Find the roots of z^3 = 1 in polar form
To find the roots of z^3 = 1 in polar form, we first note that 1 can also be written as 1 cis(0). We can then write the cube roots of 1 as:
z = (1 cis(0))^(1/3)
z = 1^(1/3) cis(0/3) = 1 cis(0)
z = (1 cis(0 + 2π))^(1/3)
z = 1^(1/3) cis(2π/3) = cos(2π/3) + i sin(2π/3) = -1/2 + i√3/2
z = (1 cis(0 + 4π/3))^(1/3)
z = 1^(1/3) cis(4π/9) = cos(4π/9) + i sin(4π/9) = √3/2 + i/2
Therefore, the roots of z^3 = 1 in polar form are:
z = 1 cis(0), -1/2 + i√3/2, √3/2 + i/2
To find the roots of a complex number in polar form, we can use De Moivre's theorem.
Step 1: Write the equation in exponential form:
z^3 = 1
1 = e^(0)
Step 2: Express 1 in polar form:
1 = e^(0) = e^(2kπ), where k is an integer.
Step 3: Apply De Moivre's theorem to find the roots:
Let's substitute z = r * e^(iθ), where r is the magnitude and θ is the argument.
(r * e^(iθ))^3 = e^(2kπ)
Taking the cube root of both sides gives:
r * e^(iθ) = e^(2kπ/3)
Step 4: Equate the magnitudes and arguments separately:
Magnitude:
r = 1^(1/3) = 1
Argument:
iθ = 2kπ/3
Dividing by i gives:
θ = 2kπ/3i
Here, k can take the values 0, 1, 2.
Thus, the roots of z^3 = 1 in polar form are:
z₁ = 1 * e^(2(0)π/3i) = e^(0)
z₂ = 1 * e^(2(1)π/3i) = e^(2π/3i)
z₃ = 1 * e^(2(2)π/3i) = e^(4π/3i)