Need help in simplifying this $y=\frac{\sqrt[3]{\sqrt{7}i-3}+\sqrt[3]{-\sqrt{7}i-3}}{2\sqrt[3]{2^2}}$

Question

Need help in simplifying this:$$y=\frac{\sqrt[3]{\sqrt{7}i-3}+\sqrt[3]{-\sqrt{7}i-3}}{2\sqrt[3]{2^2}}$$ This has real and imaginary parts, at first glance, but after a numerical evaluation in mathematica, $y$ is effectively real, as shown by: $$y=0.692184 + 0. i\Rightarrow y=0.692184 $$ But I have no idea how to simplify this.

Answer 1

Note that every complex number $z=x+iy$ can be written as $z=re^{i\varphi}$. Also, $e^{i\varphi}=\cos\varphi+i\sin\varphi$.

Thus, with $z=-3+i\sqrt{7}$ we can write:

$$y=\frac{\sqrt[3]{\sqrt{7}i-3}+\sqrt[3]{-\sqrt{7}i-3}}{2\sqrt[3]{2^2}}\\ = \frac{1}{2\sqrt[3]{2^2}}\left(z^{\frac{1}{3}}+\left(z^*\right)^{\frac{1}{3}}\right)\\ = \frac{1}{2\sqrt[3]{2^2}}\left(r^{\frac{1}{3}}e^{i\frac{\varphi}{3}}+r^{\frac{1}{3}}e^{-i\frac{\varphi}{3}}\right)\\ =\frac{r^{\frac{1}{3}}}{2\sqrt[3]{2^2}}\left(\cos\left(\frac{\varphi}{3}\right)+i\sin\left(\frac{\varphi}{3}\right)+\cos\left(\frac{\varphi}{3}\right)-i\sin\left(\frac{\varphi}{3}\right)\right)\\ =\frac{r^{\frac{1}{3}}}{\sqrt[3]{2^2}}\cos\left(\frac{\varphi}{3}\right)\in\mathbb{R}$$