How do you find all solutions to #x^3+64i=0#?

Question

Sadie Ackerman · Accepted Answer

# x = 4i,+-2sqrt(3)-2i #

Let # omega=-64i #, and then #x^3 = omega#

First let us plot the point #omega# on the Argand diagram:

# " " :. x = 4(-sqrt(3)/2-1/2i) #
# " " :. x = -2sqrt(3)-2i #
# n=0 => theta = (-pi)/6 #
# " " :. x = 4(cos ((-pi)/6)+ isin ((-pi)/6)) #
# " " :. x = 4(sqrt(3)/2-1/2i) #
# " " :. x = 2sqrt(3)-2i #
# n=1 => theta = (pi)/2 #
# " " :. x = 4(cos ((pi)/2)+ isin ((pi)/2)) #
# " " :. x = 4(0+i) #
# " " :. x = 4i #

After which the pattern continues.

We can plot these solutions on the Argand Diagram

Benjamin Achtenberg · Answer

undefined To find all solutions to the equation $x^3 + 64i = 0$, you can use the cubic formula or exploit the factorization of the equation. The given equation is in the form of a sum of cubes, so it can be factored using the sum of cubes formula:

$$ a^3 + b^3 = (a + b)(a^2 - ab + b^2) $$

Here, $a = x$ and $b = 4i$. Substituting into the formula:

$$ x^3 + (4i)^3 = (x + 4i)(x^2 - 4ix + (4i)^2) $$

$$ x^3 + 64i = (x + 4i)(x^2 - 4ix - 16) $$

Now, we set each factor equal to zero and solve for $x$:

1. $x + 4i = 0$ gives $x = -4i$.
2. $x^2 - 4ix - 16 = 0$ is a quadratic equation which can be solved using the quadratic formula:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

where $a = 1$, $b = -4i$, and $c = -16$.

$$ x = \frac{4i \pm \sqrt{(-4i)^2 - 4(1)(-16)}}{2(1)} $$

$$ x = \frac{4i \pm \sqrt{-16 + 64}}{2} $$

$$ x = \frac{4i \pm \sqrt{48}}{2} $$

$$ x = \frac{4i \pm 4\sqrt{3}i}{2} $$

$$ x = 2i \pm 2\sqrt{3}i $$

So, the solutions to the equation $x^3 + 64i = 0$ are $x = -4i$, $x = 2i + 2\sqrt{3}i$, and $x = 2i - 2\sqrt{3}i$.