Let's go through each part of this problem systematically.

Given: $z_1 = 4 \operatorname{cis}\left(-\frac{\pi}{3}\right) \quad \text{and} \quad z_2 = 3 \operatorname{cis}\left(\frac{5\pi}{6}\right)$ where $\operatorname{cis}(\theta) = \cos(\theta) + i \sin(\theta)$.

Part (a): $z_1 z_2$

To multiply two complex numbers in polar form: $z_1 z_2 = |z_1| |z_2| \operatorname{cis}(\theta_1 + \theta_2)$

1. Magnitudes: $|z_1| = 4$ and $|z_2| = 3$, so $|z_1 z_2| = 4 \times 3 = 12$.
2. Angles: $\theta_1 = -\frac{\pi}{3}$ and $\theta_2 = \frac{5\pi}{6}$, so $\theta_1 + \theta_2 = -\frac{\pi}{3} + \frac{5\pi}{6} = \frac{\pi}{2}$.

Therefore, $z_1 z_2 = 12 \operatorname{cis}\left(\frac{\pi}{2}\right)$ Converting $12 \operatorname{cis}\left(\frac{\pi}{2}\right)$ to rectangular form: $z_1 z_2 = 12 \left(\cos\frac{\pi}{2} + i \sin\frac{\pi}{2}\right) = 12 \times (0 + i \times 1) = 12i$ So, the answer is: $z_1 z_2 = 0 + 12i$

Part (b): $\left( \frac{z_1}{z_2} \right)^3$

To divide and then raise to a power:

1. Division of magnitudes: $\left| \frac{z_1}{z_2} \right| = \frac{4}{3}$.
2. Division of angles: $\theta_1 - \theta_2 = -\frac{\pi}{3} - \frac{5\pi}{6} = -\frac{7\pi}{6}$.

Thus, $\frac{z_1}{z_2} = \frac{4}{3} \operatorname{cis}\left(-\frac{7\pi}{6}\right)$

Now, raising to the power of 3: $\left( \frac{z_1}{z_2} \right)^3 = \left(\frac{4}{3}\right)^3 \operatorname{cis}\left(3 \times -\frac{7\pi}{6}\right)$

1. Magnitude: $\left(\frac{4}{3}\right)^3 = \frac{64}{27}$.
2. Angle: $3 \times -\frac{7\pi}{6} = -\frac{21\pi}{6} = -\frac{7\pi}{2}$. (Adding $2\pi$ repeatedly to make it between $-\pi$ and $\pi$, we get $\frac{\pi}{2}$).

So, $\left( \frac{z_1}{z_2} \right)^3 = \frac{64}{27} \operatorname{cis}\left(\frac{\pi}{2}\right)$ Converting to rectangular form: $\left( \frac{z_1}{z_2} \right)^3 = \frac{64}{27} \left(0 + i \cdot 1\right) = \frac{64i}{27}$ So, the answer is: $\left( \frac{z_1}{z_2} \right)^3 = 0 + \frac{64i}{27}$

Part (c): $\left( z_1^{z_2} \right)^7$

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