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. Show
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 NumbersAssume that z1 and z2 are the two complex numbers. Here z1 = a1+i b1 and z2 = a2+ib2 If both the complex numbers z1 and z2 are equal (i.e) z1 = z2, 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(z1) = Re(z2) and Im(z1) = Im(Z2) Thus, the equality of complex number states that, if a1+ib1 = a2+ib2, then a1 = a2 and b1 = b2. The basic algebraic operations on complex numbers discussed here are:
Addition of Two Complex NumbersWe know that a complex number is of the form z=a+ib where a and b are real numbers. Consider two complex numbers z1 = a1 + ib1 and z2 = a2 + ib2 Then the addition of the complex numbers z1 and z2 is defined as, z1+z2 =( a1+a2 )+i( b1+b2 ) 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(z1+z2 )= Re( z1 )+Re( z1 ) Im( z1+z2 )=Im( z1)+Im(z2) For the complex numbers, z1 = a1+ib1 z2 = a1+ib2 z3 = a3+ib3 ……….. ……….. zn = an+ibn a1+a2+a3+….+an = (a1+a2+a3+….+an )+i(b1+b2+b3+….+bn) Let’s Work Out: Example: z1 = a+3i, z2 = 4+bi, z3 = 6+10i Find the value of a and b if z3 = z1+z2 Solution: By the definition of addition of two complex numbers, Re(z3 ) = Re(z1 )+Re(z2 ) 6 = a + 4 a = 6 – 4 = 2 Im(z3 ) = Im(z1 ) + Im(z2 ) 10 = 3+b b = 10-3 =7 Conjugate of Complex numberConjugate of a complex number z=a+ib is given by changing the sign of the imaginary part of z which is denoted as \(\begin{array}{l} \bar z \end{array} \) \(\begin{array}{l}\bar z = a-ib \\ z+ \bar z =2a \\ z- \bar z =2bi\end{array} \) Properties of Addition of Complex Numbers
Difference of Two Complex NumbersConsider the complex numbers z1 = a1+ib1 and z2 = a2+ib2, then the difference of z1 and z2, z1-z2 is defined as, From the definition, it is understood that, Re(z1-z2)=Re(z1)-Re(z2) Im(z1-z2)=Im(z1)-ImRe(z2) Example: z1 =4+ai,z2=2+4i,z3 =2. Find the value of a if z3=z1-z2 Solution: By the definition of difference of two complex numbers, Im3=Im1-Im2 0 = a – 4 a = 4 Note: All real numbers are complex numbers with imaginary part as zero. Multiplication of Two Complex NumbersWe know the expansion of (a+b)(c+d)=ac+ad+bc+bd Similarly, consider the complex numbers z1 = a1+ib1 and z2 = a2+ib2 Then, the product of z1 and z2 is defined as: z1 z2=(a1+ib1)(a2+ib2) z1 z2 = a1 a2+a1 b2 i+b1 a2 i+b1 b2 i2 Since, i2 = -1, therefore, z1 z2 = (a1 a2 – b1 b2 ) + i(a1 b2 + a2 b1 ) Let us see here a solved example based on the multiplication of complex numbers. Example: z1=6-2i, z2=4+3i. Find z1 z2 Solution: z1 z2 = (6-2i) (4+3i) = 6 × 4 + 6 × 3i + (-2i) × 4 + (-2i)(3i) = 24 + 18i – 8i – 6i2 = 24 + 10i + 6 = 30 + 10i Multiplicative inverse of a complex numberDefinition: 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, \(\begin{array}{l}z^{-1} = \frac{a}{a^2 + b^2} + i \frac{(-b)}{a^2 + b^2}\\ Re(z^{-1}) = \frac{a}{a^2 + b^2}\\ Im(z^{-1}) = \frac{-b}{a^2 + b^2}\end{array} \) Example: z = 3 + 4i Solution: \(\begin{array}{l}z^{-1} ~of ~a + ib = \frac{a}{a^2 + b^2} +i \frac{(-b)}{a^2 + b^2}= \frac{(a-ib)}{a^2 + b^2}\\\end{array} \) 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 \(\begin{array}{l}z^{-1} = \frac{3-4i}{3^2 + 4^2} = \frac{3-4i}{25}\\ z^{-1}= \frac{3}{25} – \frac{4i}{25}\end{array} \) Properties of Multiplication of Complex Numbers
Division of Complex NumbersConsider the complex number z1 = a1 + ib1 and z2 = a2 + ib2, then the quotient of z1/z2 is defined as, \(\begin{array}{l}\frac{z_1}{z_2}= z_1 \times \frac{1}{z_2}\end{array} \) Therefore, to find z1/z2, we have to multiply z1 with the multiplicative inverse of z2. Now, let us discuss in detail about the division of complex numbers: Let z1 = a1+ib1 and z2 = a2+ib2, then z1/z2 is given as: z1/z2 = (a1+ib1)/(a2+ib2) Hence, (a1+ib1)/(a2+ib2) = [(a1+ib1)(a2-ib2)]/[(a2+ib2)(a2-ib2)] (a1+ib1)/(a2+ib2) = [(a1a2)-(a1b2i)+(a2b1i)+b1b2)]/[(a22+b22)] (a1+ib1)/(a2+ib2) = [(a1a2)+(b1b2) +i(a2b1-a1b2)]/(a22+b22) Hence, \(\begin{array}{l}\frac{z_{1}}{z_{2}} = \frac{a_{1}a_{2}+b_{1}b_{2}}{a_{2}^{2}+b_{2}^{2}}+ i\frac{a_{2}b_{1}-a_{1}b_{2}}{a_{2}^{2}+b_{2}^{2}}\end{array} \) Example: If z1 = 2 + 3i and z2 = 1 + i, find z1/z2. Solution: \(\begin{array}{l}\frac{z_1}{z_2} = z_1 \times \frac{1}{z_2}\end{array} \) \(\begin{array}{l}\frac{2+3i}{1+i} = (2+3i) \times \frac{1}{1+i}\end{array} \) \(\begin{array}{l}Since, \frac{1}{1+i} = \frac{1-i}{1^2 + 1^2} = \frac{1-i}{2}\end{array} \) \(\begin{array}{l}\frac{2 + 3i}{1 + i} = 2+3i \times \frac{1-i}{2} = \frac{(2+3i)(1-i)}{2}\end{array} \) \(\begin{array}{l}=\frac{2 – 2i + 3i – 3i^2}{2} = \frac{5+i}{2}\end{array} \) Algebraic Operations on Complex Numbers SummaryAssume that z1 = a1+ib1 and z2 = a2+ib2 are the two complex numbers Addition: z1+z2 =( a1+a2 )+i( b1+b2 ) Subtraction: z1-z2 = (a1-a2)+i(b1-b2) Multiplication: z1 z2 = (a1 a2 – b1 b2 ) + i(a1 b2 + a2 b1 ) Division: \(\begin{array}{l}\frac{z_{1}}{z_{2}} = \frac{a_{1}a_{2}+b_{1}b_{2}}{a_{2}^{2}+b_{2}^{2}}+ i\frac{a_{2}b_{1}-a_{1}b_{2}}{a_{2}^{2}+b_{2}^{2}}\end{array} \) Solved ExamplesQuestion 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 QuestionsSimplify the following:
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Stay tuned with BYJU’S – The Learning App and download the app to learn all Maths related concepts easily by exploring more videos. Frequently Asked Questions on Algebraic operations on complex numbersComplex numbers are the combination of real numbers and imaginary numbers. It is represented by z = a + ib, where a is real part and ib is imaginary part. To add two complex numbers, add the real part and imaginary part separately. Subtract the real part from real part and imaginary part from imaginary part, to subtract two complex numbers. The four basic algebraic
operations are addition, subtraction, multiplication and division. If a+bi and c+di are the two complex numbers, then the product is given as: What is a complex number in Algebra 2?A complex number is a combination of a Real Number and an Imaginary Number, written as a + bi (where a and/or b may equal zero). (a and b are real numbers and i is the imaginary unit) • If we have a + bi with a = 0, we have 0 + bi which gives bi, a purely imaginary number.
How do you multiply complex numbers?(x + yi) u = xu + yu i. In other words, you just multiply both parts of the complex number by the real number. For example, 2 times 3 + i is just 6 + 2i. Geometrically, when you double a complex number, just double the distance from the origin, 0.
What is complex number example?Complex numbers are the numbers that are expressed in the form of a+ib where, a,b are real numbers and 'i' is an imaginary number called “iota”. The value of i = (√-1). For example, 2+3i is a complex number, where 2 is a real number (Re) and 3i is an imaginary number (Im).
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