Answer
$(2-3i)^{-1}=\dfrac{2}{13}+\dfrac{3}{13}i$
Work Step by Step
$(2-3i)^{-1}$
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 $-1$:
$...=\dfrac{2+3i}{4-9(-1)}=\dfrac{2+3i}{4+9}=\dfrac{2+3i}{13}=\dfrac{2}{13}+\dfrac{3}{13}i$
