Find the product \( z_{1} z_{2} \) and the quotient \( \frac{z_{1}}{z_{2}} \). Express your answers in polar form. \[ z_{1}=\frac{8}{9}\left(\cos \left(40^{\circ}\right)+i \sin \left(40^{\circ}\right)\right), \quad z_{2}=\frac{1}{9}\left(\cos \left(160^{\circ}\right)+i \sin \left(160^{\circ}\right)\right) \] \( z_{1} z_{2}=\square \) \( \frac{z_{1}}{z_{2}}=\square \)

To find the product \( z_{1} z_{2} \) and the quotient \( \frac{z_{1}}{z_{2}} \), we will use the properties of multiplying and dividing complex numbers in polar form. First, let's express both complex numbers in polar form: 1. **Magnitude of \( z_1 \)**: \[ |z_1| = \frac{8}{9} \] **Argument of \( z_1 \)**: \[ \text{arg}(z_1) = 40^\circ \] 2. **Magnitude of \( z_2 \)**: \[ |z_2| = \frac{1}{9} \] **Argument of \( z_2 \)**: \[ \text{arg}(z_2) = 160^\circ \] Now we can find the product \( z_1 z_2 \): \[ z_1 z_2 = |z_1||z_2| \left( \cos(\text{arg}(z_1) + \text{arg}(z_2)) + i \sin(\text{arg}(z_1) + \text{arg}(z_2)) \right) \] Calculating the magnitude: \[ |z_1 z_2| = \frac{8}{9} \cdot \frac{1}{9} = \frac{8}{81} \] Calculating the argument: \[ \text{arg}(z_1 z_2) = 40^\circ + 160^\circ = 200^\circ \] So, we have: \[ z_1 z_2 = \frac{8}{81}\left( \cos(200^{\circ}) + i \sin(200^{\circ}) \right) \] Next, let's find the quotient \( \frac{z_1}{z_2} \): \[ \frac{z_1}{z_2} = \frac{|z_1|}{|z_2|} \left( \cos(\text{arg}(z_1) - \text{arg}(z_2)) + i \sin(\text{arg}(z_1) - \text{arg}(z_2)) \right) \] Calculating the magnitude: \[ \frac{|z_1|}{|z_2|} = \frac{8/9}{1/9} = 8 \] Calculating the argument: \[ \text{arg}\left(\frac{z_1}{z_2}\right) = 40^\circ - 160^\circ = -120^\circ \] Adjusting the argument to a positive angle: \[ -120^\circ + 360^\circ = 240^\circ \] So, we have: \[ \frac{z_1}{z_2} = 8\left( \cos(240^{\circ}) + i \sin(240^{\circ}) \right) \] Finally, the answers are: \[ z_{1} z_{2}=\frac{8}{81}\left(\cos(200^{\circ})+i \sin(200^{\circ})\right) \] \[ \frac{z_{1}}{z_{2}}=8\left(\cos(240^{\circ})+i \sin(240^{\circ})\right) \]