1. **State the problem:** We are given the expression for the division of two complex numbers in polar form: $$\frac{z_1}{z_2} = \frac{r_1}{r_2} \left( \cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2) \right)$$ We want to evaluate: $$-\frac{1}{\sqrt{2}} \left( \cos\left( \frac{5\pi}{3} - \frac{3\pi}{4} \right) + i \sin\left( \frac{5\pi}{3} - \frac{3\pi}{4} \right) \right)$$ 2. **Calculate the angle difference:** $$\theta_1 - \theta_2 = \frac{5\pi}{3} - \frac{3\pi}{4} = \frac{20\pi}{12} - \frac{9\pi}{12} = \frac{11\pi}{12}$$ 3. **Rewrite the expression:** $$-\frac{1}{\sqrt{2}} \left( \cos\left( \frac{11\pi}{12} \right) + i \sin\left( \frac{11\pi}{12} \right) \right)$$ 4. **Note on the sign:** Multiplying by $-1$ is equivalent to adding $\pi$ to the angle inside cosine and sine: $$-\frac{1}{\sqrt{2}} \left( \cos\alpha + i \sin\alpha \right) = \frac{1}{\sqrt{2}} \left( \cos(\alpha + \pi) + i \sin(\alpha + \pi) \right)$$ 5. **Apply this to our angle:** $$\alpha = \frac{11\pi}{12}$$ $$\alpha + \pi = \frac{11\pi}{12} + \frac{12\pi}{12} = \frac{23\pi}{12}$$ 6. **Final expression:** $$\frac{1}{\sqrt{2}} \left( \cos\left( \frac{23\pi}{12} \right) + i \sin\left( \frac{23\pi}{12} \right) \right)$$ This is the simplified polar form of the given expression. **Summary:** $$\boxed{\frac{z_1}{z_2} = \frac{1}{\sqrt{2}} \left( \cos\left( \frac{23\pi}{12} \right) + i \sin\left( \frac{23\pi}{12} \right) \right)}$$