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  • The number of the value of in + i-n for different n ∈ I (integer) and i = √-1Option: 1 1<div

The number of the value of in + i-n for different n ∈ I (integer) and i = √-1

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

As we have learnt in

Iota and powers of Iota -

 

 

Integral Powers of iota (i)

 

(1) If the power of iota is whole number

 

\\\mathrm{\textit{i}^0=1,\;\;\textit{i}^1=\textit{i},\;\;\textit{i}^2=(\sqrt{-1})^2=-1}\\\mathrm{\textit{i}^3=\textit{i}\cdot\textit{i}^2=\textit{i}\times-1=-\textit{i}}\\\mathrm\textit{i}^4=\textit{i}^2\cdot\textit{i}^2=-1\times -1=1\\\mathrm{\textit{i}^5=\textit{i}\cdot\textit{i}^4=\textit{i}\times1=\textit{i}}\\\mathrm{In\;general,\;\;}\\\mathrm{\mathbf{\textit{i}^{4n}=1,\;\;\;\textit{i}^{4n+1}=\textit{i},\;\;\;\textit{i}^{4n+2}=-1,\;\;\;\textit{i}^{4n+3}=-1}}

 

(2) If the power of iota is negative integer

 

\\\mathrm{\textit{i}^{-1}=\frac{1}{\textit{i}}=\frac{\textit{i}}{\textit{i}^2}=\frac{\textit{i}}{-1}=-\textit{i}}\\\mathrm{\textit{i}^{-2}=\frac{1}{\textit{i}^2}=-1}\\\mathrm{\textit{i}^{-3}=\frac{1}{\textit{i}^3}=\frac{i}{\textit{i}^4}=\textit{i}}\\\mathrm{\textit{i}^{-4}=\frac{1}{\textit{i}^4}=\frac{1}{1}=1}

 

-

\\\mathrm{\textit{i}^n+\textit{i}^{-n}=\textit{i}^n+\frac{1}{\textit{i}^n}}\\\mathrm{n=1}\\\mathrm{\Rightarrow \textit{i}^1+\frac{1}{\textit{i}^1}=\textit{i}-\textit{i}=0}\\\mathrm{n=2}\\\mathrm{\Rightarrow \textit{i}^2+\frac{1}{\textit{i}^2}=-1-1=-2}\\\mathrm{n=3}\\\mathrm{\Rightarrow \textit{i}^3+\frac{1}{\textit{i}^3}=-\textit{i}+\textit{i}=0}\\\mathrm{n=4}\\\mathrm{\Rightarrow \textit{i}^4+\frac{1}{\textit{i}^4}=1+1=2}\\\mathrm{n=5}\\\mathrm{\Rightarrow \textit{i}^5+\frac{1}{\textit{i}^5}=\textit{i}-\textit{i}=0}

 

So the values are 0, -2, and 2

Correct option is (c)

Posted by

avinash.dongre

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