## Precalculus: Mathematics for Calculus, 7th Edition

$(2-3i)^{-1}=\dfrac{2}{13}+\dfrac{3}{13}i$
$(2-3i)^{-1}$ Rewrite the expression: $(2-3i)^{-1}=\dfrac{1}{2-3i}=...$ Multiply the fraction by $\dfrac{2+3i}{2+3i}$: $...=\Big(\dfrac{1}{2-3i}\Big)\Big(\dfrac{2+3i}{2+3i}\Big)=\dfrac{2+3i}{2^{2}-(3i)^{2}}=\dfrac{2+3i}{4-9i^{2}}=...$ Substitute $i^{2}$ with $-1$: $...=\dfrac{2+3i}{4-9(-1)}=\dfrac{2+3i}{4+9}=\dfrac{2+3i}{13}=\dfrac{2}{13}+\dfrac{3}{13}i$