convert each complex number to polar form. \( \begin{array}{ll}4.9+2 i & \text { 5. }-7 \\ 6.9 i & \text { 7. }-3-3 i \\ 8 .-5+7 i & \text { 9. } \frac{\sqrt{3}}{4}-\frac{1}{4} i\end{array} \)

To convert each complex number to its polar form, we'll express them in the form: \[ r (\cos \theta + i \sin \theta) \quad \text{or} \quad r e^{i\theta} \] where: - \( r \) is the magnitude (or modulus) of the complex number. - \( \theta \) is the argument (or angle) of the complex number in degrees. Let's convert each given complex number: --- ### **5. \( 4.9 + 2i \)** **Magnitude (\( r \)):** \[ r = \sqrt{4.9^2 + 2^2} = \sqrt{24.01 + 4} = \sqrt{28.01} \approx 5.29 \] **Angle (\( \theta \)):** \[ \theta = \arctan\left(\frac{2}{4.9}\right) \approx \arctan(0.408) \approx 22.2^\circ \] **Polar Form:** \[ 5.29 \left( \cos 22.2^\circ + i \sin 22.2^\circ \right) \quad \text{or} \quad 5.29\, e^{i22.2^\circ} \] --- ### **6. \( -7 \)** **Magnitude (\( r \)):** \[ r = |-7| = 7 \] **Angle (\( \theta \)):** \[ \theta = 180^\circ \quad \text{(since it's on the negative real axis)} \] **Polar Form:** \[ 7 \left( \cos 180^\circ + i \sin 180^\circ \right) \quad \text{or} \quad 7\, e^{i180^\circ} \] --- ### **7. \( 6.9i \)** **Magnitude (\( r \)):** \[ r = |6.9| = 6.9 \] **Angle (\( \theta \)):** \[ \theta = 90^\circ \quad \text{(since it's on the positive imaginary axis)} \] **Polar Form:** \[ 6.9 \left( \cos 90^\circ + i \sin 90^\circ \right) \quad \text{or} \quad 6.9\, e^{i90^\circ} \] --- ### **8. \( -3 - 3i \)** **Magnitude (\( r \)):** \[ r = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} \approx 4.24 \] **Angle (\( \theta \)):** \[ \theta = \arctan\left(\frac{-3}{-3}\right) = \arctan(1) = 45^\circ \] Since both parts are negative, the angle is in the **third quadrant**: \[ \theta = 180^\circ + 45^\circ = 225^\circ \] **Polar Form:** \[ 4.24 \left( \cos 225^\circ + i \sin 225^\circ \right) \quad \text{or} \quad 4.24\, e^{i225^\circ} \] --- ### **9. \( \frac{\sqrt{3}}{4} - \frac{1}{4}i \)** **Magnitude (\( r \)):** \[ r = \sqrt{\left(\frac{\sqrt{3}}{4}\right)^2 + \left(-\frac{1}{4}\right)^2} = \sqrt{\frac{3}{16} + \frac{1}{16}} = \sqrt{\frac{4}{16}} = \frac{2}{4} = \frac{1}{2} \] **Angle (\( \theta \)):** \[ \theta = \arctan\left(\frac{-\frac{1}{4}}{\frac{\sqrt{3}}{4}}\right) = \arctan\left(-\frac{1}{\sqrt{3}}\right) = -30^\circ \] Since the real part is positive and the imaginary part is negative, the angle is in the **fourth quadrant**: \[ \theta = 360^\circ - 30^\circ = 330^\circ \] **Polar Form:** \[ \frac{1}{2} \left( \cos 330^\circ + i \sin 330^\circ \right) \quad \text{or} \quad \frac{1}{2}\, e^{i330^\circ} \] --- **Summary of Polar Forms:** 1. **\( 4.9 + 2i \):** \[ 5.29\, e^{i22.2^\circ} \] 2. **\( -7 \):** \[ 7\, e^{i180^\circ} \] 3. **\( 6.9i \):** \[ 6.9\, e^{i90^\circ} \] 4. **\( -3 - 3i \):** \[ 4.24\, e^{i225^\circ} \] 5. **\( \frac{\sqrt{3}}{4} - \frac{1}{4}i \):** \[ \frac{1}{2}\, e^{i330^\circ} \] --- Feel free to reach out if you need further clarification on any of these conversions!