 maths > complex-number

Addition and Subtraction of Complex numbers

what you'll learn...

Overview

»  $\left(a+ib\right)+\left(c+id\right)$$\left(a + i b\right) + \left(c + i d\right)$
by associative, commutative, distributive laws of real numbers and by considering $i$$i$ as a variable
→  $=\left(a+c\right)+i\left(b+d\right)$$= \left(a + c\right) + i \left(b + d\right)$

Complex Number Subtraction

»  $\left(a+ib\right)-\left(c+id\right)$$\left(a + i b\right) - \left(c + i d\right)$
by associative, commutative, distributive laws of real numbers and by considering $i$$i$ as a variable
→  $=\left(a-c\right)+i\left(b-d\right)$$= \left(a - c\right) + i \left(b - d\right)$

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}$${z}_{1} + {z}_{2}$ is '$\left({a}_{1}+{a}_{2}\right)+i\left({b}_{1}+{b}_{2}\right)$$\left({a}_{1} + {a}_{2}\right) + i \left({b}_{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}+{z}_{2}$${z}_{1} + {z}_{2}$
$\quad \quad = {a}_{1} + i {b}_{1} + {a}_{2} + i {b}_{2}$
$\quad \quad = {a}_{1} + {a}_{2} + i {b}_{1} + i {b}_{2}$ (associative law of addition)
$\quad \quad = {a}_{1} + {a}_{2} + i \left({b}_{1} + {b}_{2}\right)$ (distributive law of multiplication over addition)
$\quad \quad = \left({a}_{1} + {a}_{2}\right) + i \left({b}_{1} + {b}_{2}\right)$ (real and imaginary parts of result)

Given two complex numbers $3+2i$$3 + 2 i$ and $-1+i$$- 1 + i$, what is the sum?

The answer is '$2+3i$$2 + 3 i$'

summary

Addition of 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}\right)+i\left({b}_{1}+{b}_{2}\right)$${z}_{1} + {z}_{2} = \left({a}_{1} + {a}_{2}\right) + i \left({b}_{1} + {b}_{2}\right)$

Addition of two complex numbers is the addition of real and imaginary parts individually.

Subtraction

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

Complex number Subtraction:
${z}_{1}-{z}_{2}$${z}_{1} - {z}_{2}$
$\quad \quad = {a}_{1} + i {b}_{1} - \left({a}_{2} + i b 2\right)$
$\quad \quad = {a}_{1} - {a}_{2} + i {b}_{1} - {i}_{b} 2$ (associative law)
$\quad \quad = {a}_{1} - {a}_{2} + i \left({b}_{1} - b 2\right)$ (distributive law)
$\quad \quad =$(a_1-a_2) + i (b_1-b_2) (real and imaginary parts of result)`

Given ${z}_{1}=2.1+i$${z}_{1} = 2.1 + i$ and ${z}_{2}=-2.1+i$${z}_{2} = - 2.1 + i$ What is ${z}_{1}-{z}_{2}$${z}_{1} - {z}_{2}$?

The answer is '$4.2$$4.2$'

summary

Subtraction of 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}\right)+i\left({b}_{1}-{b}_{2}\right)$${z}_{1} - {z}_{2} = \left({a}_{1} - {a}_{2}\right) + i \left({b}_{1} - {b}_{2}\right)$

Subtraction of two complex numbers is the subtraction of real and imaginary parts individually.

Outline