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