1. **State the problem:** We are given a complex number $z$ with modulus $8$ and argument $\frac{2\pi}{3}$. We need to find $z^{1/z}$ in Cartesian form. 2. **Express $z$ in polar form:** $$z = 8 \left(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}\right)$$ 3. **Write $z$ in exponential form:** $$z = 8 e^{i \frac{2\pi}{3}}$$ 4. **Calculate $1/z$:** $$\frac{1}{z} = \frac{1}{8 e^{i \frac{2\pi}{3}}} = \frac{1}{8} e^{-i \frac{2\pi}{3}}$$ 5. **Calculate $z^{1/z}$:** $$z^{1/z} = \left(8 e^{i \frac{2\pi}{3}}\right)^{\frac{1}{8} e^{-i \frac{2\pi}{3}}} = e^{\frac{1}{8} e^{-i \frac{2\pi}{3}} \ln \left(8 e^{i \frac{2\pi}{3}}\right)}$$ 6. **Simplify the logarithm:** $$\ln \left(8 e^{i \frac{2\pi}{3}}\right) = \ln 8 + i \frac{2\pi}{3}$$ 7. **Substitute back:** $$z^{1/z} = e^{\frac{1}{8} e^{-i \frac{2\pi}{3}} (\ln 8 + i \frac{2\pi}{3})}$$ 8. **Calculate $e^{-i \frac{2\pi}{3}}$:** $$e^{-i \frac{2\pi}{3}} = \cos \left(-\frac{2\pi}{3}\right) + i \sin \left(-\frac{2\pi}{3}\right) = \cos \frac{2\pi}{3} - i \sin \frac{2\pi}{3} = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$$ 9. **Multiply:** $$\left(-\frac{1}{2} - i \frac{\sqrt{3}}{2}\right) (\ln 8 + i \frac{2\pi}{3}) = -\frac{1}{2} \ln 8 - \frac{1}{2} i \frac{2\pi}{3} - i \frac{\sqrt{3}}{2} \ln 8 - i^2 \frac{\sqrt{3}}{2} \frac{2\pi}{3}$$ 10. **Simplify $i^2 = -1$:** $$= -\frac{1}{2} \ln 8 - i \frac{\pi}{3} - i \frac{\sqrt{3}}{2} \ln 8 + \frac{\sqrt{3} \pi}{3}$$ 11. **Group real and imaginary parts:** Real part: $$-\frac{1}{2} \ln 8 + \frac{\sqrt{3} \pi}{3}$$ Imaginary part: $$- \frac{\pi}{3} - \frac{\sqrt{3}}{2} \ln 8$$ 12. **Multiply by $\frac{1}{8}$:** Real: $$\frac{1}{8} \left(-\frac{1}{2} \ln 8 + \frac{\sqrt{3} \pi}{3}\right) = -\frac{\ln 8}{16} + \frac{\sqrt{3} \pi}{24}$$ Imaginary: $$\frac{1}{8} \left(- \frac{\pi}{3} - \frac{\sqrt{3}}{2} \ln 8\right) = -\frac{\pi}{24} - \frac{\sqrt{3} \ln 8}{16}$$ 13. **Rewrite exponent:** $$z^{1/z} = e^{\left(-\frac{\ln 8}{16} + \frac{\sqrt{3} \pi}{24}\right) + i \left(-\frac{\pi}{24} - \frac{\sqrt{3} \ln 8}{16}\right)}$$ 14. **Convert to Cartesian form:** $$z^{1/z} = e^{-\frac{\ln 8}{16} + \frac{\sqrt{3} \pi}{24}} \left[\cos \left(-\frac{\pi}{24} - \frac{\sqrt{3} \ln 8}{16}\right) + i \sin \left(-\frac{\pi}{24} - \frac{\sqrt{3} \ln 8}{16}\right)\right]$$ 15. **Final answer:** $$\boxed{z^{1/z} = e^{-\frac{\ln 8}{16} + \frac{\sqrt{3} \pi}{24}} \left[\cos \left(-\frac{\pi}{24} - \frac{\sqrt{3} \ln 8}{16}\right) + i \sin \left(-\frac{\pi}{24} - \frac{\sqrt{3} \ln 8}{16}\right)\right]}$$ This is the Cartesian form expressed as $a + bi$ where $a$ and $b$ are the cosine and sine terms multiplied by the exponential factor.