Question

Evaluate ${{i}^{37}}+\dfrac{1}{{{i}^{67}}}$ .

Answer 1

Hint: To evaluate ${{i}^{37}}+\dfrac{1}{{{i}^{67}}}$ , we will express the powers of i in terms of 4 so that we can easily simplify the even powers of i. We will write 37 as $\left( 4\times 9 \right)+1$ and 67 as $\left( 4\times 16 \right)+3$ and substitute it in the given expression. We will then use the exponential rules ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ and ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ to simplify.

Complete step by step solution:
We have to evaluate ${{i}^{37}}+\dfrac{1}{{{i}^{67}}}$ . We have to express the powers of i in terms of 4 so that we can easily simplify the even powers of i.
Let us write 37 as $\left( 4\times 9 \right)+1$ and 67 as $\left( 4\times 16 \right)+3$ .
\[\Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}={{i}^{\left( 4\times 9 \right)+1}}+\dfrac{1}{{{i}^{\left( 4\times 16 \right)+3}}}\]
We know that ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ . Therefore, we can write the terms in i as
\[\Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}={{i}^{\left( 4\times 9 \right)}}\times i+\dfrac{1}{{{i}^{\left( 4\times 16 \right)}}\times {{i}^{3}}}\]
We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ . Therefore, we can write the above equation as
\[\Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}={{\left( {{i}^{4}} \right)}^{9}}\times i+\dfrac{1}{{{\left( {{i}^{4}} \right)}^{16}}\times {{i}^{3}}}...\left( i \right)\]
We know that $i=\sqrt{-1}$ . Let us square both the sides.
 $\begin{align}
  & {{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1 \\
 & \Rightarrow {{i}^{4}}={{\left( {{i}^{2}} \right)}^{2}}={{\left( -1 \right)}^{2}}=1 \\
\end{align}$
Let us substitute the above value in equation (i).
\[\begin{align}
  & \Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}={{1}^{9}}\times i+\dfrac{1}{{{1}^{16}}\times {{i}^{3}}} \\
 & \Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=1\times i+\dfrac{1}{1\times {{i}^{3}}} \\
 & \Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=i+\dfrac{1}{{{i}^{3}}} \\
\end{align}\]
We know that ${{i}^{3}}=\left( {{i}^{2}} \right)\times i=-i$ . Therefore, we can write the above equation as
\[\begin{align}
  & \Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=i+\dfrac{1}{-i} \\
 & \Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=i-\dfrac{1}{i} \\
\end{align}\]
Let us multiply and dive the second term of RHS by i.
\[\begin{align}
  & \Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=i-\dfrac{1}{i}\times \dfrac{i}{i} \\
 & \Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=i-\dfrac{i}{{{i}^{2}}} \\
\end{align}\]
We know that ${{i}^{2}}=-1$ . The, we can write the above equation as
\[\begin{align}
  & \Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=i-\dfrac{i}{-1} \\
 & \Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=i+\dfrac{i}{1} \\
 & \Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=i+i \\
\end{align}\]
Let us add the RHS.
\[\Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=2i\]

Therefore, the value of ${{i}^{37}}+\dfrac{1}{{{i}^{67}}}$ is $2i$.

Note: Students must know the value of I to find the powers of i. They must firstly change the big exponents to a power that can be easily evaluated with i. We can also evaluate ${{i}^{37}}+\dfrac{1}{{{i}^{67}}}$ in the following manner.
Let us write 37 as $36+1$ and 67 as $66+1$ .
$\Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}={{i}^{36+1}}+\dfrac{1}{{{i}^{66+1}}}$
We know that ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ . Therefore, we can write the terms in i as
$\Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}={{i}^{36}}\times i+\dfrac{1}{{{i}^{66}}\times i}$
Now, we can write 36 as $4\times 9$ and 66 as $2\times 33$ .
$\Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}={{i}^{4\times 9}}\times i+\dfrac{1}{{{i}^{2\times 33}}\times i}$
We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ . Therefore, we can write the above equation as
$\Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}={{\left( {{i}^{4}} \right)}^{9}}\times i+\dfrac{1}{{{\left( {{i}^{2}} \right)}^{33}}\times i}$
We know that ${{i}^{2}}=-1$ and ${{i}^{4}}=1$ .
$\Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}={{1}^{9}}\times i+\dfrac{1}{{{\left( -1 \right)}^{33}}\times i}$
We know that -1 raised to an odd power is always -1.
$\begin{align}
  & \Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=1\times i+\dfrac{1}{-1\times i} \\
 & \Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=i-\dfrac{1}{i} \\
\end{align}$
Let us multiply and dive the second term of RHS by i.
\[\begin{align}
  & \Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=i-\dfrac{1}{i}\times \dfrac{i}{i} \\
 & \Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=i-\dfrac{i}{{{i}^{2}}} \\
\end{align}\]
We know that ${{i}^{2}}=-1$ . The, we can write the above equation as
\[\begin{align}
  & \Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=i-\dfrac{i}{-1} \\
 & \Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=i+\dfrac{i}{1} \\
 & \Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=i+i \\
\end{align}\]
Let us add the RHS.
\[\Rightarrow {{i}^{37}}+\dfrac{1}{{{i}^{67}}}=2i\]