Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.1 Complex Numbers - 8.1 Exercises - Page 358: 78

$$-5i(4-3i)^2=-120-35i$$

$$A=-5i(4-3i)^2$$ For $(4-3i)^2$, remember that $$(a-b)^2=a^2+b^2-2ab$$ which means $$(4-3i)^2$$ $$=16+9i^2-24i$$ $$=16+9(-1)-24i$$ $$=16-9-24i$$ $$=7-24i$$ Now we can apply the result just found to $A$ $$A=-5i(7-24i)$$ $$A=-35i+120i^2$$ $$A=120(-1)-35i$$ $$A=-120-35i$$