 maths > complex-number

Multiplication, Conjugate, & Division in Complex numbers

what you'll learn...

Overview

Complex number multiplication

»  $\left(a+ib\right)×\left(c+id\right)$$\left(a + i b\right) \times \left(c + i d\right)$
by associative, commutative, distributive laws of real numbers, considering $i$$i$ as a variable, and applying ${i}^{2}=-1$${i}^{2} = - 1$
→  $=\left(ac-bd\right)+i\left(cb+ad\right)$$= \left(a c - b d\right) + i \left(c b + a d\right)$

Complex Number Conjugate

»  Conjuage of $a+ib$$a + i b$
Conjugate means "coupled or related". Conjugate of a complex number makes the number real by addition or multiplication.

→  $=\overline{a+ib}$$= \overline{a + i b}$
→  $=a-ib$$= a - i b$

Complex Number Division

»  $\left(a+ib\right)÷\left(c+id\right)$$\left(a + i b\right) \div \left(c + i d\right)$
by multiplicative identity, multiplicative inverse, distributive properties.

→  $=\frac{a+ib}{c+id}$$= \frac{a + i b}{c + i d}$$×\frac{c-id}{c-id}$$\times \frac{c - i d}{c - i d}$

→  $=\frac{\left(a+ib\right)×\left(c-id\right)}{{c}^{2}+{d}^{2}}$$= \frac{\left(a + i b\right) \times \left(c - i d\right)}{{c}^{2} + {d}^{2}}$

multiplication

Consider two complex numbers ${z}_{1}={a}_{1}+i{b}_{1}$${z}_{1} = {a}_{1} + i {b}_{1}$ and ${z}_{2}={a}_{2}+i{b}_{2}$${z}_{2} = {a}_{2} + i {b}_{2}$. Then ${z}_{1}×{z}_{2}$${z}_{1} \times {z}_{2}$ is '$\left({a}_{1}{a}_{2}-{b}_{1}{b}_{2}\right)+i\left({a}_{2}{b}_{1}+{a}_{1}{b}_{2}\right)$$\left({a}_{1} {a}_{2} - {b}_{1} {b}_{2}\right) + i \left({a}_{2} {b}_{1} + {a}_{1} {b}_{2}\right)$'. This is from the associative and distributive laws of real numbers extended to numbers with $\sqrt{-1}$$\sqrt{- 1}$.

${z}_{1}={a}_{1}+i{b}_{1}$${z}_{1} = {a}_{1} + i {b}_{1}$ and ${z}_{2}={a}_{2}+i{b}_{2}$${z}_{2} = {a}_{2} + i {b}_{2}$. What is ${z}_{1}×{z}_{2}$${z}_{1} \times {z}_{2}$?
Solution :
${z}_{3}={z}_{1}×{z}_{2}$${z}_{3} = {z}_{1} \times {z}_{2}$
$\quad \quad = \left({a}_{1} + i {b}_{1}\right) \times \left({a}_{2} + i {b}_{2}\right)$
$\quad \quad = {a}_{1} \times \left({a}_{2} + i {b}_{2}\right)$
$\quad \quad \quad \quad + i {b}_{1} \times \left({a}_{2} + i {b}_{2}\right)$
$\quad \quad = {a}_{1} {a}_{2} + i {a}_{1} {b}_{2} + i {b}_{1} {a}_{2}$
$\quad \quad \quad \quad + {i}^{2} {b}_{1} {b}_{2}$
$\quad \quad = \left({a}_{1} {a}_{2} - {b}_{1} {b}_{2}\right)$
$\quad \quad \quad \quad + i \left({b}_{1} {a}_{2} + {a}_{1} {b}_{2}\right)$

Given ${z}_{1}=1+2i$${z}_{1} = 1 + 2 i$ and ${z}_{2}=3-i$${z}_{2} = 3 - i$ what is ${z}_{1}×{z}_{2}$${z}_{1} \times {z}_{2}$?

The answer is '$5+5i$$5 + 5 i$'

summary

Multiplication of two complex numbers : For any complex number ${z}_{1}={a}_{1}+i{b}_{1}\in ℂ$${z}_{1} = {a}_{1} + i {b}_{1} \in \mathbb{C}$ and ${z}_{2}={a}_{2}+i{b}_{2}\in ℂ$${z}_{2} = {a}_{2} + i {b}_{2} \in \mathbb{C}$
${z}_{1}×{z}_{2}=\left({a}_{1}{a}_{2}-{b}_{1}{b}_{2}\right)$${z}_{1} \times {z}_{2} = \left({a}_{1} {a}_{2} - {b}_{1} {b}_{2}\right)$
$\quad \quad \quad \quad \quad + i \left({a}_{2} {b}_{1} + {a}_{1} {b}_{2}\right)$

Multiplication of two complex numbers follows numerical expression laws and properties with $\sqrt{-1}$$\sqrt{- 1}$ handled as per the property ${\left(\sqrt{-1}\right)}^{2}=-1$${\left(\sqrt{- 1}\right)}^{2} = - 1$

conjugate

In numerical expressions or algebraic expressions, we can manipulate the expressions without modifying the value of the expression
•  add and subtract : eg $3=3+1-1$$3 = 3 + 1 - 1$
•  multiply and divide : eg : $3=3×\frac{2}{2}$$3 = 3 \times \frac{2}{2}$
•  etc.

. These manipulations help to arrive at a different form of expressions or help to solve.

In case of complex numbers, $a+ib$$a + i b$, one modification stands out "convert the complex number to real number". This can be achieved either by addition or multiplication with the number $a-ib$$a - i b$.
•  addition $\left(a+ib\right)+\left(a-ib\right)=2a$$\left(a + i b\right) + \left(a - i b\right) = 2 a$
•  multiplication $\left(a+ib\right)×\left(a-ib\right)={a}^{2}+{b}^{2}$$\left(a + i b\right) \times \left(a - i b\right) = {a}^{2} + {b}^{2}$

For a given complex number $z=a+ib$$z = a + i b$, the connected number that give a real number on multiplication is $a-ib$$a - i b$. It is named as conjugate of z and represented as $\overline{z}$$\overline{z}$ or $\overline{a+ib}=a-ib$$\overline{a + i b} = a - i b$

The word 'conjugate' means 'coupled; joined ; related in reciprocal or complementary'

Find $\overline{1-3i}$$\overline{1 - 3 i}$.

summary

Conjugate of a Complex NumberFor a complex number $z=a+ib\in ℂ$$z = a + i b \in \mathbb{C}$ the conjugate of $z$$z$ is given as $\overline{z}=a-ib$$\overline{z} = a - i b$.

Conjugate of a complex number is the number with the same real part and negative of imaginary part.

division

Consider two complex numbers ${z}_{1}={a}_{1}+i{b}_{1}$${z}_{1} = {a}_{1} + i {b}_{1}$ and ${z}_{2}={a}_{2}+i{b}_{2}$${z}_{2} = {a}_{2} + i {b}_{2}$. Then, $\frac{{z}_{1}}{{z}_{2}}$${z}_{1} / {z}_{2}$ in $a+bi$$a + b i$ form is '$\frac{\left({a}_{1}{a}_{2}+{b}_{1}×{b}_{2}\right)+i\left({a}_{2}{b}_{1}-{a}_{1}{b}_{2}\right)}{{a}_{2}^{2}+{b}_{2}^{2}}$$\frac{\left({a}_{1} {a}_{2} + {b}_{1} \times {b}_{2}\right) + i \left({a}_{2} {b}_{1} - {a}_{1} {b}_{2}\right)}{{a}_{2}^{2} + {b}_{2}^{2}}$'

${z}_{1}={a}_{1}+i{b}_{1}$${z}_{1} = {a}_{1} + i {b}_{1}$ and ${z}_{2}={a}_{2}+i{b}_{2}$${z}_{2} = {a}_{2} + i {b}_{2}$. What is $\frac{{z}_{1}}{{z}_{2}}$${z}_{1} / {z}_{2}$?
Solution :
${z}_{3}=\frac{{z}_{1}}{{z}_{2}}$${z}_{3} = {z}_{1} / {z}_{2}$
$\quad \quad = \frac{{a}_{1} + i {b}_{1}}{{a}_{2} + i {b}_{2}}$
$\quad \quad = \frac{{a}_{1} + i {b}_{1}}{{a}_{2} + i {b}_{2}} \times \frac{{a}_{2} - i {b}_{2}}{{a}_{2} - i {b}_{2}}$
$\quad \quad = \frac{\left({a}_{1} + i {b}_{1}\right) \times \left({a}_{2} - i {b}_{2}\right)}{\left({a}_{2} + i {b}_{2}\right) \times \left({a}_{2} - i {b}_{2}\right)}$
$\quad \quad = \frac{\left({a}_{1} {a}_{2} + {b}_{1} {b}_{2}\right) + i \left({a}_{2} {b}_{1} - {a}_{1} {b}_{2}\right)}{{a}_{2}^{2} + {b}_{2}^{2}}$

Given ${z}_{1}=10+5i$${z}_{1} = 10 + 5 i$ and ${z}_{2}=3-4i$${z}_{2} = 3 - 4 i$ then $\frac{{z}_{1}}{{z}_{2}}$${z}_{1} / {z}_{2}$ is '$\frac{10+11i}{5}$$\frac{10 + 11 i}{5}$'.

summary

Division of two complex numbers : For any complex number ${z}_{1}={a}_{1}+i{b}_{1}\in ℂ$${z}_{1} = {a}_{1} + i {b}_{1} \in \mathbb{C}$ and ${z}_{2}={a}_{2}+i{b}_{2}\in ℂ$${z}_{2} = {a}_{2} + i {b}_{2} \in \mathbb{C}$
$\frac{{z}_{1}}{{z}_{2}}=$${z}_{1} / {z}_{2} =$ $\frac{\left({a}_{1}{a}_{2}+{b}_{1}{b}_{2}\right)+i\left({a}_{2}{b}_{1}-{a}_{1}{b}_{2}\right)}{{a}_{2}^{2}+{b}_{2}^{2}}$$\frac{\left({a}_{1} {a}_{2} + {b}_{1} {b}_{2}\right) + i \left({a}_{2} {b}_{1} - {a}_{1} {b}_{2}\right)}{{a}_{2}^{2} + {b}_{2}^{2}}$

Division of two complex numbers follows numerical expression laws and properties with $i$$i$ handled as ${\left(\sqrt{-1}\right)}^{2}=-1$${\left(\sqrt{- 1}\right)}^{2} = - 1$ to arrive at $a+bi$$a + b i$ form.

Outline