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Operations on Complex Numbers | Algebra - Mathematics PDF Download

The Complex Algebra
In basic algebra of numbers, we have four operations namely – addition, subtraction, multiplication and division. As we will see in a bit, we can combine complex numbers with them. Let z1 and z2 be any two complex numbers and let, z1 = a+ib and z2 = c+id.
Operations on Complex Numbers | Algebra - Mathematics
Example: Schrodinger Equation which governs atoms is written using complex numbers
Addition and Subtraction of Imaginary Numbers

The addition of two complex numbers is defined as:
z1+z2 = (a+ib) ± (c+id) = (a+c) ± i(b+d)
Which gives another complex number whose real part is Re(z1) + Re(z2) = a + c and imaginary part of the new complex number = Im(z1) + Im(z2) = b + d. For example, on adding 2 and 3 + 4i, we can write 2 as 2 + 0*i and therefore, 2 + (3+4i) = (2+3) + i(0+4) = 5 + i.4

Properties of Addition and Subtraction
The addition or subtraction of complex numbers always results in a complex number. This operation follows the following rules *:
• The Closure law or the closure property: Addition or subtraction of complex numbers yields a complex number.
• The Commutative Law: Any two complex numbers commute with respect to addition or subtraction i.e. z1+z2=z2+z1,
• The Associative Law: If z1, z2, z3 are any three complex numbers then we have  (z1+z2)+z3 = z1+(z2+z3)
• Additive Identity: Additive identity is such a complex number that will return the original number upon addition. For example for real numbers, 0 is the additive identity. Similarly, we see for complex numbers 0 + i.0 is the additive identity (We will just denote it by 0).
• Additive Inverse: For each operation involving the combination of complex numbers through addition or subtraction, there exists an inverse such that the addition or subtraction of a complex number with it, yields the additive identity. For every complex number z = a + ib, there exists a complex number – z = -a + i(-b) such that z + (-z) = 0 or the additive identity.
*These properties are very important. Any collection of mathematical objects (set) that follows the above-mentioned properties under an operation (here ‘+’ and ‘-‘) is said to be a Group. You will learn about Groups and Fields in higher algebra. Groups are used in the development of almost all modern scientific theories.
Operations on Complex Numbers | Algebra - Mathematics
Complex Numbers are used to study the shapes of atoms and molecules

Multiplication and Division of Imaginary Numbers
The multiplication of two complex numbers is defined as:
(z1×z2) = (a+ib) × (c+id) = (ac–bd)+i(ad + bc)
Similar z1/z2 = z1 × 1/z2; we can use cross multiplication and the multiplication of complex numbers for division. The multiplication and division also form a group i.e. they have similar properties as addition and subtraction.

Properties of Multiplication And Addition
• The Closure law or the closure property: (z1×z2) is always a complex number.
• The Commutative Law: If z1 and z2 are two  complex numbers the, (z,×z2) = (z2×z1)
• The Associative Law: If z1, z2, z3 are any three complex numbers then we have  (z1×z2) × z3 = z1 × (z2×z3).
• Multiplicative Identity: For each complex number there exists a number 1 + i.0 such that z×( 1 + i.0) = z ; where z is a complex number.
• Multiplicative Inverse: For each complex number z there exists a number 1/z such that z×(1/z) = 1 + i.0. This is known as the Multiplicative Inverse.
• The Distributive Law: If z1, z2, z3 are any three complex numbers then we have z1 × (z2 + z3) = z1 z2 + z1 z3 (Left distributive law) and ( z1 + z2 ) × z3 = z1 z3 + z1 z2
The division also follows the same properties.

Solved Examples For You
Question: Find the smallest integer n such that Operations on Complex Numbers | Algebra - Mathematics is
A) 16
B) 12
C) 8
D) 4
Solution:(D). Here we use a little trick. We have Operations on Complex Numbers | Algebra - Mathematics
Remember that i×i = -1 and therefore we have, Operations on Complex Numbers | Algebra - Mathematics. Hence using it in the denominator of the given equation, we have:Operations on Complex Numbers | Algebra - MathematicsHence, for (i)n = 1, n should be atleast 4.

The document Operations on Complex Numbers | Algebra - Mathematics is a part of the Mathematics Course Algebra.
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FAQs on Operations on Complex Numbers - Algebra - Mathematics

1. What are complex numbers?
Ans. Complex numbers are numbers that consist of a real part and an imaginary part. They are written in the form a + bi, where a is the real part and bi is the imaginary part. The imaginary unit i is defined as the square root of -1.
2. How do you add complex numbers?
Ans. To add complex numbers, you simply add the real parts together and add the imaginary parts together. For example, to add (3 + 2i) and (1 + 4i), you add 3 + 1 to get 4, and add 2i + 4i to get 6i. So the sum is 4 + 6i.
3. Can you subtract complex numbers?
Ans. Yes, you can subtract complex numbers. To subtract complex numbers, you subtract the real parts and subtract the imaginary parts separately. For example, to subtract (5 + 3i) from (2 + 8i), you subtract 5 - 2 to get -3, and subtract 3i - 8i to get -5i. So the difference is -3 - 5i.
4. How do you multiply complex numbers?
Ans. To multiply complex numbers, you use the distributive property and the fact that i^2 is equal to -1. For example, to multiply (2 + 3i) and (-4 + 5i), you multiply 2 by -4 to get -8, multiply 2 by 5i to get 10i, multiply 3i by -4 to get -12i, and multiply 3i by 5i to get 15i^2. Simplifying these terms, you get -8 + 10i - 12i + 15i^2. Since i^2 is equal to -1, this becomes -8 + 10i - 12i - 15. Combining like terms, the product is -23 - 2i.
5. How do you divide complex numbers?
Ans. To divide complex numbers, you multiply the numerator and denominator by the conjugate of the denominator. The conjugate of a complex number a + bi is a - bi. For example, to divide (6 + 4i) by (2 - 3i), you multiply the numerator and denominator by the conjugate of (2 - 3i), which is (2 + 3i). This gives you (6 + 4i)(2 + 3i) / (2 - 3i)(2 + 3i). Expanding this expression, you get (12 + 18i + 8i + 12i^2) / (4 - 9i^2). Simplifying further, you get (12 + 26i - 12) / (4 + 9). This becomes -12 + 26i / 13, which simplifies to -12/13 + 2i.
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