What is a fifth root of the complex number represented by 32cos(3π) + isin(3π)?

To find the fifth roots of the complex number represented by 32cos(3π) + isin(3π), we first need to express this number in polar form.

The given complex number can be written as:

z = r(cos(θ) + isin(θ))

Here, r is the modulus (magnitude) of the complex number and θ is the argument (angle).

1. **Calculate the modulus (r):**
The modulus r of the complex number is given by:

r = |z| = sqrt(x² + y²)

In our case, 32cos(3π) + isin(3π) translates to:

r = 32

2. **Determine the argument (θ):**
The angle or argument is given by:

θ = 3π

Now substituting the values, we get:

z = 32(cos(3π) + isin(3π))

3. **Calculate the fifth roots:**
To find the fifth roots, we can use the formula for the n-th roots of a complex number:

w_k = r^(1/n) * (cos((θ + 2kπ)/n) + isin((θ + 2kπ)/n))

where k takes integer values from 0 to n-1. In our case, n = 5:

w_k = 32^(1/5) * (cos((3π + 2kπ)/5) + isin((3π + 2kπ)/5))

The value of 32^(1/5) is 2, since 2⁵ = 32.

4. **Calculate roots for k = 0, 1, 2, 3, 4:**

• For k = 0:

w_0 = 2(cos(3π/5) + isin(3π/5))

• For k = 1:

w_1 = 2(cos((3π + 2π)/5) + isin((3π + 2π)/5)) = 2(cos(5π/5) + isin(5π/5))

• For k = 2:

w_2 = 2(cos((3π + 4π)/5) + isin((3π + 4π)/5)) = 2(cos(7π/5) + isin(7π/5))

• For k = 3:

w_3 = 2(cos((3π + 6π)/5) + isin((3π + 6π)/5)) = 2(cos(9π/5) + isin(9π/5))

• For k = 4:

w_4 = 2(cos((3π + 8π)/5) + isin((3π + 8π)/5)) = 2(cos(11π/5) + isin(11π/5))

Thus, the fifth roots of the complex number 32cos(3π) + isin(3π) are:

• 2(cos(3π/5) + isin(3π/5))
• 2(cos(π) + isin(π))
• 2(cos(7π/5) + isin(7π/5))
• 2(cos(9π/5) + isin(9π/5))
• 2(cos(11π/5) + isin(11π/5))

This gives us a complete set of the fifth roots of the given complex number.