Bringen Sie die in der Polarform vorliegenden komplexen Zahlen $$ \begin{array}{ll} z_{1}=4(\cos 1+\mathrm{j} \cdot \sin 1), & z_{2}=3 \cdot \mathrm{e}^{\mathrm{j} 30^{\circ}}, \quad z_{3}=5 \cdot \mathrm{e}^{j 135^{\circ}} \\ z_{4}=5\left[\cos \left(-60^{\circ}\right)+\mathrm{j} \cdot \sin \left(-60^{\circ}\right)\right], & z_{5}=2 \cdot \mathrm{e}^{-\mathrm{j} \frac{3}{2} \pi}, \quad z_{6}=\mathrm{e}^{\mathrm{j} 240^{\circ}} \\ z_{7}=2\left(\cos 210^{\circ}+\mathrm{j} \cdot \sin 210^{\circ}\right), & z_{8}=\cos (-0,5)+\mathrm{j} \cdot \sin (-0,5) \end{array} $$ in die kartesische Form und bestimmen Sie die jeweilige konjugiert komplexe Zahl.

Recall that for a complex number in polar form, represented as \[ z = re^{jθ} = r(\text{cos}(θ) + j \text{sin}(θ)) \], the Cartesian form is given by \[ z = a + jb \] where \[ a = r \text{cos}(θ) \] and \[ b = r \text{sin}(θ) \]. The conjugate \[ \overline{z} \] is \[ a - jb \].

1. \[ z_1 = 4(\text{cos}(1) + j \text{sin}(1)) \] which gives \[ a = 4\text{cos}(1) \] and \[ b = 4\text{sin}(1) \]. Calculate and get \[ z_1 = 4\text{cos}(1) + 4j\text{sin}(1) \] and \[ \overline{z_1} = 4\text{cos}(1) - 4j\text{sin}(1) \]. 2. \[ z_2 = 3 \text{e}^{j 30^\text{o}} \] which simplifies to \[ 3(\text{cos}(30^\text{o}) + j \text{sin}(30^\text{o})) \]. Therefore, \[ a = 3\text{cos}(30^\text{o}) \] and \[ b = 3\text{sin}(30^\text{o}) \]. Thus, \[ z_2 = 3\text{cos}(30^\text{o}) + 3j\text{sin}(30^\text{o}) \] and \[ \overline{z_2} = 3\text{cos}(30^\text{o}) - 3j\text{sin}(30^\text{o}) \].

3. \[ z_3 = 5 \text{e}^{j 135^\text{o}} \] which simplifies to \[ 5(\text{cos}(135^\text{o}) + j \text{sin}(135^\text{o})) \]. Therefore, \[ a = 5\text{cos}(135^\text{o}) \] and \[ b = 5\text{sin}(135^\text{o}) \], so \[ z_3 = 5\text{cos}(135^\text{o}) + 5j\text{sin}(135^\text{o}) \] and \[ \overline{z_3} = 5\text{cos}(135^\text{o}) - 5j\text{sin}(135^\text{o}) \]. 4. \[ z_4 = 5(\text{cos}(-60^\text{o}) + j \text{sin}(-60^\text{o})) \] gives \[ a = 5\text{cos}(-60^\text{o}) \] and \[ b = 5\text{sin}(-60^\text{o}) \], leading to \[ z_4 = 5\text{cos}(-60^\text{o}) + 5j\text{sin}(-60^\text{o}) \] and \[ \overline{z_4} = 5\text{cos}(-60^\text{o}) - 5j\text{sin}(-60^\text{o}) \].

5. \[ z_5 = 2 \text{e}^{-j \frac{3}{2} \text{π}} \] which simplifies to \[ 2(\text{cos}(-\frac{3}{2} \text{π}) + j \text{sin}(-\frac{3}{2} \text{π})) \]. Thus, \[ a = 2\text{cos}(-\frac{3}{2} \text{π}) \] and \[ b = 2\text{sin}(-\frac{3}{2} \text{π}) \]. Therefore, \[ z_5 = 2\text{cos}(-\frac{3}{2} \text{π}) + 2j\text{sin}(-\frac{3}{2} \text{π}) \] and \[ \overline{z_5} = 2\text{cos}(-\frac{3}{2} \text{π}) - 2j\text{sin}(-\frac{3}{2} \text{π}) \]. 6. \[ z_6 = \text{e}^{j 240^\text{o}} \] converts to \[ \text{cos}(240^\text{o}) + j \text{sin}(240^\text{o}) \] giving \[ a = \text{cos}(240^\text{o}) \] and \[ b = \text{sin}(240^\text{o}) \]. Therefore, \[ z_6 = \text{cos}(240^\text{o}) + j \text{sin}(240^\text{o}) \] and \[ \overline{z_6} = \text{cos}(240^\text{o}) - j \text{sin}(240^\text{o}) \].

7. \[ z_7 = 2(\text{cos}(210^\text{o}) + j \text{sin}(210^\text{o})) \] gives \[ a = 2\text{cos}(210^\text{o}) \] and \[ b = 2\text{sin}(210^\text{o}) \]. Thus, \[ z_7 = 2\text{cos}(210^\text{o}) + 2j\text{sin}(210^\text{o}) \] and \[ \overline{z_7} = 2\text{cos}(210^\text{o}) - 2j\text{sin}(210^\text{o}) \]. 8. \[ z_8 = (\text{cos}(-0.5) + j \text{sin}(-0.5)) \], so \[ a = \text{cos}(-0.5) \] and \[ b = \text{sin}(-0.5) \]. Therefore, \[ z_8 = \text{cos}(-0.5) + j \text{sin}(-0.5) \] and \[ \overline{z_8} = \text{cos}(-0.5) - j \text{sin}(-0.5) \].