How to show that $e^{-(-2)^\frac{2}{3}}$ is complex.

Question

I am baffled with showing that $e^{-(-2)^\frac{2}{3}}$ is complex.

My understanding is:

$$ e^{-(-2)^\frac{2}{3}}= e^{-(4)^\frac{1}{3}}$$

since ${-(4)^\frac{1}{3}} $ is negative real number, so $ e^{-(4)^\frac{1}{3}}$ is always a positive real number.

Answer 1

Let $$Z=e^{-(-2)^{2/3}}$$Note that

$$-2=2e^{(2k+1)i\pi}$$

let $$z=(-2)^{2/3}=\sqrt[3]{4}e^{i\pi/3},\sqrt[3]{4}e^{i\pi},\sqrt[3]{4}e^{5i\pi/3}$$

$$e^{i\theta}=\cos\theta+i\sin\theta$$

Now $z$ has $3$ values, $$z=-\sqrt[3]{4} \,\, ,\,\, \sqrt[3]{4}\left[\cos\left(\frac{i\pi}{3}\right)+i\sin\left(\frac{i\pi}{3}\right)\right] \,\, ,\,\, \sqrt[3]{4}\left[\cos\left(\frac{5i\pi}{3}\right)+i\sin\left(\frac{5i\pi}{3}\right)\right]$$

Now writing $Z=e^{-z}$ in the form $a+bi$ is lengthy task but you can notice that as you pointed out there is a real value of $Z$ given as

$$Z=e^{-(-\sqrt[3]{4})}=e^{\sqrt[3]{4}}$$ but we should not ignore other values of $z$ and which will yield complex(pun intended here!) values of $Z=e^{-z}$

Answer 2

Recall that $e^{i\pi}=-1$, so $-2=2e^{i\pi}=e^{\log(2)+i\pi}$. You may use this to compute $(-2)^{\frac{2}{3}}=e^{\frac{2}{3}(\log(2)+i\pi)}$. You may then use Euler's formula (from which you can also figure out that $e^{i\pi}=-1$) to express $(-2)^{\frac{2}{3}}$ as $a+bi$ for some real numbers $a$ and $b$. A second application of Euler's formula (this time applied to $e^{-(-2)^{\frac{2}{3}}}$) will solve your problem.