m}}.{i^r}$
But we obtained earlier that ${i^4} = 1$ so from this, we get:
${i^{4m}}.{i^r} = {\left( {{i^4}} \right)^m}.{i^r} = {1^m}.{i^r} = {i^r}$
That means, ${i^{4m + r}} = {i^r}$.

So, applying this equation, we would solve the question we are given that is ${\left[ {{i^{17}} + \dfrac{1}{{{i^{315}}}}} \right]^9}$.
Solving each operand separately, initiating with ${i^{17}}$:
From the obtained formula above, ${i^{17}}$ can be written as:
$
  {i^{17}} = {i^{16 + 1}} = {i^{4 \times 4}}.i \\
    \Rightarrow {i^{17}} = i \\
 $
Similarly, solving for the second operand $\dfrac{1}{{{i^{315}}}}$, we get:
$\dfrac{1}{{{i^{315}}}} = \dfrac{1}{{{i^{312 + 3}}}} = \dfrac{1}{{{i^{4 \times 78 + 3}}}} = \dfrac{1}{{{i^3}}}$
Now, substituting these values in the original equation and, we get:
$
  {\left[ {{i^{17}} + \dfrac{1}{{{i^{315}}}}} \right]^9} \\
    \Rightarrow {\left[ {i + \dfrac{1}{{{i^3}}}} \right]^9} \\
 $
Since, we know that ${i^3} = \left( { - i} \right)$, so the equation becomes:
${\left[ {i + \dfrac{1}{{{i^3}}}} \right]^9} = {\left[ {i - \dfrac{1}{i}} \right]^9}$
Simplifying the values inside the parenthesis by multiplying and dividing the first operand with iota to have a common denominator and, we get:
$
  {\left[ {i + \dfrac{1}{{{i^3}}}} \right]^9} \\
    \Rightarrow {\left[ {\dfrac{{i \times i}}{i} - \dfrac{1}{i}} \right]^9} \\
    \Rightarrow {\left[ {\dfrac{{{i^2}}}{i} - \dfrac{1}{i}} \right]^9} \\
    \Rightarrow {\left[ {\dfrac{{{i^2} - 1}}{i}} \right]^9} \\
 $
Since, we know that ${i^2} = - 1$, so replacing $ - 1$ with ${i^2}$from the above equation, and we get:
$
  {\left[ {\dfrac{{{i^2} - 1}}{i}} \right]^9} \\
    \Rightarrow {\left[ {\dfrac{{{i^2} + \left( { - 1} \right)}}{i}} \right]^9} \\
    \Rightarrow {\left[ {\dfrac{{{i^2} + {i^2}}}{i}} \right]^9} \\
    \Rightarrow {\left[ {\dfrac{{2{i^2}}}{i}} \right]^9} \\
 $
Cancelling the common iota terms and we get:
\[{\left[ {\dfrac{{2{i^2}}}{i}} \right]^9} = {\left[ {2i} \right]^9}\]
Expanding the brackets:
${\left[ {2i} \right]^9} = {\left( 2 \right)^9}.{i^9} = 512{i^9}$
Again, using ${i^{4m + r}} = {i^r}$, we can write ${i^9}$ as ${i^9} = {i^{8 + 1}} = {i^{4.2}}.i = i$.
And, our equation becomes:
$512{i^9} = 512i$, which is our required value and matches with option 4.
Therefore, ${\left[ {{i^{17}} + \dfrac{1}{{{i^{315}}}}} \right]^9}$ is equal to option 4 that is $512i$.

Note:
> It’s important to use ${i^{4m + r}} = {i^r}$, for $n > 4$. As it’s not possible to multiply iota to each term, especially for long terms.
> Always prefer solving step by step rather than solving at once, otherwise it would lead to error for larger numbers.