Question

If $i=\sqrt{-1}$,${{i}^{2}}=-1$,${{i}^{3}}=-i$,${{i}^{4}}=1$ and so on, then the value of$\dfrac{({{i}^{368}}+{{i}^{370}})}{({{i}^{399}}+{{i}^{400}})}$ is
1. ${{i}^{368}}+{{i}^{399}}$
2. ${{i}^{372}}+{{i}^{400}}$
3. ${{i}^{368}}+{{i}^{370}}$
4.${{i}^{399}}+{{i}^{400}}$

Answer 1

Hint: In this problem, all the powers are for $i$. So, first we will write all these powers in the form of ${{i}^{4}}$ and then we will evaluate the given expression. In this way we can easily solve this problem. We will be using the following properties of complex numbers to solve this problem.

Complete Step-by-Step solution:
$\begin{align}
  & \Rightarrow i=\sqrt{-1} \\
 & \Rightarrow {{i}^{2}}=-1 \\
 & \Rightarrow {{i}^{3}}=({{i}^{2}}\times i)=(-1\times i) \\
 & \Rightarrow {{i}^{3}}=-i \\
 & \Rightarrow {{\left( {{i}^{2}} \right)}^{2}}={{i}^{4}}=1 \\
\end{align}$

In this problem we have,
$\Rightarrow \dfrac{({{i}^{368}}+{{i}^{370}})}{({{i}^{399}}+{{i}^{400}})}.......(i)$
First, we will evaluate ${{i}^{368}}+{{i}^{370}}$.
We know that, when 368 is divided by 4 we get 92 as quotient and 0 as remainder.
$\Rightarrow 368=(4\times 92)+0.......(ii)$
Similarly, when 370 is divided by 4 we get 92 as quotient and 2 as remainder.
$\Rightarrow 370=(4\times 92)+2.......(iii)$
Then we can rewrite ${{i}^{368}}+{{i}^{370}}$ as,
\[\Rightarrow {{i}^{368}}+{{i}^{270}}={{\left( {{i}^{4}} \right)}^{92}}+{{\left( {{i}^{4}} \right)}^{92}}\times {{i}^{2}}..........(iv)\]
From the properties of complex numbers, we know that ${{i}^{4}}=1$and ${{i}^{2}}=-1$.
Then, equation (iv) changes to,
\[\begin{align}
  & \Rightarrow {{i}^{368}}+{{i}^{270}}={{\left( 1 \right)}^{92}}+\left( {{\left( 1 \right)}^{92}}\times -1 \right) \\
 & \Rightarrow {{i}^{368}}+{{i}^{270}}=1+-1 \\
 & \Rightarrow {{i}^{368}}+{{i}^{270}}=0..........(v) \\
\end{align}\]
Next, we will evaluate, ${{i}^{399}}+{{i}^{400}}$.
We know that, when 399 is divided by 4 we get 99 as quotient and 3 as remainder.
$\Rightarrow 399=(4\times 99)+3.......(vi)$
Similarly, when 400 is divided by 4 we get 100 as quotient and 0 as remainder.
$\Rightarrow 400=(4\times 100)+0.......(vii)$
Then we can rewrite ${{i}^{399}}+{{i}^{400}}$ as,
\[\Rightarrow {{i}^{399}}+{{i}^{400}}={{\left( {{i}^{4}} \right)}^{99}}\times {{i}^{3}}+{{\left( {{i}^{4}} \right)}^{100}}..........(viii)\]
From the properties of complex numbers, we know that ${{i}^{4}}=1$and ${{i}^{3}}=-i$.
Then, equation (viii) changes to,
\[\begin{align}
  & \Rightarrow {{i}^{399}}+{{i}^{400}}=\left( {{\left( 1 \right)}^{99}}\times -i \right)+{{1}^{100}} \\
 & \Rightarrow {{i}^{399}}+{{i}^{400}}=-i+1 \\
 & \Rightarrow {{i}^{399}}+{{i}^{400}}=-i+1..........(ix) \\
\end{align}\]
Now, substituting equations (v) and (ix) in equation (i) we get,
$\Rightarrow \dfrac{({{i}^{368}}+{{i}^{370}})}{({{i}^{399}}+{{i}^{400}})}=\dfrac{0}{-i+1}=0$
$\therefore $the answer is 0.
Now we have to find the correct option. For this we have to take each option and check whether its value is equal to 0.
Now looking at the options we can say that, option 3 is the correct answer because from equation (v) we know that ${{i}^{368}}+{{i}^{370}}$ is 0.
Hence, option 3 is the correct answer.

Note: The main idea of this problem is the usage of properties of complex numbers. The properties used in this problem are:
$\begin{align}
  & \Rightarrow i=\sqrt{-1} \\
 & \Rightarrow {{i}^{2}}=-1 \\
 & \Rightarrow {{i}^{3}}=({{i}^{2}}\times i)=(-1\times i) \\
 & \Rightarrow {{i}^{3}}=-i \\
 & \Rightarrow {{\left( {{i}^{2}} \right)}^{2}}={{i}^{4}}=1 \\
\end{align}$
Here, all the powers of $i$ are greater than 4. According to another property of complex numbers, the value of ${{i}^{n}}$ for $n > 4$ is ${{i}^{r}}$. Here, r is the remainder when n is divided by 4. This way also the problem can be solved easily.