Simplify the coefficient of z on the left side. We do this by rationalizing the denominators and multiplying them by their complex conjugates:
[tex]\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{3-2i}{1+i}\cdot\dfrac{1-i}{1-i} - \dfrac{5+3i}{1+2i}\cdot\dfrac{1-2i}{1-2i}[/tex]
[tex]\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{(3-2i)(1-i)}{1-i^2} - \dfrac{(5+3i)(1-2i)}{1-(2i)^2}[/tex]
[tex]\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{3 - 2i - 3i + 2i^2}{1-(-1)} - \dfrac{5 + 3i - 10i - 6i^2}{1-4(-1)}[/tex]
[tex]\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{3 - 5i + 2(-1)}2 - \dfrac{5 - 7i - 6(-1)}5[/tex]
[tex]\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{1 - 5i}2 - \dfrac{11 - 7i}5[/tex]
[tex]\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{1 - 5i}2\cdot\dfrac55 - \dfrac{11 - 7i}5\cdot\dfrac22[/tex]
[tex]\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = \dfrac{5 - 25i - 22 + 14i}{10}[/tex]
[tex]\dfrac{3-2i}{1+i} - \dfrac{5+3i}{1+2i} = -\dfrac{17 + 11i}{10}[/tex]
So, the equation is simplified to
[tex]-\dfrac{17+11i}{10} z = \dfrac12 - \dfrac{2i}5[/tex]
Let's combine the fractions on the right side:
[tex]\dfrac12 - \dfrac{2i}5 = \dfrac12\cdot\dfrac55 - \dfrac{2i}5\cdot\dfrac22[/tex]
[tex]\dfrac12 - \dfrac{2i}5 = \dfrac{5-4i}{10}[/tex]
Then
[tex]-\dfrac{17+11i}{10} z = \dfrac{5-4i}{10}[/tex]
reduces to
[tex]-(17+11i) z = 5-4i[/tex]
Multiply both sides by -1/(17 + 11i) :
[tex]\dfrac{-(17+11i)}{-(17+11i)} z = \dfrac{5-4i}{-(17+11i)}[/tex]
[tex]z = -\dfrac{5-4i}{17+11i}[/tex]
Finally, simplify the right side:
[tex]-\dfrac{5-4i}{17+11i} = -\dfrac{5-4i}{17+11i} \cdot \dfrac{17-11i}{17-11i}[/tex]
[tex]-\dfrac{5-4i}{17+11i} = -\dfrac{(5-4i)(17-11i)}{17^2-(11i)^2}[/tex]
[tex]-\dfrac{5-4i}{17+11i} = -\dfrac{85 - 68i - 55i + 44i^2}{289-121(-1)}[/tex]
[tex]-\dfrac{5-4i}{17+11i} = -\dfrac{85 - 68i - 55i + 44(-1)}{410}[/tex]
[tex]-\dfrac{5-4i}{17+11i} = -\dfrac{41 - 123i}{410}[/tex]
[tex]-\dfrac{5-4i}{17+11i} = -\dfrac{41 - 41\cdot3i}{410}[/tex]
[tex]-\dfrac{5-4i}{17+11i} = -\dfrac{1 - 3i}{10}[/tex]
So, the solution to the equation is
[tex]z = -\dfrac{1-3i}{10} = \boxed{-\dfrac1{10} + \dfrac3{10}i}[/tex]
