Question

Find all solutions to the following equations:

(a) z⁶ = 1
(b) z⁴ = −16
(c) z⁶ = −9
(d) z⁶−z³−2=0

Answer 1

The solutions for each equation: The solutions z⁶ = 1 are z =
`e^(i(0)), e^{i((\pi)/(3))}, e^{i((2\pi)/(3))}, e^(i(\pi)), e^{i((4\pi)/(3))}, e^{i((5\pi)/(3))}` . The solutions z⁴ = −16 are z = 2i and z = -2i. The solutions z⁶ = −9 are z = ∛i and z = -∛i. The solutions z⁶ − z³ − 2 = 0 are z = ∛2 and z = -1 .

Step-by-step explanation:

Let's find the solutions for each equation:

(a) z⁶ = 1

The solutions for this equation are the 6th roots of unity. The complex roots can be expressed as:

z =
`e^(i\theta)`

where θ is an angle in radians. The solutions are:

z =
`e^(i(0)), e^{i((\pi)/(3))}, e^{i((2\pi)/(3))}, e^(i(\pi)), e^{i((4\pi)/(3))}, e^{i((5\pi)/(3))}`

(b) z⁴ = -16

Take the 4th root of both sides:

z = ± 2i

The solutions are z = 2i and z = -2i .

(c) z⁶ = -9

Take the 6th root of both sides:

z = ± ∛i

The solutions are z = ∛i and z = -∛i.

(d) z⁶ - z³ - 2 = 0

Let w = z³, then the equation becomes a quadratic equation:

w² - w - 2 = 0

Factorize:

(w - 2)(w + 1) = 0

So, w = 2 or w = -1 .

Now, substitute back z³ for w:

For w = 2, z³ = 2 gives z = ∛2.

For w = -1 , z³ = -1 gives z = -1 .

The solutions are z = ∛2 and z = -1 .