Algebra of Complex Numbers
Algebraic Operations on Complex Numbers
The basic algebraic operations on complex numbers discussed here are:
- Addition of Two Complex Numbers
- Subtraction(Difference) of Two Complex Numbers
- Multiplication of Two Complex Numbers
- Division of Two Complex Numbers.
Addition of Two Complex Numbers
We know that a complex number is of the form z=a+ib where a and b are real numbers.
Let two complex numbers z_{1} = a_{1} + ib_{1} and z_{2} = a_{2} + ib_{2}
Then the addition of the complex numbers z_{1} and z_{2} is defined as,
z_{1}+z_{2} =( a_{1}+a_{2} )+i( b_{1}+b_{2} )
The real part of the resulting complex number is the sum of the real part of each complex number and the imaginary part of the resulting complex number is equal to the sum of the imaginary part of each complex number.
That is, Re(z_{1}+z_{2} )= Re( z_{1} )+Re( z_{1} )
Im( z_{1}+z_{2} )=Im( z_{1})+Im(z_{2})
Example:
z_{1} = a+3i, z_{2} = 4+bi, z_{3} = 6+10i
Find the value of a and b if z_{3} = z_{1}+z_{2}
Solution:
By the definition of addition of two complex numbers,
Re(z_{3} ) = Re(z_{1} )+Re(z_{2} )
6 = a + 4
a = 6 – 4 = 2
Im(z_{3} ) = Im(z_{1} ) + Im(z_{2} )
10 = 3+b
b = 10-3 =7
Properties of Addition of Complex Numbers
Name of the Property |
Description |
Expression |
Closure property |
Addition of two complex numbers is a complex number |
z_{1} + z_{2} = z |
Commutative property |
Order of addition of two complex numbers, does not change the result |
z_{1} + z_{2} = z_{2} + z_{1} |
Associative property |
Regrouping three complex numbers, while adding them, does not change the result |
(z_{1}+z_{2})+z_{3} = z_{1}+(z_{2}+z_{3}) |
Additive inverse property |
If z = a+ib is a complex number, then its additive inverse will be -z = -a – ib |
z+(-z) = 0 |
Additive identity |
If a value added to complex number results in the same complex number, then it becomes the additive identity |
(a+ib) + (0 + i0) = a + ib |
Difference of Two Complex Numbers
Let the complex numbers z_{1} = a_{1}+ib_{1} and z_{2} = a_{2}+ib_{2}, then the difference of z_{1} and z_{2}, z_{1}-z_{2} is defined as,
z_{1}-z_{2} = (a_{1}-a_{2})+i(b_{1}-b_{2})
Re(z_{1}-z_{2})=Re(z_{1})-Re(z_{2})
Im(z_{1}-z_{2})=Im(z_{1})-ImRe(z_{2})
Example:
z_{1} =8+ai,z_{2}=6+4i,z_{3} =2. Find the value of a if z_{3}=z_{1}-z_{2}
Solution:
By the definition of difference of two complex numbers,
Im_{3}=Im_{1}-Im_{2}
0 = a – 4
a = 4
Note: All real numbers are complex numbers with imaginary part as zero.
Multiplication of Two Complex Numbers
We know the expansion of (a+b)(c+d)=ac+ad+bc+bd
Similarly, Let the complex numbers z_{1} = a_{1}+ib_{1} and z_{2} = a_{2}+ib_{2}
Then, the product of z_{1} and z_{2} is defined as:
z_{1} z_{2}=(a_{1}+ib_{1})(a_{2}+ib_{2})
z_{1} z_{2} = a_{1} a_{2}+a_{1} b_{2} i+b_{1} a_{2} i+b_{1} b_{2} i_{2}
Since, i^{2} = -1, therefore,
z_{1} z_{2} = (a_{1} a_{2} – b_{1} b_{2} ) + i(a_{1} b_{2} + a_{2} b_{1} )
Example:
z_{1}=6-2i, z_{2}=4+3i. Find z_{1} z_{2}
Solution:
z_{1} z_{2} = (6-2i) (4+3i)
= 6 × 4 + 6 × 3i + (-2i) × 4 + (-2i)(3i)
= 24 + 18i – 8i – 6i^{2}
= 24 + 10i + 6
= 30 + 10i
Multiplicative inverse of a complex number
For any non-zero complex number z=a+ib(a≠0 and b≠0) there exists another complex number z^{-1} or 1/z, which is known as the multiplicative inverse of z such that zz^{-1} = 1.
z = a+ib, then,
Example:
z = 3 + 4i
Solution The numerator of z^{-1} is conjugate of z, that is a – ib
Denominator of z^{-1} is sum of squares of the Real part and imaginary part of z
Here, z = 3 + 4i
Properties of Multiplication of Complex Numbers
Name of the Property |
Description |
Expression |
Closure property |
Product of two complex number is a complex number only |
z_{1} x z_{2} = z |
Commutative property |
Change of order of complex numbers, does not change the result of their product |
z_{1}.z_{2} = z_{2}.z_{1} |
Associative property |
Regrouping of complex numbers, does not change the result of their product |
z_{1}(z_{2}.z_{3}) = (z_{1}.z_{2})z_{3} |
Distributive property |
Multiplication of a complex number with the sum of two complex numbers is given by: |
z_{1}(z_{2}+z_{3}) = z_{1}.z_{2} + z_{1}.z_{3} |
Division of Complex Numbers
Let the complex number z_{1} = a_{1} + ib_{1} and z_{2} = a_{2} + ib_{2}, then the quotient of z_{1}/z_{2} is defined as, z_{1}/z_{2=} z_{1(}z_{2}^{-1})
Therefore, to find z_{1}/z_{2}, we have to multiply z_{1} with the multiplicative inverse of z_{2}.
Now, let us discuss in detail about the division of complex numbers:
Let z_{1} = a_{1}+ib_{1} and z_{2} = a_{2}+ib_{2}, then z_{1}/z_{2} is given as:
z_{1}/z_{2} = (a_{1}+ib_{1})/(a_{2}+ib_{2})
Hence, (a_{1}+ib_{1})/(a_{2}+ib_{2}) = [(a_{1}+ib_{1})(a_{2}-ib_{2})]/[(a_{2}+ib_{2})(a_{2}-ib_{2})]
(a_{1}+ib_{1})/(a_{2}+ib_{2}) = [(a_{1}a_{2})-(a_{1}b_{2}i)+(a_{2}b_{1}i)+b_{1}b_{2})]/[(a_{2}^{2}+b_{2}^{2})]
(a_{1}+ib_{1})/(a_{2}+ib_{2}) = [(a_{1}a_{2})+(b_{1}b_{2}) +i(a_{2}b_{1}-a_{1}b_{2})]/(a_{2}^{2}+b_{2}^{2})
Hence,
Example:
If z_{1} = 2 + 3i and z_{2} = 1 + i, find $\frac{z_1}{z_2}$z_{1}z_{2}