(3+2i) z+ (5+2i) * (1-3i) =7-10i ezt a feladatot hogy kell megoldani?

(5+2i)(1-3i) = 5-15i+2i-6i² = 5-13i+6 = 11-13i

(3+2i)z+11-13i = 7-10i /-11+13i

(3+2i)z = -4+3i /:(3+2i)

z = (-4+3i):(3+2i)=-6/13+17i/13

Ez trigonometrikus alakban:

r=√((6/13)²+(17/13)²)=√(325)/13=5√(13)/13=5/√(13)

tgφ=-17/6

φ=-70,56°+180°=109,44°

z = 5/√(13)∙(cos109,44°+i∙sin109,44°)