Algebraic Operations On Complex Numbers
In Mathematics, algebraic operations on complex numbers are given by four basic arithmetic operations which include addition, subtraction, multiplication, and division. A complex number is the combination of a real number and an imaginary number.
The algebraic operations on complex numbers are defined purely by the algebraic methods. Some basic algebraic laws like associative, commutative, and distributive law are used to explain the relationship between the number of operations. By the use of these laws, the algebraic expressions are solved in a simple way. Since algebra is a concept based on known and unknown values (variables), its own rules are created to solve the problems. In this article, let us discuss the basic algebraic operations on complex numbers with examples.
Table of Contents: |
What are Complex Numbers?
In Maths, basically, a complex number is defined as the combination of a real number and an imaginary number. Real numbers are the numbers that we usually work on to do mathematical calculations. But the imaginary numbers are not generally used for calculations but only in the case of complex numbers.
Equality of Complex Numbers
Assume that z_{1} and z_{2} are the two complex numbers.
Here z_{1} = a_{1}+i b_{1} and z_{2} = a_{2}+ib_{2}
If both the complex numbers z_{1} and z_{2} are equal (i.e) z_{1} = z_{2}, then we can say that the real part of the first complex number is equal to the real part of the second complex number, whereas the imaginary part of the first complex number is equal to the imaginary part of the second complex number.
(i.e) Re(z_{1}) = Re(z_{2}) and Im(z_{1}) = Im(Z_{2})
Thus, the equality of complex number states that,
if a_{1}+ib_{1} = a_{2}+ib_{2}, then a_{1} = a_{2} and b_{1} = b_{2}.
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.
Consider 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} ) |
We can see that 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})
For the complex numbers,
z_{1} = a_{1}+ib_{1}
z_{2} = a_{1}+ib_{2}
z_{3} = a_{3}+ib_{3}
………..
………..
z_{n} = a_{n}+ib_{n}
a_{1}+a_{2}+a_{3}+….+a_{n} = (a_{1}+a_{2}+a_{3}+….+a_{n} )+i(b_{1}+b_{2}+b_{3}+….+b_{n})
Let’s Work Out:
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 |
Conjugate of Complex number
Conjugate of a complex number z=a+ib is given by changing the sign of the imaginary part of z which is denoted as \( \bar z \)
\(\bar z = a-ib \\ z+ \bar z =2a \\ z- \bar z =2bi\)
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
Consider 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}) |
From the definition, it is understood that,
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} =4+ai,z_{2}=2+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, consider 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} ) |
Let us see here a solved example based on the multiplication of complex numbers.
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
Definition: 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,
\(z^{-1} = \frac{a}{a^2 + b^2} + i \frac{(-b)}{a^2 + b^2}\\ Re(z^{-1}) = \frac{a}{a^2 + b^2}\\ Im(z^{-1}) = \frac{-b}{a^2 + b^2}\)
Example:
z = 3 + 4i Solution: \(z^{-1} ~of ~a + ib = \frac{a}{a^2 + b^2} +i \frac{(-b)}{a^2 + b^2}= \frac{(a-ib)}{a^2 + b^2}\\\) 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 \(z^{-1} = \frac{3-4i}{3^2 + 4^2} = \frac{3-4i}{25}\\ z^{-1}= \frac{3}{25} – \frac{4i}{25}\) |
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
Consider 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,
\(\frac{z_1}{z_2}= z_1 \times \frac{1}{z_2}\)
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, \(\frac{z_{1}}{z_{2}} = \frac{a_{1}a_{2}+b_{1}b_{2}}{a_{2}^{2}+b_{2}^{2}}+ i\frac{a_{2}b_{1}-a_{1}b_{2}}{a_{2}^{2}+b_{2}^{2}}\)
Example:
If z_{1} = 2 + 3i and z_{2} = 1 + i, find z_{1}/z_{2}. Solution: \(\frac{z_1}{z_2} = z_1 \times \frac{1}{z_2}\) \(\frac{2+3i}{1+i} = (2+3i) \times \frac{1}{1+i}\) \(Since, \frac{1}{1+i} = \frac{1-i}{1^2 + 1^2} = \frac{1-i}{2}\) \(\frac{2 + 3i}{1 + i} = 2+3i \times \frac{1-i}{2} = \frac{(2+3i)(1-i)}{2}\) \(=\frac{2 – 2i + 3i – 3i^2}{2} = \frac{5+i}{2}\) |
Algebraic Operations on Complex Numbers Summary
Assume that z_{1} = a_{1}+ib_{1} and z_{2} = a_{2}+ib_{2} are the two complex numbers
Addition: z_{1}+z_{2} =( a_{1}+a_{2} )+i( b_{1}+b_{2} )
Subtraction: z_{1}-z_{2} = (a_{1}-a_{2})+i(b_{1}-b_{2})
Multiplication: z_{1} z_{2} = (a_{1} a_{2} – b_{1} b_{2} ) + i(a_{1} b_{2} + a_{2} b_{1} )
Division: \(\frac{z_{1}}{z_{2}} = \frac{a_{1}a_{2}+b_{1}b_{2}}{a_{2}^{2}+b_{2}^{2}}+ i\frac{a_{2}b_{1}-a_{1}b_{2}}{a_{2}^{2}+b_{2}^{2}}\)
Solved Examples
Question 1:
Add 2+4i and -1+3i.
Solution:
Given the two complex numbers are:
2+4i and -1+3i
(2+4i)+(-1+3i)
⇒ (2-1)+(4i+3i)
⇒ 1 + 7i
Question 2:
Simplify: 7 + i + 4 + 4.
Solution:
7 + i + 4 + 4
⇒ (7+4+4) + i
⇒ 15 + i
Example 3:
Multiply the complex numbers: (5+3i). (3+4i)
Solution:
Given:(5+3i). (3+4i)
(5+3i). (3+4i) = 15+20i+9i-12
(5+3i). (3+4i) = (15-12) + i(20+9)
(5+3i). (3+4i) = 3+29i
Hence, the product of (5+3i) and (3+4i) is 3+29i.
Example 4:
Subtract (2+5i) from (7+15i).
Solution:
We know that (a+bi) – (c+di) = (a-c) + i(b-d).
Hence,
(7+15i) – (2+5i) = (7-2)+i(15-5)
(7+15i) – (2+5i) =5+10i
Hence, (7+15i) – (2+5i) is 5+10i
Practice Questions
Simplify the following:
- −3 + 6i − (−5 − 3i) − 8i
- 4i(−2 − 8i)
- (−2 − i)(4 + i)
- (−2 − 2i)(−4 − 3i)(7 + 8i)
- −3i ⋅ 6i − 3(−7 + 6i)
- (6i)^{3}
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Frequently Asked Questions on Algebraic operations on complex numbers
What are complex numbers?
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What are the four algebraic operations?
How do we multiply two complex numbers?
(a+bi).(c+di) = (ac-bd) + i(bc+ad).