a) Ta có \((3 - 2i)z + (4 + 5i) = 7 + 3i \Leftrightarrow (3 - 2i)z = 7 + 3i - 4 - 5i\)
\(\Leftrightarrow (3-2i)z=3-2i \Leftrightarrow z = \dfrac{3-2i}{3-2i} \Leftrightarrow z = 1\).
Vậy \(z = 1\).
b) Ta có \((1 + 3i)z - (2 + 5i) = (2 + i)z \\ \Leftrightarrow (1 + 3i)z -(2 + i)z = (2 + 5i)\)
\(\Leftrightarrow (1 + 3i - 2 - i)z = 2 + 5i \\ \Leftrightarrow (-1 + 2i)z = 2 + 5i\)
\(\Leftrightarrow z = \dfrac{2 + 5i}{-1+2i} \\ \Leftrightarrow z=\dfrac{(2+5i)(-1-2i)}{1^2+2^2}\\\Leftrightarrow z=\dfrac{-2-4i-5i-10i^{2}}{5} \\ \Leftrightarrow z=\dfrac{8-9i}{5} =\dfrac{8}{5}-\dfrac{9}{5}i\)
Vậy \(z =\dfrac{8}{5}-\dfrac{9}{5}i.\)
\(\begin{array}{l}c)\;\;\dfrac{z}{{4 - 3i}} + 2 - 3i = 5 - 2i\\ \Leftrightarrow \;\dfrac{z}{{4 - 3i}} = 5 - 2i - 2 + 3i\\ \Leftrightarrow \;\dfrac{z}{{4 - 3i}} = 3 + i\\ \Leftrightarrow z = \left( {3 + i} \right)\left( {4 - 3i} \right)\\ \Leftrightarrow z = 12 - 5i - 3{i^2}\\ \Leftrightarrow z = 15 - 5i.\end{array}\)
Vậy \(z=15-5i.\)
