\(\displaystyle{\dfrac{z_D-z_C}{z_B-z_A} = \dfrac{7-\left(4+3i\right)}{1+3i-\left(-1+i\right)} = \dfrac{3-3i}{2+2i} }\)
D'où :
\(\displaystyle{\dfrac{z_D-z_C}{z_B-z_A} = \dfrac{3-3i}{2+2i} }\)
\(\displaystyle{\dfrac{z_D-z_C}{z_B-z_A} = \dfrac{3-3i}{2+2i} \times \dfrac{2-2i}{2-2i}}\)
\(\displaystyle{\dfrac{z_D-z_C}{z_B-z_A} = \dfrac{6-6i-6i+6i^2}{4+4} }\)
Finalement :
\(\displaystyle{\dfrac{z_D-z_C}{z_B-z_A} = -\dfrac{12}{8} i= -\dfrac{3}{2} i}\)