The square of any real number, whether it is positive or negative or zero is always non-negative, i.e. $x^2 \geq 0$ for all $x \in R$.
Hence, the equation $x^2+1=0$ is not satisfied for any real value of $x$ or not solvable in real number system..
Thus, the equation $x^2+1=0$ has an imaginary solution. 'Eular' was the first mathematician to introduce the symbol $i$ (read as 'iota'). The imaginary number i is defined as the square root of -1 .

$
\begin{aligned}
& \mathrm{x}^2+1=0 \\
& \Rightarrow \quad \mathrm{x}^2=-1 \\
& \text { or, } \mathrm{x}= \pm \sqrt{-1}= \pm i
\end{aligned}
$

Equation, $\mathrm{x}^2+1=0$ has two solution, $\mathrm{x}=i$ and $\mathrm{x}=-i$.

$
\begin{aligned}
\sqrt{-1} & =\mathrm{i} \\
1^2 & =(\sqrt{-1})^2=-1
\end{aligned}
$

We can write the square root of any negative number as a multiple of i. Consider the square root of -25

$
\begin{aligned}
\sqrt{-25} & =\sqrt{25(-1)} \\
& =\sqrt{25} \sqrt{-1} \\
& =5 \mathrm{i}
\end{aligned}
$

Integral Powers of iota (i)

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

$
\begin{aligned}
& \mathrm{i}^0=1, \quad \mathrm{i}^1=\mathrm{i}, \quad \mathrm{i}^2=(\sqrt{-1})^2=-1 \\
& \mathrm{i}^3=\mathrm{i} \cdot \mathrm{i}^2=\mathrm{i} \times-1=-\mathrm{i} \\
& \mathrm{i}^4=\mathrm{i}^2 \cdot \mathrm{i}^2=-1 \times-1=1 \\
& \mathrm{i}^5=\mathrm{i} \cdot \mathrm{i}^4=\mathrm{i} \times 1=\mathrm{i}
\end{aligned}
$

In general,

$
\mathrm{i}^{4 \mathrm{n}}=1, \quad \mathrm{i}^{4 \mathrm{n}+1}=\mathrm{i}, \quad \mathrm{i}^{4 \mathrm{n}+2}=-1, \quad \mathrm{i}^{4 \mathrm{n}+3}=-1
$

(2) If the power of iota is the negative integer

$
\begin{aligned}
i^{-1} & =\frac{1}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i \\
i^{-2} & =\frac{1}{i^2}=-1 \\
i^{-3} & =\frac{1}{i^3}=\frac{i}{i^4}=i \\
i^{-4} & =\frac{1}{i^4}=\frac{1}{1}=1
\end{aligned}
$

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

$\mathrm{n} \in \mathbb{I}$ and $\mathrm{i}=\sqrt{-1}$, then

$
\begin{aligned}
i^{\mathrm{n}}+i^{\mathrm{n}+1}+i^{\mathrm{n}+2}+i^{\mathrm{n}+3} & =i^{\mathrm{n}}\left(1+i+i^2+i^3\right) \\
& =i(1+i-1-i)=0
\end{aligned}
$