Complex numbers are the numbers of the form x+iy where $ i=\sqrt{(-1)}$ and x and y are real numbers.
Definition:- Complex numbers are defined as an ordered pair of real numbers like (x,y) where
$z=(x,y)=x+iy$
Both x and y are real numbers and x is known as real part of complex number and y is known as imaginary part of the complex number.
We also write that as Re(z) =x and Im(z) =y
Two complex Number z1 = a + ib and z2 = c + id are equal if a = c and b = d.
Example
z1= 2+4i
z1= 8-4i
Algebra of Complex Number
Just like real numbers, Addition ,subtraction,multiplication ,division is also performed on Complex numbers
Addition of complex numbers
Let z1=x1+iy1 and z2=x2+iy2 then
z1+z2=(x1+x2)+i(y1+y2)
Therefore , Real part of both the complex number are added and similarly imaginary part of both the complex numbers are added Important Notes
Addition of Complex numbers also follows the closure, commutative,associative Laws.I will leave the proof of them to you.
We also have additive identity (0) for Complex numbers.
Additive inverse also exists for the complex number for z=x+iy, Additive inverse = -x-iy such that z+(-z) =0
Subtraction of complex numbers
z1-z2=(x1-x2)+i(y1-y2)
Here the real part are subtracted and imaginary part is also subtracted
Mulitiplication of complex numbers
(z1.z2)=(x1+iy1).(x2+iy2)
=x1x2 -y1y2 + i(x1y2 + x2y1)
Multiplication is similar to multiplication of algebraic expression and we just need put out i2=-1 wherever it comes Important Notes
Multiplication of Complex numbers also follows the closure, commutative,associative Laws.I will leave the proof of them to you.
We also have Multiplicative identity (0) for Complex numbers.
Multiplicative inverse also exists for the complex number .Details are given below
Multiplicative Inverse
for $z=a+ib$
z-1 is given by
=$(\frac{a}{a^{2}+b^{2}})+i(\frac{-b}{a^{2}+b^{2}})$
Division of Complex Numbers
To divide complex number by another complex number, first write quotient as a fraction. Then reduce the denominator complex number to multiplicative Inverse and then simple multiplication applies
Example -1
Find the value of
$\frac {1+i}{1+2i}$ Solution:
The multiplicative inverse of (1+2i) is given as
$=(\frac{a}{a^{2}+b^{2}})+i(\frac{-b}{a^{2}+b^{2}})$
$=(\frac{1}{5}-i \frac{2}{5})$
So $\frac {1+i}{1+2i}$
$=(1+i)(\frac{1}{5}-i \frac{2}{5})$
=$\frac {1}{5} -i \frac {2}{5}+ i \frac {1}{5}+\frac {2}{5}$
=$ \frac {=i}{5} + \frac {3}{5}$
Example -2
Find the value of (1+3i)(2-5i) Solution:
(1+3i)(2-5i) = 2-5i+ 6i-15i2
Now i2=-1
Therefore,
=17+i
