Question

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

Answer 1

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

Answer 2

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

Answer 3

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)