Trigonometric form of writing a complex number

Every complex number z=a+bi that is different from zero can be written in the form

z = r(cosφ + i sinφ),

where r - is the modulus of the number, and φ - is one (any) of its arguments. This notation is called the trigonometric form of a complex number. In order to move from the algebraic form of writing a number to the trigonometric form, it is enough to find modulus of a complex number and one of its arguments. Two numbers written in trigonometric form are equal if and only if their moduli are equal and the difference of their arguments is a multiple of 2π.

Example 1.
Write the number z = -5 in trigonometric form.
Solution.
z = -5 ⇔ a=-2; b=0 ⇔ r=5; cosφ = -5/5=-1; sinφ = 0. These cosine and sine values correspond to the value of the argument φ=π. Therefore, z=5(cos(π) + i sin(π)).
Answer: z=5(cos(π) + i sin(π)).

Example 2.
Write the number z = i in trigonometric form.
Solution.
z = i ⇔ a=0; b=1 ⇔ r=1; cosφ = 0; sinφ = 1. These cosine and sine values correspond to the value of the argument φ=π/2. Therefore, z=cos(π/2) + i sin(π/2)).
Answer: z=cos(π/2) + i sin(π/2)).

Example 3.
Write the number z = -1-i in trigonometric form.
Solution.
z = -1-i ⇔ a=-1; b=-1 ⇔ |z₁|=r=√((-1)²+(-1)²)=√2; cosφ = a/r = -1/√2; sinφ = b/r = -1/√2. These cosine and sine values correspond to the value of the argument φ=-3π/4. Therefore, z₁=√2(cos(-3π/4) + i sin(-3π/4)). Since any of its arguments can be used for the trigonometric form of notation, this form of notation is also possible: z₁=√2(cos(5π/4) + i sin(5π/4)).
Answer: z=√2(cos(-3π/4) + i sin(-3π/4)).

Example 4.
Write the number z = 3cos(-9π/4) - 3i sin(π/4) in trigonometric form.
Solution.
In this case, there is no need to find the modulus and argument z. It is enough to note that z = 3(cos(-9π/4) - i sin(π/4)) and cos(-9π/4)=cos(-π/4), and -i sin(π/4)=i sin(-π/4). Therefore, z = 3(cos(-π/4) + i sin(-π/4)).
Answer: z=3(cos(-π/4) + i sin(-π/4)).

Example 5.
Write the number z = cos(π) + i sin(π/2) in trigonometric form.
Solution.
To write this number in trigonometric form, we first write it in algebraic form: z = cos(π) + i sin(π/2) = -1 + i. Therefore, z = -1+i ⇔ a=-1; b=1 ⇔ r=√2; cosφ = -1/√2; sinφ = 1/√2. These cosine and sine values correspond to the value of the argument φ=3π/4. Thus, z = √2(cos(3π/4) + i sin(3π/4)).
Answer: z = √2(cos(3π/4) + i sin(3π/4)).

The trigonometric form of writing complex numbers turns out to be very simple and convenient when multiplying and dividing complex numbers.

Multiplication of numbers written in trigonometric form

Let z₁ = r₁(cosφ₁+i sinφ₁), z₂ = r₂(cosφ₂+i sinφ₂). We obtain the formula for their product in trigonometric form:

z₁z₂ = r₁r₂(cosφ₁cosφ₂-sinφ₁sinφ₂+i sinφ₁cosφ₂+icosφ₁sinφ₂),

z₁z₂ = r₁r₂(cos(φ₁+φ₂) + i sin(φ₁+φ₂)).

Thus,

|z₁z₂| = r₁r₂,     arg(z₁z₂) = φ₁+φ₂ + 2πk, k∈Z.

The following statement is true: the modulus of the product of two complex numbers is equal to the product of the moduli of these numbers, the argument of the product is equal to the sum of the arguments of the factors.

Example 6.
Find the product of numbers z₁ = √2(cos11π/4+i sin11π/4), z₂ = √8(cos3π/8+i sin3π/8).
Solution.
According to the rule of multiplication of numbers written in trigonometric form, |z₁z₂|=√2√8=4. Argument of the product z₁z₂ equal to the sum of the arguments of the factors φ₁+φ₂ = 11π/4+3π/8=25π/8.
Therefore, z₁z₂=4(cos(25π/8) + i sin(25π/8))=4(cos(9π/8) + i sin(9π/8)).
Answer: z₁z₂ = 4(cos(9π/8) + i sin(9π/8)).

Example 7.
Find the product of numbers z₁ = 4(cos2π/5+i sin2π/5), z₂ = 3(cos5π/7+i sin5π/7).
Solution.
According to the rule of multiplication of numbers written in trigonometric form, |z₁z₂|=4*3=12. Argument of the product z₁z₂ equal to the sum of the arguments of the factors φ₁+φ₂ = 2π/5+5π/7=39π/35.
Therefore, z₁z₂=12(cos(39π/35) + i sin(39π/35))=12(cos(-31π/35) + i sin(-31π/35)).
Answer: z₁z₂ = 12(cos(-31π/35) + i sin(-31π/35)).

Division of numbers written in trigonometric form

We obtain a formula for the quotient of two complex numbers z₁ = r₁(cosφ₁+i sinφ₁) and z₂ = r₂(cosφ₂+i sinφ₂) in trigonometric form. Multiply the numerator and denominator of the quotient z₁/z₂ by the complex conjugate of the denominator cosφ₂-i sinφ₂:

z₁/z₂

=

r₁(cosφ₁+isinφ₁)(cosφ₂-isinφ₂)/r₂(cosφ₂+i sinφ₂)(cosφ₂-i sinφ₂)

=

=

r₁(cosφ₁cosφ₂+ sinφ₁sinφ₂ +isinφ₁cosφ₂-i cosφ₁sinφ₂)/r₂(cos²φ₂+sin²φ₂)

.

Therefore,

z₁/z₂

=

r₁/r₂

(cos(φ₁-φ₂) + i sin(φ₁-φ₂)).

That is,

|z₁/z₂| = r₁/r₂,     arg(z₁/z₂) = φ₁-φ₂ + 2πk, k∈Z.

Thus, the modulus of the quotient of two complex numbers is equal to the quotient of the moduli of these numbers, and the argument of the quotient is equal to the difference between the arguments of the dividend and the divisor.

Example 8.
Find the quotient of numbers z₁ = i-1, z₂ = cosπ/3+i sinπ/3.
Solution.
The number z₂ is written in trigonometric form. Let us represent the number z₁ = i-1 in trigonometric form: z₁ = i-1 ⇔ a=-1; b=1 ⇔ |z₁|=r=√2; cosφ = a/r = -1/√2; sinφ = b/r = 1/√2. These values ​​of cosine and sine correspond to the value of the argument φ=3π/4. Therefore, z₁ = √2(cos(3π/4) + i sin(3π/4)). According to the rule of division of numbers written in trigonometric form,|z₁/z₂|=√2, and argument of the quotient z₁/z₂ will be the difference between the arguments of the dividend and the divisor 3π/4-π/3 = 5π/12.
Thus, z₁/z₂=√2(cos(5π/12) + i sin(5π/12)).
Answer: z₁/z₂ = √2(cos(5π/12) + i sin(5π/12)).

Example 9.
Find the quotient of numbers z₁ = 6(cos7π/10+i sin7π/10), z₂ = 2(cosπ/5+i sinπ/5).
Solution.
According to the rule of division of numbers written in trigonometric form, |z₁/z₂|=6/2=3, and argument of the quotient z₁/z₂ will be the difference between the arguments of the dividend and the divisor 7π/10-π/5 = π/2.
Thus, z₁/z₂=3(cos(π/2) + i sin(π/2)) = 3(0+i) = 3i.
Answer: z₁/z₂ = 3(cos(π/2) + i sin(π/2)) = 3i.