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# Algebra of complex numbers multiplication

Multiplication of two complex numbers generates a product that is also a complex number. In this article, we will know about what the algebra of complex numbers is.

Multiplying complex numbers is a basic operation on complex numbers that concerns multiplying two or more complex numbers. When compared to complex number addition and subtraction, it is a more difficult operation. A complex is stated as a + ib where i is an imaginary concept and a and b are real numbers. The mechanism of complex number multiplication is similar to that of binomial multiplication using the distributive property. In this article, we will learn about what the algebra of complex numbers is and the algebra of complex numbers multiplication.

## Multiplying Complex Numbers

Complex number multiplication is the process of multiplying two or more complex numbers using the distributive property. If we have two complex numbers, z = a + ib and w = c + id, then multiplication of z and w is written as zw = (a + ib) (c + id). To find the product of complex numbers, we use the distributive property of multiplication.

### What is the formula for multiplying Complex Numbers?

Complex numbers are multiplied in the same way that polynomials are multiplied.

The formula for multiplication of complex numbers is given as:

(a + ib) (c + id) = ac + iad + ibc + i2bd

⇒ (a + ib) (c + id) = (ac – bd) + i(ad + bc)

[Because i2 = -1]

### How Can You Multiply Complex Numbers In Polar Form?

A complex number in polar form is written as z = r (cos + I sin), where r is the complex number’s modulus and is its argument.

You can multiply complex numbers using the following formula

z1= r1 (cos θ1+ i sin θ2) and z2 = r2(cos θ1+i sin θ2) in polar form is given as:

z1z2 = [r1 (cos θ1 + i sin θ2)] [r2 (cos θ1 + i sin θ2)]

= r1 r2 (cos θ1 cos θ2 + i cos θ1 sin θ2 + i sin θ1 cos θ2 + i2 sin θ1 sin θ2)

= r1 r2 (cos θ1 cos θ2 + i cos θ1 sin θ2 + i sin θ1 cos θ2 – sin θ1 sin θ2)

{Because i2 = -1}

= r1 r2 [cos θ1 cos θ2 – sin θ1 sin θ2 + i (cos θ1 sin θ2 + sin θ1 cos θ2 )]

= r1 r2 [cos (θ1 + θ2) + i sin (θ1 + θ2)] {Because cos a cos b – sin a sin b = cos (a + b) and sin a cos b + sin b cos a = sin (a + b)}

Therefore, the formula for multiplying complex is in polar form is

[r1 (cos θ1 + i sin θ1)] [r2 (cos θ2 + i sin θ2)] = r1 r2 [cos (θ1 + θ2) + i sin (θ1 + θ2)]

### Multiplying Pure Imaginary And Real Numbers

If you want to multiply pure imaginary and real numbers, then the formula is

(a + ib) (c + id) = (ac – bd) + i(ad + bc).

If the b = 0, then you can say that the two complex numbers are ‘a’ and ‘c + id’.

Hence, the formula for multiplying complex numbers with real numbers becomes

(c + id) = ac + iad. For example, we multiply 2 with 1 + 3i as:

2 × (1 + 3i) = 2 + 6i

Then, if we do multiplication of a purely imaginary number of the form bi with a complex number, then the formula becomes (bi) (c + id) = ibc – bd. For example, if we multiply a complex number 2 + 3i with -5i, we have:

(-5i) (2 + 3i) = -10i -15i2

= -10i + 15

Solved Examples

1. Question

Two complex numbers are given, multiply them to find the answer

z = 3 − 2i and w = − 4 + 3i.

Solution

You can use the following formula if you want to multiply complex numbers z and w

(a + ib) (c + id) = (ac – bd) + i(ad + bc).

The values of variables are a = 3, b = -2, c = -4, d = 3

(3 − 2i) (− 4 + 3i) = [3 × (-4) – (-2) × 3) + i(3 × 3 + (-2) × (-4))

= (-12 + 6) + i (9 + 8)

= -6 + 17i

1. Question

Find the product of the given complex number

(1 + 4i) and (3 + 5i).

Solution

(1 + 4i) (3 + 5i)

= (3 + 12i) + (5i + 20i2)

= 3 + 17i − 20

= −17 + 17i

1. Question

Determine the product of 5 and (4 + 7i).

Solution

5 ∗ (4+7i) can be written as (5 + 0i) ∗ (4 + 7i)

= 5 (4+7i)

= 20 +35i

1. Question

Find the product of the given equation

3i and (2 + 6i).

Solution

3i ∗ (2 + 6i) can be viewed as (0 + 3i) ∗ (2 + 6i)

= 3i ∗(2 + 6i)

= 6i + 18i2

= 6i − 18

= −18 + 6i

1. Question

Find the product of mentioned complex numbers

(5 + 3i)  and  (3 + 4i).

Solution

(5+3i) ∗ (3+4i) = (5 + 3i) ∗ 3 + (5 + 3i) ∗ 4i

= (15 + 9i) + (20i + 12i2)

= (15 − 12) + (20 + 9)i

= 3 + 29i

1. Question

Find the product of two complex numbers (-2 + √3i) and (-3 + 2√3i) and express the result in standard form A + iB.

Solution:

(-2 + √3i)(-3 + 2√3i)

= -2(-3 + 2√3i) + √3i(-3 + 2√3i)

= 6 – 4√3i – 3√3i + 2(√3i)2

= 6 – 7√3i – 6

= 6 – 6 – 7√3i

= 0 – 7√3i, which is the required form A + iB, where A = 0 and B = – 7√3

### Conclusion

A complex number is a combination of an imaginary and a real number. When multiplying complex numbers, keep in mind that the same properties that we use when performing arithmetic with real numbers also apply to complex numbers.

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