Question

Find the value of $ 2{{i}^{15}}+3{{i}^{6}} $.

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,
 $ \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} $

Complete step-by-step answer:
In this problem we have,
 $ \Rightarrow 2{{i}^{15}}+3{{i}^{6}}.......(i) $
First we will write all powers of $ i $ that is, $ {{i}^{15}} $ and $ {{i}^{6}} $ in the terms of $ {{i}^{4}} $ .
We know that, when 15 is divided by 4 we get 3 as quotient and 3 as remainder.
 $ \Rightarrow 15=(4\times 3)+3.......(ii) $
Similarly, when 6 is divided by 4 we get 1 as quotient and 2 as remainder.
\[\Rightarrow 6=(4\times 1)+2.....(iii)\]
From equation (i) and equation (ii) we can write $ {{i}^{15}} $ and $ {{i}^{6}} $ as,
 $ \begin{align}
  & \Rightarrow {{i}^{15}}={{({{i}^{4}})}^{3}}\times {{i}^{3}}........(iv) \\
 & \Rightarrow {{i}^{6}}={{i}^{4}}\times {{i}^{2}}.......(v) \\
\end{align} $
Substituting the above equations (iv) and (v) in equation (i) we get,
 $ \Rightarrow 2{{i}^{15}}+3{{i}^{6}}=2{{({{i}^{4}})}^{3}}\times {{i}^{3}}+3({{i}^{4}})\times {{i}^{2}}........(vii) $
Now, we know that,
 $ \begin{align}
  & \Rightarrow i=\sqrt{-1} \\
 & \Rightarrow {{i}^{2}}=-1........(viii) \\
 & \Rightarrow {{i}^{3}}=({{i}^{2}}\times i)=(-1\times i) \\
\end{align} $

\[\begin{align}
  & \Rightarrow {{i}^{3}}=-i.......(ix) \\
 & \Rightarrow {{({{i}^{2}})}^{2}}={{i}^{4}}=1.......(x) \\
\end{align}\]
Now, substituting values from equation (viii), equation (ix) and equation (x) in equation (vii), we get,
 $ \Rightarrow 2{{i}^{15}}+3{{i}^{6}}=2({{(1)}^{3}}\times -i)+3(1\times -1) $
Simplifying the above equation, we get,
 $ \begin{align}
  & \Rightarrow 2{{i}^{15}}+3{{i}^{6}}=(2\times -i)+3(1\times -1) \\
 & \Rightarrow 2{{i}^{15}}+3{{i}^{6}}=-2i-3......(xi) \\
\end{align} $
In equation (xi) we get the answer as $ -2i-3 $ . A complex number $ Z $ is represented as, $ Z=a+ib $ .
Here, $ a $ is the real part and $ b $ is the imaginary part. Now, comparing this with equation (xi), we can rewrite the answer as,
 $ \Rightarrow -3-2i......(xii) $
Hence, the correct answer is $ -3-2i $.

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, in this question all the powers of $ i $ are greater than 4. Then this problem can be solved in another way. We can represent the value of $ {{i}^{n}} $ for $ n>4 $ , as $ {{i}^{r}} $ . Here, $ r $ is the remainder when $ n $ is divided by 4. This can be written as,
 $ \begin{align}
  & \Rightarrow 2{{i}^{15}}+3{{i}^{6}}=2{{i}^{3}}+3{{i}^{2}} \\
 & \Rightarrow 2{{i}^{15}}+3{{i}^{6}}=2\left( -i \right)+3\left( -1 \right) \\
 & \Rightarrow 2{{i}^{15}}+3{{i}^{6}}=-2i-3=-3-2i. \\
\end{align} $