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Powers of Iota - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Iota and powers of Iota is considered one of the most asked concept.

  • 9 Questions around this concept.

Concepts Covered - 1

Iota and powers of Iota

The square of any real number, whether it is positive or negative or zero is always non-negative, i.e. x2 ≥ 0 for all x ∈ R. 

Hence, the equation x2 + 1 = 0 is not satisfied for any real value of x or not solvable in real  number system..

Thus, the equation  x2 + 1 = 0 has imaginary solution. ‘Eular’ was the first mathematician to introduced the symbol i (read as ‘iota’).  The imaginary number i is defined as the square root of −1 .  

Hence, the equation

 

\\\mathrm{x^2+1=0}\\\mathrm{\Rightarrow \;\;x^2=-1}\\\mathrm{or,\;\;x=\pm\sqrt{-1}=\pm \mathit{i} }\\\mathrm{Equation,\;x^2+1=0\;\;has\;two\;solution,\;\;x=\textit{i}\;\;and\;\;x=-\textit{i}.}

 

\\\mathrm{\sqrt{-1}=i}\\\mathrm{\;\;\;\;\;\i^2=(\sqrt{-1})^2=-1}\\\\\mathrm{We\:can\:write\:the\:square\:root\:of\:any\:negative\:number\:as\:a\:multiple\:of\:i.}\\\mathrm{Consider\:the\:square\:root\:of\;\;-25}\\\\\mathrm{\sqrt{-25}=\sqrt{25\left(-1\right)}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;=\sqrt{25}\sqrt{-1}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;=5i}

 

 

Integral Powers of iota (i)

 

(1) If the power of iota is the whole number

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

(2) If the power of iota is the 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}

 

The sum of four consecutive powers of iota (i) is zero

 

\\\mathrm{n\in\mathbb{I}\;\;and\;\;i=\sqrt{-1}, \;then}\\\mathrm{\textit{i}^{n}+\textit{i}^{n+1}+\textit{i}^{n+2}+\textit{i}^{n+3}=\textit{i}^n(1+\textit{i}+\textit{i}^2+\textit{i}^3)}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \textit{i}(1+\textit{i}-1-\textit{i})=0}

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Iota and powers of Iota

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Powers of Iota Current Topic

Powers of Iota
Complex number
Algebraic operation on Complex Numbers
Multiplication of Complex Numbers
Conjugates of Complex Numbers
Modulus of complex number and its Properties
Argument of complex number
Polar form of complex numbers
Euler form of complex number
Square root of complex numbers

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