Find all solutions to the following equation: $x^3=-8i$ Find all solutions to the following equation: $$x^3=-8i$$
I found the modulus, $$r=8$$
$$\operatorname{arg}(x)=\arctan(-8/0)=-π/2+2πk$$
By De Moivre's Theorem: $$2[\cos(-π/6+2/3πk)+i\sin(-π/6+2/3πk)]$$
First solution I got is: $$\sqrt3-i$$Is my answer correct?
and is there any easier way to get the other solutions?
 A: Your answer is correct.
Notice also that $(2i)^3=-8i$, so $x^3-(-8i) = (x-2i)(\cdots\cdots)$, and the other factor can be found by long division.  You get
$$
x^3+8i = (x-2i)(x^2 +2ix - 4).
$$
If you try to solve $x^2+2ix-4=0$ by the usual formula, you need the square roots of the discriminant $b^2-4ac = -4-4(1)(-4)=12=4\cdot3$ and no knowledge of complex numbers is needed for that.
A: $$(x-2i)(x^2+2i-4)=0$$
the first root is
$$x=2i$$
the other roots by using quadratic formula
A: Yes, there is one. Hint: use the exponential form: $z^3 = -8i$ will be replaced by $z^3 = 8 \exp(-\pi/2 + 2k\pi)$, so $z = 2 \exp(-\pi/6 + 2k\pi/3)$ for $k \in \{0,1,2\}$.
A: $$x^3=-8i$$
$$x^3=8e^{-(\pi/2)i}$$
Solutions:
$$x=2i$$
$$x=-\sqrt{3}-i$$
$$x=\sqrt{3}-i$$
A different way to work it out:
$$x^3=-8i<=>$$
$$x^3=|-8i|e^{arg(-8i)i}<=>$$
$$x^3=8e^{\left(-\frac{1}{2}\pi +2\pi k\right)i}$$
(with k is the element of Z)
$$x=\left(8e^{\left(-\frac{1}{2}\pi +2\pi k\right)i}\right)^{\frac{1}{3}}<=>$$
$$x=2e^{\left(\frac{1}{3}\pi +\frac{2}{3}\pi k\right)i}$$
(with k goes from 0 to 2 -> k=0-2)
