What are the solutions to the equation iz^3+8=0 ?

The solutions to the equation iz^3+8=0 are z=sqrt(3)+i,z=-sqrt(3)+i,z=-2i

Find the zeros of iz^3+8=0

The zeros of iz^3+8=0 are z=sqrt(3)+i,z=-sqrt(3)+i,z=-2i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

i z^{3} + 8 = 0

z = 3 + i, z = - 3 + i, z = - 2 i

Solution steps

i z^{3} + 8 = 0

Move

8

to the right side

i z^{3} = - 8

Divide both sides by

i

z^{3} = 8 i

For

z^{n} = a

the solutions are

z_{k} = n ∣ a ∣ (cos (\frac{ar g ( a ) + 2 kπ}{n}) + i sin (\frac{ar g ( a ) + 2 kπ}{n})),

k = 0, 1, \dots, n - 1

For

n = 3, a = 8 i

∣ a ∣ = 8

ar g (a) = \frac{π}{2}

z = 38 (cos (\frac{\frac{π}{2} + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 0 π}{3})), z = 38 (cos (\frac{\frac{π}{2} + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 1 π}{3})), z = 38 (cos (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}))

Simplify

38 (cos (\frac{\frac{π}{2} + 2 \cdot 0 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 0 π}{3})) : 3 + i

Simplify

38 (cos (\frac{\frac{π}{2} + 2 \cdot 1 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 1 π}{3})) : - 3 + i

Simplify

38 (cos (\frac{\frac{π}{2} + 2 \cdot 2 π}{3}) + i sin (\frac{\frac{π}{2} + 2 \cdot 2 π}{3})) : - 2 i

z = 3 + i, z = - 3 + i, z = - 2 i

- 2 x^{4} + 13 x^{3} - 21 x^{2} + 2 x + 8 = 0

25 x^{3} - 100 x^{2} - 16 x + 64 = 0

- 5 x^{2} = 36 - x^{4}

x^{3} - 15 x - 1 = 0

2 + 3 x^{2} - x^{3}

What are the solutions to the equation iz^3+8=0 ?
The solutions to the equation iz^3+8=0 are z=sqrt(3)+i,z=-sqrt(3)+i,z=-2i
Find the zeros of iz^3+8=0
The zeros of iz^3+8=0 are z=sqrt(3)+i,z=-sqrt(3)+i,z=-2i