1. **State the problem:** We are given two complex numbers in polar form: $$z = 2 \operatorname{cis} \left(-\frac{\pi}{4}\right)$$ and $$w = \sqrt{3} \operatorname{cis} \left(\frac{\pi}{6}\right)$$ We need to find the real part of the product $$zw$$, denoted as $$\operatorname{Re}(zw)$$, and express it as a decimal rounded to 3 significant figures. 2. **Recall the formula for multiplication of complex numbers in polar form:** If $$z = r_1 \operatorname{cis} \theta_1$$ and $$w = r_2 \operatorname{cis} \theta_2$$, then $$zw = r_1 r_2 \operatorname{cis} (\theta_1 + \theta_2)$$ where $$\operatorname{cis} \theta = \cos \theta + i \sin \theta$$. 3. **Calculate the magnitude and argument of $$zw$$:** $$r = 2 \times \sqrt{3} = 2\sqrt{3}$$ $$\theta = -\frac{\pi}{4} + \frac{\pi}{6} = -\frac{3\pi}{12} + \frac{2\pi}{12} = -\frac{\pi}{12}$$ 4. **Express $$zw$$ in rectangular form:** $$zw = 2\sqrt{3} \left( \cos \left(-\frac{\pi}{12}\right) + i \sin \left(-\frac{\pi}{12}\right) \right)$$ 5. **Find the real part:** Since $$\cos(-\theta) = \cos \theta$$, $$\operatorname{Re}(zw) = 2\sqrt{3} \cos \left(\frac{\pi}{12}\right)$$ 6. **Calculate $$\cos \left(\frac{\pi}{12}\right)$$:** Using the half-angle formula or known values, $$\cos \left(\frac{\pi}{12}\right) = \cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4} \approx 0.9659$$ 7. **Calculate the final value:** $$\operatorname{Re}(zw) = 2 \times 1.732 \times 0.9659 \approx 3.346$$ 8. **Round to 3 significant figures:** $$\boxed{3.35}$$ Thus, the real part of $$zw$$ is approximately 3.35.