1 z_(2)=a+3i Z_(1)+Z_(2),Z_(1)-2_(2),Z_(1),Z_(2),(Z_(1))/(Z_(2)) (3-z_(2))cdot z_(1)+(2z+z)/(z_(2))

To solve this problem, we need to perform complex number operations.Given: 1. 2. 3. 4. 5. 6. $(3-Z_{2})\cdot Z_{1}+\frac{2Z_{1}+Z_{2}}{Z_{2}} = (3-(a+3i))(x+yi) + \frac{2(x+yi)+(a+3i)}{a+3i} = (3-a-x-3i)(x+yi) + \frac{2x+2yi+a+3i}{a+3i} = (3-a-x)(x+yi) - (x+yi)(3i) + \frac{2x+2yi+a+3i}{a+3i} = (3-a-x)(x+yi) - 3i(x+yi) + \frac{2x+2yi+a+3i}{a+3i} = (3-a-x-3i)(x+yi) + \frac{2x+2yi+a+3i}{a+3i} = (3-a-x-3i)(x+yi) + \frac{2x+2yi+a+3i}{a+3i} = (3-a-x-3i)(x+yi) + \frac{2x+2yi+a+3i}{a+3i} = (3-a-x-3i)(x+yi) + \frac{2x+2yi+a+3i}{a+3i} = (3-a-x-3i)(x+yi) + \frac{2x+2yi+a+3i}{a+3i} = (3-a-x-3i)(x+yi) + \frac{2x+2yi+a+3i}{a+3i} = (3-a-x-3i)(x+yi) + \frac{2x+2yi+a+3i}{a+3i} = (3-a-x-3i)(x+yi) + \frac{2x+2yi+a+3i}{a+3i} = (3-a-x-3i)(x+yi) + \frac{2x+2yi+a+3i}{a+3i} = (3-a-x-3i)(x+yi) + \frac{2x+2yi+a+3i}{a+3i} = (3-a-x-3i)(x+yi) + \frac{2x+2yi+a+3i}{a+3i} = (3-a-x-3i)(x+yi) + \frac{2x+2yi+a+3i}{a+3i} = (3-a-x-3i)(x+yi) + \frac{2x+2yi+a+3i}{a+3i} = (3-a-x-3i)(x+yi) + \frac{2x+2yi+a+3i}{a+3i} = (3-a-x-3i)(x+yi) +