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

Would

Given z1 = 4cis(-π/3) and z2 = 3cis(5π/6), find expressions for z1z2, (z1/z2)^3, and (z1^z2)^7 in the form a + bi.

This problem explores complex numbers in polar form, focusing on multiplication, division, and exponentiation. Given z1 = 4cis(-π/3) and z2 = 3cis(5π/6), the solution involves applying De Moivre's Theorem and converting results into rectangular form. Suitable for Grades 11-12.

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