Operations of raising to a power and extracting a root for complex numbers written in trigonometric form

For any complex numbers z≠0 it is true z⁰ = 1. For complex numbers in trigonometric form, the formulas for operations of raising to a power with a natural exponent and extracting the root are very simple and easy to use. Therefore, for complex numbers written in algebraic form, you must first represent them in trigonometric form, then raise to a power or extract the root, and again switch to algebraic form.

Raising a complex number to a power with a natural exponent

If the formula for the product of two complex numbers in trigonometric form is generalized to several factors, then it is easy to obtain a formula for raising to a power with a natural exponent:

(r(cosφ + i sinφ))ⁿ = rⁿ(cos(nφ) + i sin(nφ)).

Thus, to raise a complex number to a natural power, its modulus must be raised to this power, and its argument must be multiplied by the exponent of this power.

Examples of raising complex numbers to a power with a natural exponent

Example 1.
Calculate (-1-i)¹⁰.
Solution.
First, let us represent the number z=-1-i in trigonometric form: z = a+bi = -1-i ⇔ a=-1; b=-1 => tgφ = b/a=1; φ=π/4. But since the number z is located in the fourth quadrant of the complex plane, φ=-3π/4. Thus,

z = -1-i = √2(cos(-3π/4)+i sin(-3π/4)).

In accordance with the rule of exponentiation of numbers written in trigonometric form, we have

z¹⁰ = (√2)¹⁰(cos(-30π/4)+i sin(-30π/4)) = 32(cos(-15π/2)+i sin(-15π/2)) = 32(0+i) = 32i.

Answer: (-1-i)¹⁰ = 32i.

Example 2.
Calculate (1/(1+i))⁸.
Solution.
Let's represent the number 1/(1+i) in trigonometric form. To do this, we first represent the numerator and denominator of the fraction in trigonometric form:
a+bi = 1 ⇔ a=1; b=0 ⇔ 1 = cos(0)+i sin(0).
a+bi = 1+i ⇔ a=1; b=1 => |1+i|=√2; tgφ = b/a=1; φ=π/4. Since the number 1+i is located in the first quadrant of the complex plane, φ=π/4. Thus, 1+i = √2(cos(π/4)+i sin(π/4)).
In accordance with the rule of division of complex numbers, we obtain:
1/(1+i) = 1/√2(cos(0-π/4)+i sin(0-π/4)) = 1/√2(cos(-π/4)+i sin(-π/4)).
Then
(1/(1+i))⁸ = (1/√2)⁸(cos(-8π/4)+i sin(-8π/4)) = (1/16)(cos(-2π)+i sin(-2π)) = 1/16.
Answer: (1/(1+i))⁸ = 1/16.

Example 3.
Calculate (-√3/ 2 + 1/2 i)⁶.
Solution.
Let's represent the number z=-√3/ 2 + 1/2 i in trigonometric form:
z = a+bi = -√3/ 2 + 1/2 i ⇔ a=-√3/ 2; b=1/2 ⇔ |z| = √ ¾ + ¼ =1; tgφ = b/a=-1/√3. Since the number z is located in the second quadrant of the complex plane, φ=5π/6.Thus,
z = cos(5π/6) + i sin(5π/6).
Then
z⁶ = cos(5π)+i sin(5π) = -1.
Answer: (-√3/ 2 + 1/2 i)⁶ = -1.

Example 4.
Calculate (i + √3)⁹.
Solution.
Let's represent the number i + √3 in trigonometric form:
z = a+bi = √3 + i ⇔ a=√3; b=1 ⇔ |z| = √(3+1) = 2; tgφ = b/a=1/√3. Since the number z is located in the first quadrant of the complex plane, φ=π/6. Thus,
z = 2(cos(π/6) + i sin(π/6)).
Then
z⁹ = 2⁹(cos(9π/6)+i sin(9π/6)) = 2⁹(cos(3π/2)+i sin(3π/2)) = 512(-i) = -512 i.
Answer: (i + √3)⁹ = -512 i.

Extracting the root of a complex number written in trigonometric form

The root of degree n∈N of a complex number z is a number w whose n-th degree is equal to z. It follows from the definition that each solution of the equation wⁿ = z s a root of the n-th degree of z. On the set of complex numbers, this equation has exactly n solutions. For example, for the square root of z=-1, we have w₁=√(-1)=i and w₂=√(-1)=-i. And only for z=0 for any natural n, it has a unique solution w=0.
Let z = r(cosφ+i sinφ), w = ρ(cosα+i sinα). Let's solve the equation wⁿ = z and we get the formula for the root of the n-th degree of a complex number.

wⁿ = z ⇔ (ρ(cosα+i sinα))ⁿ = r(cosφ+i sinφ) ⇔ρⁿ=r; α⋅n=φ+2πk, k∈Z.

Therefore,

ρ = r^1/n;    α=(φ+2πk)/n, n=0,1,2,…n-1.

Then, in trigonometric form, all solutions of the equation wⁿ = z can be written as follows:

w_k=r^1/n(cos(φ+2πk)/n )+i sin(φ+2πk)/n )), k∈Z.

It is easy to see that for k=0,1,2,…n-1 we will get different values of w_k, for the remaining values of k, there will be no new values of w. Thus, if z≠0, then there are exactly n roots of the n-th degree of z, and the formula is valid for them:

w_k = r^1/n(cos(φ+2πk)/n ) + i sin(φ+2πk)/n )),   k=0,1,2,…n-1.

All roots of degree n have the same modulus, and the arguments of the roots w_k and w_k+1 differ by the same number 2π/n. Therefore,The points of the complex plane corresponding to all the roots of w_k, k=0,1,2,…n-1, are located on a circle of radius r^1/n with the center (0;0) at the vertices of a regular n-gon.

Examples of extracting the root of a complex number written in trigonometric form

Example 5.
Find all the values of the root of the sixth degree of the number z = 1.
Solution.
Let's write the number z in trigonometric form: z = 1 ⇔ a=1; b=0 => |z|=r=√1; tgφ = b/a = 0. Since the number z is located in the first quadrant of the complex plane, φ=0. Thus, 1 = cos(0) + i sin(0).
According to the formula for extracting the root of the sixth degree, we have:

1^1/6 = cos(2πk/6) + i sin(2πk/6),   k=0,1,2,3,4,5.

Therefore, for all six roots we get
z₀ = cos(0) + i sin(0) = 1,
z₁ = cos(2π/6) + i sin(2π/6) = 1/2 + i√3/ 2,
z₂ = cos(4π/6) + i sin(4π/6) = -1/2 + i√3/ 2,
z₃ = cos(6π/6) + i sin(6π/6) = -1,
z₄ = cos(8π/6) + i sin(8π/6) = -1/2 - i√3/ 2,
z₅ = cos(10π/6) + i sin(10π/6) = 1/2 - i√3/ 2.
Answer:{1, 1/2 + i√3/ 2, -1/2 + i√3/ 2, -1, -1/2 - i√3/ 2, 1/2 - i√3/ 2}

Example 6.
Find all the values of the cubic root of a number z = -64.
Solution.
Let's write the number z in trigonometric form: z = -64 ⇔ a=-64; b=0 => |z|=r=√64; tgφ = b/a = 0. Since the number z is located in the third quadrant of the complex plane, φ=π. Thus, -64 = cos(π) + i sin(π).
According to the formula for extracting the root of the third degree, we have:

1^1/3 = cos(2πk/3) + i sin(2πk/3),   k=0,1,2.

Therefore, for all three roots we get
z₀ = 4(cos(π/3) + i sin(π/3)) = 4(1/2 + i√3/ 2) = 2 + i 2√3,
z₁ = 4(cos(3π/3) + i sin(3π/3)) = 4(-1) = -4,
z₂ = 4(cos(5π/3) + i sin(5π/3)) = 4(1/2 - i√3/ 2) = 2 - i 2√3.
Answer:{2+i 2√3, -4, 2-i 2√3}