I for the life of me cannot seem to figure this out. Anything would help!

Find all solutions of the equation. Leave answers in trigonometric form.
x^3 + 8i = 0

x^3 = 8i = 0 + 8i

argument = 8
tanθ = 8/0 , which is undefined, so θ = 90°
x^3 = 8(cos90 + sin90 i)
primary root:
x = 8^(1/3)(cos30 + i sin30)
= 2(√3/2 + i(1/2)
= √3 + i

two more roots:
1) 2(cos (30+120) + i sin(30+120)°
= 2(-√3/2 + (1/2) i) = -√3 + i

2) 2(cos(30+240) + i sin(30+240))
= 2(0 - i)
= -2i

just noticed we were to leave answers in trig form, no big deal, just go back
e.g.
2nd root would be 2cos150 + i sin150 , sometimes written as 2cis150° or 2cis 5π/6

x^3 = 2^3 * (-i )

so 2 *( -i)^1/3
so lets find the cube root of -i , well -i works
convert to polar 0 - 1 i = cos 270 + sin 270
In other words straight down
360 / 3 = 120
so go clockwise 120 which is 150d eg, 30 above -y axis
and counter clockwise 120 which is 30 above x ais
so cos 30 + i sin 30 = .5 sqrt 3 + .5 i
and cos 150 + i sin 150 = -.5 sqrt 3 + .5 i
so
-2 i
sqrt 3 + i
-sqrt 3 + i
------------------------
check
(sqrt 3 +i)(sqrt 3+i ) = 2 + 2 i sqrt 3
(sqrt 3 + i)(2 + 2 i sqrt 3) = 2 sqrt 3 + 6 i + 2 i - 2 sqrt 3
= 8 i sure enough

Or may be

Recall that
a³+b³=(a+b)(a²-ab+b²)

z³+(2i)³=(z+2i)(z²-2iz-4)

Z=-2i

(Z²-2iz-4)=0

Z=(2i±√[(-2i)²-(-4(1)(4)]/2

Z=(2i+√[-4+16]/2

Z=(2i±√(12)/2

Z=(2i±2√3)/2

Z=i±√3
Which set out to be
Z=(I+√3)(i-√3)(-2i)

To find the solutions of the equation x^3 + 8i = 0, we can start by setting x^3 = -8i, since we want to solve for x.

We can rewrite -8i in polar form as follows:
-8i = 8 * (-i)
= 8 * e^(i(3π/2)) (using Euler's formula: e^(iθ) = cos(θ) + i*sin(θ))

In polar form, -8i has a magnitude of 8 and an argument of 3π/2.

Now, let's express x^3 also in polar form. We can write x^3 as r^3 * e^(i(θ + 2kπ)), where r is the magnitude of x and θ is its argument. Here, k is an integer representing the number of full rotations around the unit circle.

So, we have:
x^3 = r^3 * e^(i(θ + 2kπ))

To find the solutions, we need x^3 to equal -8i, or r^3 * e^(i(θ + 2kπ)) = 8 * e^(i(3π/2)). By comparing the polar forms, we can see that:
r^3 = 8 (the magnitudes are equal)
θ + 2kπ = 3π/2 (the arguments are equal)

Let's first solve for r:
Taking the cube root of both sides, we get:
r = 2

Now let's solve for θ:
θ + 2kπ = 3π/2

To find the values of θ, we need to find the possible values of k that satisfy the equation. Since k is an integer, we can solve for k by subtracting 3π/2 from both sides and dividing by 2π:

(θ + 2kπ) - (3π/2) = 0
θ + 2kπ = 3π/2

Subtracting (3π/2) from both sides, we get:
θ = -3π/2

So one possible value of θ is -3π/2. To find additional values, we need to add multiples of 2π to k.

Therefore, the set of solutions for θ is:
θ = -3π/2 + 2kπ (where k is an integer)

In summary, the solutions of the equation x^3 + 8i = 0, expressed in trigonometric form, are given by:
x = 2 * e^(i(θ + 2kπ)) (where θ = -3π/2 + 2kπ and k is an integer)