Show that $(-1+\sqrt{3}i)^{3/2}= ±2\sqrt{2}$ using the fact that $z^c=e^{c \log(z)}$

Question

I want to show that $(-1+\sqrt{3}i)^{3/2}= ±2\sqrt{2}$ using the fact that $z^c=e^{c \log(z)}$

Here's my attempt:

$(-1+\sqrt{3}i)^{3/2} = e^{\dfrac{3}{2}log(-1+i\sqrt{3})} = e^{\log(2\sqrt{2}e^{ 2i \pi/3^{3/2}}} =\Big(2\sqrt{2}e^{i2 \pi/3}\Big)^{3/2} = (2\sqrt{2})^{3/2}$

Which is obviously the wrong result

where did I go wrong?

Answer 1

$-1 + \sqrt{3}i = 2e^{(2\pi/3)i}$, so its cube is $8e^{2\pi i} = 8$, whose square roots are $\pm 2 \sqrt{2}$. Thus $$(-1 + \sqrt{3} i)^{3/2} = \pm 2 \sqrt{2}.$$ Or, going about it the way you'd like, $$(2e^{2\pi i/3})^{3/2} = e^{(3/2) \log(2e^{2\pi i/3})} = e^{(3/2) \log 2 + (3/2) \log(e^{2\pi i/3})} = 2^{3/2} \cdot e^{i \pi} = -2^{3/2}$$ is one value of the root. Be careful to note I am simply using one value of the logarithm to make this calculation, which is why only one of the two answers pops out. Of course $2^{3/2}$ is the other one, yielding the result.

Answer 2

Since$$\log\left(-1+\sqrt3i\right)=\log\left(2\left(-\frac12+\frac{\sqrt3}2i\right)\right)=\log(2)+\left(\frac{2\pi}3+2k\pi\right)i$$($k\in\mathbb Z$), you have\begin{align}\exp\left(\frac32\log\left(-1+\sqrt3i\right)\right)&=\exp\left(\frac32\left(\log(2)+\left(\frac{2\pi}3+2k\pi\right)i\right)\right)\\&=2^{\frac32}\exp\bigl((\pi+3k\pi)i\bigr)\\&=\pm2\sqrt2.\end{align}