Multiplication of Complex Numbers

Four fundamental arithmetic operations—addition, subtraction, multiplication, and division—provide the algebraic operations on complex numbers in mathematics. The algebraic methods define the algebraic operations on complex numbers. To explain the link between the number of operations, some fundamental algebraic laws are employed, such as distributive, commutative, and associative laws. Algebra has its own set of principles designed to solve problems since it is a concept based on known and unknown values or variables.

This Story also Contains

1. Complex Number
2. Multiplication Of Complex Numbers
3. Properties Of Multiplication Of Complex Numbers
4. Solved Examples

In this article, we will cover the concept of multiplication of complex numbers. This concept falls under the broader category of complex numbers. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.

Complex Number

The number which has no real meaning then these numbers are represented in complex forms. The general form of complex numbers are $a+i b$ where i is iota or$\sqrt{-1}$.

A number of the form$a+i b$ is called a complex number (where a and b are real numbers and i is iota). We usually denote a complex number by the letter z, z₁, z₂, etc

For example,$z=5+2 i$ is a complex number.

5 here is called the real part and is denoted by Re(z), and 2 is called the imaginary part and is denoted by Im(z)

Multiplication Of Complex Numbers

Multiplying any two complex numbers is equal to the multiplication of two binomials.

Let $z_1={a}+\mathrm{ib}$ and $\mathrm{z}_2=\mathrm{c}+\mathrm{id}$ be any two complex numbers. Then, the multiplication $\mathrm{z}_1 \cdot \mathrm{z}_2$ is defined as

$
\begin{aligned}
z_1 \cdot z_2 & =(a+i b) \cdot(c+i d) \\
& =a c+i a d+i b c+i^2 b d \\
& =a c+i(a d+b c)-b d \\
& =(a c-b d)+i(a d+b c)
\end{aligned}
$

For example, $z_1=(4+3 i)$ and $z_2=(2-5 i)$, then $z_1 \cdot z_2$ is

$
\begin{aligned}
(4+3 i)(2-5 i) & =4(2)-4(5 i)+3 i(2)-(3 i)(5 i) \\
& =8-20 i+6 i-15\left(i^2\right) \\
& =(8+15)+(-20+6) i \\
& =23-14 i
\end{aligned}
$

Properties Of Multiplication Of Complex Numbers

Multiplicative identity: if the multiplication of a complex number z₁ with another complex number z₂ is z₁, then z₂ is called the multiplicative identity. We have z・1 = z = 1・z, so 1 is the multiplicative identity.

Multiplicative inverse: For every non-zero complex $z=a+i b,(a \neq 0, b \neq 0)$ we have the complex number $\frac{a}{a^2+b^2}+i \frac{-b}{a^2+b^2}\left(\right.$ denoted by $\frac{1}{z}$ or $\left.z^{-1}\right)$ called the multiplicative inverse of z.

Recommended Video Based on Multiplication of Complex Numbers:

Solved Examples

Example 1: The multiplicative inverse of the complex number $z=\frac{5+i \sqrt{2}}{1-i \sqrt{2}}$ is:

Solution:

As we have learnt, The multiplicative inverse of $z$ is $1 / z$
Now,

$
\begin{aligned}
& z=\frac{5+i \sqrt{2}}{1-i \sqrt{2}} \\
& \text { So } \frac{1}{z}=\frac{1-i \sqrt{2}}{5+i \sqrt{2}} \\
& =\frac{1-i \sqrt{2}}{5+i \sqrt{2}} \cdot \frac{5-i \sqrt{2}}{5-i \sqrt{2}} \\
& =\frac{5-i \sqrt{2}-i 5 \sqrt{2}-2}{25+2} \\
& =\frac{3-6 \sqrt{2} i}{27} \\
& =\frac{3(1-2 \sqrt{2} i)}{27} \\
& =\frac{(1-2 \sqrt{2} i)}{9}
\end{aligned}
$
So the multiplication inverse is $\frac{1-2 i \sqrt{2}}{9}$.
Hence, the answer is the option 2.

Example 2: Let $z_1=2+3 i$ and $z_2=-1-5 i$ then $z_1 z_2$ equals

Solution:

As we learned in

Multiplication of Complex Numbers -

(a+ib)(c+id)=(ac-bd)+i(bc+ad)

$z_1 z_2=(2+3 i)(-1-5 i)=\{(2)(-1)-(3)(-5)\}+i\{(3)(-1)+(2)(-5)\}$

$\Rightarrow z_1 z_2=13-13 i$

Hence, the answer is 13-13i

Example 3: For two non-zero complex numbers $z_1$ and $z_2$ if $\operatorname{Re}\left(z_1 z_2\right)=0$ and $\operatorname{Re}\left(z_1+z_2\right)=0$, then which of the following are possible?

A. $\operatorname{lm}\left(z_1\right)>0$ and $\operatorname{Im}\left(z_2\right)>0$
B. $\operatorname{lm}\left(z_1\right)<0$ and $\operatorname{Im}\left(z_2\right)>0$
C. $\operatorname{lm}\left(z_1\right)>0$ and $\operatorname{Im}\left(z_2\right)<0$
D. $\operatorname{lm}\left(z_1\right)<0$ and $\operatorname{Im}\left(z_2\right)<0$

Choose the correct answer from the options given below:
1) $B$ and $D$
2) A and B
3) $B$ and $C$
4) A and C

Solution:

$\operatorname{Re}\left(z_1 z_2\right)=0$ and $\operatorname{Re}\left(z_1+z_2\right)=0$
let $z_1=a_1+i h_1 a_{n t} z_n=a_2+i_2$
$a_1 z_2=\left(a_1 a_2-b_1 b_2\right)+i\left(a_1 b_2+b_1 a_2\right)$
$\because R e\left(z_1 z_2\right)=a_1 a_2-b_1 b_2=0$
$\therefore a_1 a_2=b_1 b_2 \ldots \ldots \ldots(1)$
and $\operatorname{Re}\left(z_1+z_2\right)=0 \Rightarrow a_1+a_2=0$
$E a_2=-a_1 \ldots \ldots \ldots \ldots \ldots(2)$
rom (1) and (2)
$-b_1 b_2=-1_1^2<1$
The product of $b_1 b_2$ is a Negative
- Im (z) and Im (z) are also of opposite sign

Hence, the answer is the option 3.