# Properties of Complex Numbers

Properties of complex numbers are given as follows:

## Theorem 1: The operations of addition and multiplication of complex numbers satisfy the commutative, associative and distributive laws.

Let $z_1=(a,b)$, $z_2=(c,d)$ and $z_3=(e,f)$ be three complex numbers.

### i. Commutative Law

$\begin{array}{l}z_1+z_2&=(a,b)+(c,d)\\&=(a+c,b+d)\\&=(c+a,d+b)\\&=(c,d)+(a,b)\\&=z_2+z_1 \end{array}$ $\therefore$ Addition of complex numbers is commutative.

$\begin{array}{l}z_1z_2&=(a,b)(c,d)\\&=(ac-bd,ad+bc)\\&=(ca-db,cb+da)\\&=(c,d)(a,b)\\&=z_2z_1 \end{array}$ $\therefore$ Multiplication of complex numbers is commutative.

### ii. Associative Law

$\begin{array}{l}(z_1+z_2)+z_3&=(a+c,b+d)+(e,f)\\&=(a+c+e,b+d+f)\\&=(a,b)+(c+e,d+f)\\&=z_1+(z_2+z_3)\end{array}$ $\therefore$ Addition of complex numbers is associative.

$(z_1z_2)z_3=(ac-bd,ad+bc)(e,f)$ $=(ace-bde-adf-bcf,acf-bdf+ade+bce)$ also, $z_1(z_2z_3)=(a,b)(ce-df,cf+de)$ $=(ace-adf-bcf-bde,acf+ade+bce-bdf)$ $\therefore(z_1z_2)z_3=z_1(z_2z_3)$ Hence, multiplication of complex numbers is associative.

### iii. Distributive Law

$\begin{array}{l}z_1(z_2+z_3)&=(a,b)[(c,d)+(e,f)]\\&=(a,b)(c+e,d+f)\end{array}$ $=(ac+ae-bd-bf,ad+af+bc+be)$ Also, $z_1z_2+z_1z_3=(a,b)(c,d)+(a,b)(e,f)$ $=(ac-bd,ad+bc)+(ae-bf,af+be)$ $=(ac+ae-bd-bf,ad+af+bc+be)$ $\therefore z_1(z_2+z_3)=z_1z_2+z_1z_3$ $\therefore$ Addition and multiplication of complex numbers are distributive.

## Theorem 2

Let $z=(a,b)$ be a complex number. Then, $\begin{array}{l}1.&(a,b)+(0,0)&=(a,b)\\2.&(a,b)(0,0)&=(0,0) \\3.&(a,b)(1,0)&=(a,b)\\4.&\frac{(a,b)}{(1,0)}&=(a,b)\\5.&(a,b)+(-a,-b)&=(0,0)\end{array}$ These theorems can be proved easily from the definitions.

In above theorems, we observe that the complex numbers $(0,0)$ and $(1,0)$ have properties similar to those of the real numbers $0$ and $1$. Hence, we shall take them as the real numbers $0$ and $1$. Also, the complex number $(-a,-b)$ acts as the additive inverse of $z=(a,b)$.

Thus, by writing $(0,0)=0$, $(1,0)=1$ and $(-a,-b)=-z$, the above theorems can be restated as $\begin{array}{l}1.&z+0&=z\\2.&z\cdot 0&=0\\3.&z\cdot 1&=z\\4.&\frac{z}{1}&=z\\5.&z+(-z)&=z-z=0\end{array}$ Here, $-z$ is said to be the additive inverse of $z$.

For any non zero real numbers $(a,b)$, $\begin{array}{l}1.&(a,0)+(b,0)&=(a+b,0)\\2.&(a,0)(b,0)&=(ab,0)\\3.&\frac{(a,0)}{(b,0)}&=\left(\frac{a}{b},0\right)\end{array}$

Hence, a complex number of the form $(a,0)$ has properties similar to the real number $a$. Thus, we shall identify the complex number $(a,0)$ as the real number $a$.

In other words, a complex number $(a,0)$ with the imaginary part zero is the real number $a$ itself. Hence, the real number system is a special case of the complex number system. The set of complex number system, denoted by the letter $C$, obviously contains the set of real numbers $R$, i.e. $R\subset C$.

## Theorem 3: A real number cannot be equal to an imaginary number unless each is zero.

Let a real number $a$ be equal to an imaginary number $ib$. $i$ is known as the imaginary unit. $\begin{array}{l}\text{i.e.}&a&=ib\\ \Rightarrow & a^2 &= i^2b^2 \\ \Rightarrow & a^2 &= -b^2 \\ \Rightarrow & a^2+b^2 &= 0\end{array}$ $\therefore a=0\;\;\text{and}\;\; b=0$

$\therefore$ Real number $=a=0$.

$\therefore$ Imaginary number $=ib=i×0=0$.

## Theorem 4: If $a+ib=0$, then $a=b=0$.

Here, $\begin{array}{l}&a+ib&=0\\ \Rightarrow & a &=ib \\ \Rightarrow & a^2 &= i^2b^2 \\ \Rightarrow & a^2 &= -b^2 \\ \Rightarrow & a^2+b^2 &=0\end{array}$ $\therefore a=0\;\;\text{and}\;\; b=0$

## Theorem 5: If $a+ib=c+id$, then $a=c$ and $b=d$.

Here, $\begin{array}{l} & a+ib &= c+id \\ \Leftrightarrow & a-c &= i(d-b) \\ \Leftrightarrow & (a-c)^2 &= i^2(d-b)^2 \\ \Leftrightarrow & (a-c)^2 &= -(d-b)^2 \\ \Leftrightarrow & (a-c)^2+(d-b)^2 &=0 \end{array}$

$\begin{array}{c} \therefore & (a-c)&=0 &\text{and}& (d-b) &=0 \\ \therefore & a&=c &\text{and}& d&=b\end{array}$