$$ \omega = \frac{(1-i)(9-5i)}{(9+5i)(9-5i)} $$ $$ \omega = \frac{9 - 5i - 9i + 5i^2}{9^2 - (5i)^2} $$ $$ \omega = \frac{9 - 14i - 5}{81 - 25i^2} $$ $$ \omega = \frac{4 - 14i}{81 + 25} $$ $$ \omega = \frac{4 - 14i}{106} $$ $$ \omega = \frac{4}{106} - \frac{14}{106}i $$ $$ \omega = \frac{2}{53} - \frac{7}{53}i $$ So, $\boxed{\omega = \frac{2}{53} - \frac{7}{53}i}$. Step 5: Substitute $\omega$ into Equation (1) to solve for $z$. $$ (1-i)z + (1+i)\omega = 3 $$ $$ (1-i)z = 3 - (1+i)\omega $$ $$ (1-i)z = 3 - (1+i)\left(\frac{2}{53} - \frac{7}{53}i\right) $$ $$ (1-i)z = 3 - \frac{1}{53}(1+i)(2-7i) $$ $$ (1-i)z = 3 - \frac{1}{53}(2 - 7i + 2i - 7i^2) $$ $$ (1-i)z = 3 - \frac{1}{53}(2 - 5i + 7) $$ $$ (1-i)z = 3 - \frac{1}{53}(9 - 5i) $$ $$ (1-i)z = \frac{3 \times 53}{53} - \frac{9 - 5i}{53} $$ $$ (1-i)z = \frac{159 - (9 - 5i)}{53} $$ $$ (1-i)z = \frac{159 - 9 + 5i}{53} $$ $$ (1-i)z = \frac{150 + 5i}{53} $$ Step 6: Solve for $z$. $$ z = \frac{150 + 5i}{53(1-i)} $$ $$ z = \frac{150 + 5i}{53-53i} $$ Multiply the numerator and denominator by the conjugate of the denominator, $53+53i$: $$ z = \frac{150 + 5i}{53-53i} \times \frac{53+53i}{53+53i} $$ $$ z = \frac{(150+5i)(53+53i)}{53^2 - (53i)^2} $$ $$ z = \frac{150 \times 53 + 150 \times 53i + 5i \times 53 + 5i \times 53i}{53^2 - 53^2 i^2} $$ $$ z = \frac{7950 + 7950i + 265i + 265i^2}{2809 + 2809} $$ $$ z = \frac{7950 + 8215i - 265}{5618} $$ $$ z = \frac{7685 + 8215i}{5618} $$ $$ z = \frac{7685}{5618} + \frac{8215}{5618}i $$ Both numerator and denominator are divisible by $53$: $7685 \div 53 = 145$ $8215 \div 53 = 155$ $5618 \div 53 = 106$ $$ z = \frac{145}{106} + \frac{155}{106}i $$ So, $\boxed{z = \frac{145}{106} + \frac{155}{106}i}$. That's 2 down. 3 left today — send the next one.