Question

The value of $\dfrac{{{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}}{1+i}$ is
1. $\dfrac{1}{2}(1+i)$
2. $\dfrac{1}{2}(1-i)$
3. $1$
4. $\dfrac{1}{2}$
5. None of these.

Answer 1

Hint: In this problem, all the elements in the numerator are powers of $i$. 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,
$\Rightarrow \dfrac{{{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}}{1+i}.......(i)$
First, we will evaluate ${{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}$.
We know that, when 5 is divided by 4 we get 1 as quotient and 1 as remainder.
$\Rightarrow 5=(4\times 1)+1.......(ii)$
Similarly, when 6 is divided by 4 we get 1 as quotient and 2 as remainder.
$\Rightarrow 6=(4\times 1)+2.......(iii)$
Similarly, when 7 is divided by 4 we get 1 as quotient and 3 as remainder.
$\Rightarrow 7=(4\times 1)+3.......(iv)$
Similarly, when 8 is divided by 4 we get 2 as quotient and 0 as remainder.
$\Rightarrow 8=(4\times 2)+0.......(v)$
Thus, from equations (ii), (iii), (iv) and (v) we can rewrite ${{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}$as,
$\Rightarrow {{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}=\left( {{i}^{4}}\cdot i \right)+\left( {{i}^{4}}\cdot {{i}^{2}} \right)+\left( {{i}^{4}}\cdot {{i}^{3}} \right)+{{\left( {{i}^{4}} \right)}^{2}}+\left( {{\left( {{i}^{4}} \right)}^{2}}\cdot i \right)..........(vi)$
We also know that,
$\begin{align}
  & \Rightarrow i=\sqrt{-1} \\
 & \Rightarrow {{i}^{2}}=-1 \\
 & \Rightarrow {{i}^{3}}=({{i}^{2}}\times i)=(-1\times i) \\
 & \Rightarrow {{i}^{3}}=-i \\
 & \Rightarrow {{\left( {{i}^{2}} \right)}^{2}}={{i}^{4}}=1 \\
\end{align}$
Now, rewriting equation (vi) by substituting values for ${{i}^{2}},{{i}^{3}}$ and ${{i}^{4}}$ we get,
$\begin{align}
  & \Rightarrow {{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}=\left( 1\times i \right)+\left( 1\times -1 \right)+\left( 1\times -i \right)+{{\left( 1 \right)}^{2}}+\left( {{\left( 1 \right)}^{2}}\times i \right)..........(vii) \\
 & \Rightarrow {{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}=(i-1-i+1+i)........(viii) \\
\end{align}$
Simplifying equation (viii) we get,
$\begin{align}
  & \Rightarrow {{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}=({i}{-1}{-i}+{1}+i) \\
 & \Rightarrow {{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}=i.........(ix) \\
\end{align}$
Now, substituting this in equation (i) we get,
$\Rightarrow \dfrac{{{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}}{1+i}=\dfrac{i}{1+i}..........(x)$
We can further simplify equation (x) by multiplying the numerator and denominator by $(1-i)$.
Then equation (x) can be written as,
$\begin{align}
  & \Rightarrow \dfrac{{{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}}{1+i}=\dfrac{i}{1+i}\times \left( \dfrac{1-i}{1-i} \right) \\
 & \Rightarrow \dfrac{{{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}}{1+i}=\dfrac{i-{{i}^{2}}}{{{1}^{2}}-{{i}^{2}}}.......(xi) \\
\end{align}$
We know that, ${{i}^{2}}=-1$. Substituting this in equation (xi) we get,
$\begin{align}
  & \Rightarrow \dfrac{{{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}}{1+i}=\dfrac{i-(-1)}{1-(-1)}.......(xii) \\
 & \Rightarrow \dfrac{{{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}}{1+i}=\dfrac{i+1}{2}...........(xiii) \\
\end{align}$
Thus, $\dfrac{{{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}}{1+i}=\dfrac{1}{2}\left( 1+i \right)$.
$\therefore $ the correct answer is option 1.

Note: Here, the student might commit a mistake by stopping calculating at equation (x) and they might choose the correct answer as option 5. Therefore, equation (x) must be further simplified by multiplying the numerator and denominator by $(1-i)$, to get the correct answer as $\dfrac{1}{2}\left( 1+i \right)$, which is option 1.