1. **State the problem:** We are given two complex numbers in polar form: $z = 2 \operatorname{cis}(-\frac{\pi}{4})$ and $w = \sqrt{3} \operatorname{cis}(\frac{\pi}{6})$. We need to find the imaginary part of the quotient $\frac{z}{w}$ and express it as a decimal rounded to 3 significant figures. 2. **Recall the formula for division of complex numbers in polar form:** $$\frac{r_1 \operatorname{cis}(\theta_1)}{r_2 \operatorname{cis}(\theta_2)} = \frac{r_1}{r_2} \operatorname{cis}(\theta_1 - \theta_2)$$ where $r_1, r_2$ are magnitudes and $\theta_1, \theta_2$ are arguments. 3. **Apply the formula:** $$\frac{z}{w} = \frac{2}{\sqrt{3}} \operatorname{cis}\left(-\frac{\pi}{4} - \frac{\pi}{6}\right) = \frac{2}{\sqrt{3}} \operatorname{cis}\left(-\frac{3\pi}{12} - \frac{2\pi}{12}\right) = \frac{2}{\sqrt{3}} \operatorname{cis}\left(-\frac{5\pi}{12}\right)$$ 4. **Express $\operatorname{cis}(\theta)$ in rectangular form:** $$\operatorname{cis}(\theta) = \cos(\theta) + i \sin(\theta)$$ So, $$\frac{z}{w} = \frac{2}{\sqrt{3}} \left( \cos\left(-\frac{5\pi}{12}\right) + i \sin\left(-\frac{5\pi}{12}\right) \right)$$ 5. **Use the odd/even properties of sine and cosine:** $$\cos(-x) = \cos x, \quad \sin(-x) = -\sin x$$ Therefore, $$\frac{z}{w} = \frac{2}{\sqrt{3}} \left( \cos\left(\frac{5\pi}{12}\right) - i \sin\left(\frac{5\pi}{12}\right) \right)$$ 6. **Find the imaginary part:** $$\operatorname{Im}\left(\frac{z}{w}\right) = - \frac{2}{\sqrt{3}} \sin\left(\frac{5\pi}{12}\right)$$ 7. **Calculate $\sin\left(\frac{5\pi}{12}\right)$:** Note that $\frac{5\pi}{12} = \frac{\pi}{4} + \frac{\pi}{6}$. Using the sine addition formula: $$\sin(a+b) = \sin a \cos b + \cos a \sin b$$ So, $$\sin\left(\frac{\pi}{4} + \frac{\pi}{6}\right) = \sin\frac{\pi}{4} \cos\frac{\pi}{6} + \cos\frac{\pi}{4} \sin\frac{\pi}{6}$$ 8. **Substitute known values:** $$\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}, \quad \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \sin\frac{\pi}{6} = \frac{1}{2}$$ 9. **Calculate:** $$\sin\left(\frac{5\pi}{12}\right) = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$$ 10. **Substitute back to find imaginary part:** $$\operatorname{Im}\left(\frac{z}{w}\right) = - \frac{2}{\sqrt{3}} \times \frac{\sqrt{6} + \sqrt{2}}{4} = - \frac{\sqrt{6} + \sqrt{2}}{2 \sqrt{3}}$$ 11. **Simplify denominator:** Multiply numerator and denominator by $\sqrt{3}$: $$- \frac{(\sqrt{6} + \sqrt{2}) \sqrt{3}}{2 \sqrt{3} \sqrt{3}} = - \frac{(\sqrt{6} + \sqrt{2}) \sqrt{3}}{2 \times 3} = - \frac{(\sqrt{6} + \sqrt{2}) \sqrt{3}}{6}$$ 12. **Calculate numerical value:** $$\sqrt{6} \approx 2.449, \quad \sqrt{2} \approx 1.414, \quad \sqrt{3} \approx 1.732$$ $$ (\sqrt{6} + \sqrt{2}) \sqrt{3} = (2.449 + 1.414) \times 1.732 = 3.863 \times 1.732 \approx 6.691 $$ $$\operatorname{Im}\left(\frac{z}{w}\right) \approx - \frac{6.691}{6} = -1.115$$ **Final answer:** $$\boxed{-1.12}$$ (rounded to 3 significant figures)