The **imaginary unit** denoted by $i$ is a complex number $(0,1)$ with the real part zero and the imaginary part $1$.

We know that the complex number $(a,0)$ has properties similar to the real number $a$ (**See:** How?). So, the real number $a$ can be treated as the complex number $(a,0)$ with the imaginary part zero. Thus, the complex number $(1,0)$ is same as the real number $1$. Now, a question arises – “*What if the real part is zero?*“.

The complex number $(0,1)$ with the real part zero and the imaginary part $1$ is called the imaginary unit. It is denoted by the letter $i$ (the “*iota*” in the Greek alphabet and the first letter of the word “*imaginary*” in the English alphabet). The symbol $i$ was introduced by Euler in 1777.

## Theorem 1: $i^2=-1$

Proof: \[\begin{array}{l} i^2 &= i\cdot i\\ &= (0,1)\cdot(0,1) \\ &= (0-1,0+0) \\ &= (-1,0) \\ &= -(1,0) \\ &= -1 \end{array}\]

## Theorem 2: If $a$ and $b$ are real numbers then the complex number $(a,b)$ can be written as $a+ib$.

Proof: \[\begin{array}{l} a+ib &=(a,0)+(0,1)(b,0) \\ &=(a,0)+(0,b) \\ &=(a,b) \end{array}\] So, if $a$ and $b$ are real numbers then $a+ib$ is said to be a complex number.

Hence, now we can solve an equation of the type $x^2+4=0$. \[\begin{array}{l} & x^2+4 &= 0 \\ \Rightarrow & x^2 &= -4 \\ \Rightarrow & x^2 &= 4i^2 \\ \therefore & x &= \pm 2i \end{array}\]

Hence, the complex number $(a,0)$ with the imaginary part zero is written as the real number $a$, and it is said to be a purely real number (or simply a real number). And the complex number $(0,b)$ with the real part zero is written as $bi$ and it is said to be a purely imaginary number. The complex number $(a,b)$ is written as $a+ib$.

The set of complex numbers, denoted by the letter $C$, obviously contains the set of real numbers $R$, i.e. $R\subset C$.

## Powers of $i$

We have defined the imaginary unit $i$ as $i^2=-1$. Any real number multiple of $i$ is an imaginary number. Now,

\[\begin{array}{l} i^2 &= 1 \\ i^3 &= i^2\cdot i &= (-1)i &= -i \\ i^4 &= (i^2)^2 &= (-1)^2 &= 1 \\ i^5 &= (i^4)i &= (1)i &= i \\ i^{10} &= (i^2)^5 &= (-1)^5 &= -1 \\ i^{-1} &= \frac{1}{i} &= \frac{i}{i^2} &= -i \\ i^{-10} &= \frac{1}{i^10} &= \frac{1}{-1} &= -1 &\text{etc.} \end{array}\]

## Square Roots of Complex Numbers

Let $z=(a,b)=a+ib$ be a complex number. Let the square root of $z=a+ib$ be $x+iy$. Then, \[\begin{array}{l} &(x+iy)^2 &= a+ib \\ \text{or,} & x^2+2xyi+i^2y^2 &= a+ib \\ \text{or,} & x^2-y^2+2xyi &= a+ib \end{array}\] \[\therefore x^2-y^2=a \;\;\text{and}\;\; 2xy=b\]

Solving these two equations, we get two sets of values of $x$ and $y$, thus giving us two square roots of $a+ib$.

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