**Complex Numbers Questions for NDA, CDS, AFCAT **

Complex numbers play a vital role in mathematics and have significant applications in various fields, including engineering, physics, and computer science. For aspirants preparing for competitive exams such as NDA (National Defence Academy), CDS (Combined Defence Services), and AFCAT (Air Force Common Admission Test), a thorough understanding of complex numbers is essential. This article aims to provide a curated set of practice on Complex numbers questions to help you enhance your problem-solving skills and excel in these prestigious exams.

**Complex Numbers in Mathematics**

Complex numbers are a type of numbers that can be expressed in the form of a+ib, where ‘a’ and ‘b’ are real numbers and ‘i’ represents the imaginary unit, also known as “iota.” The value of ‘i’ is equal to the square root of -1.

**Examples of Complex Numbers:
**1+i

33+3i

0.6-2.2i

**Complex Numbers Definition**

A complex number is a mathematical concept that combines a real number and an imaginary number. It can be represented in the form of a+ib, where ‘a’ is a real number, ‘b’ is a real number multiplied by the imaginary unit ‘i’, and ‘i’ is equal to the square root of -1. The real part, ‘a’, represents the purely real component of the complex number, while the imaginary part, ‘ib’, represents the purely imaginary component. The real and imaginary parts are distinct and separate, creating a unique combination that forms the complex number.

Examples:

Complex Number | Real Number | Imaginary Number |

1+3i | 1 | 3i |

8-9i | 8 | -9i |

-5i | 0 | -5i |

In the realm of complex numbers, every real number can be considered as a complex number with a zero imaginary component. Hence, the number 0, being a real number, can also be regarded as a complex number. In the standard form of complex numbers, 0 can be represented as 0+0i, where the real part is 0 and the imaginary part is 0 multiplied by ‘i’. Thus, 0 can be viewed as a specific instance of a complex number with no imaginary contribution.

## Complex Number Formulas

When performing arithmetic operations on complex numbers, it is important to combine similar terms. This involves combining the real numbers with other real numbers and the imaginary numbers with other imaginary numbers. By grouping like terms together, we can simplify the expression and obtain the final result.

**Addition**

When adding complex numbers, (a + ib) and (c + id), we combine the real parts (a and c) and the imaginary parts (b and d) separately. The result is obtained by adding the real parts and adding the imaginary parts, which gives us (a + c) + i(b + d).

**Subtraction**

When subtracting complex numbers, (a + ib) and (c + id), we also combine the real parts (a and c) and the imaginary parts (b and d) separately. The result is obtained by subtracting the real parts and subtracting the imaginary parts, which gives us (a – c) + i(b – d).

**Multiplication**

For multiplication of two complex numbers, (a + ib) and (c + id), we use the distributive property, also known as the FOIL method. This involves multiplying each term of the first complex number with each term of the second complex number. The result is obtained by combining the real parts (ac – bd) and the imaginary parts (ad + bc), resulting in (ac – bd) + i(ad + bc).

**Division****
**Division of complex numbers involves multiplying the numerator and denominator by the conjugate value of the denominator. The conjugate of (c + id) is (c – id). By applying the FOIL method to the numerator and denominator, we obtain [(ac + bd) + i(bc – ad)] / (c^2 + d^2).

## Complex Number: Solved Examples

**Question 1:** (3 + 2i)(1 + 7i)

**Solution:** To multiply the complex numbers (3 + 2i) and (1 + 7i), we can use the distributive property of multiplication.

(3 + 2i)(1 + 7i) can be expanded as follows:

(3)(1) + (3)(7i) + (2i)(1) + (2i)(7i)

Simplifying further:

3 + 21i + 2i + 14i^2

Since i^2 is equal to -1, we can substitute it in the equation:

3 + 21i + 2i – 14

Combining like terms, we get:

-11 + 23i

Therefore, the product of (3 + 2i) and (1 + 7i) is -11 + 23i.

**Question 2:** (1 + i)^2

**Solution:** To calculate (1 + i) squared, we can multiply (1 + i) by itself:

(1 + i)(1 + i) can be expanded as follows:

(1)(1) + (1)(i) + (i)(1) + (i)(i)

Simplifying further:

1 + i + i + i^2

Since i^2 is equal to -1, we can substitute it in the equation:

1 + i + i – 1

Combining like terms, we get:

2i

Therefore, (1 + i) squared is equal to 2i.

**Question 3: **(4-5i)(12+11i)

**Solution: ** To solve (4 – 5i)(12 + 11i), we can use the distributive property:

(4 – 5i)(12 + 11i) = 4 * 12 + 4 * 11i – 5i * 12 – 5i * 11i

Simplifying further:

= 48 + 44i – 60i – 55i^2

Since i^2 is equal to -1, we can substitute it in the equation:

= 48 + 44i – 60i – 55(-1)

= 48 + 44i – 60i + 55

Combining like terms, we get:

= 103 – 16i

Therefore, (4 – 5i)(12 + 11i) is equal to 103 – 16i.

**Question 4: **(2+7i)(8+3i)

**Solution:** To solve (2 + 7i)(8 + 3i), we can use the distributive property:

(2 + 7i)(8 + 3i) = 2 * 8 + 2 * 3i + 7i * 8 + 7i * 3i

Simplifying further:

= 16 + 6i + 56i + 21i^2

Since i^2 is equal to -1, we can substitute it in the equation:

= 16 + 6i + 56i + 21(-1)

= 16 + 6i + 56i – 21

Combining like terms, we get:

= -5 + 62i

Therefore, (2 + 7i)(8 + 3i) is equal to -5 + 62i.