Finding inverse of congruence class involving complex number How do you find the inverse of $[2+i]_{3}$? I changed it into solving for x in $(2+i)x \equiv 1 \pmod{3}$, and tried to solve for x with extended euclid algorithm, but with no luck. Am I supposed to do something different with complex numbers?
Thanks!

Use $(2+i)(2-i)=5$, so $(2+i)(2-i)5^{-1} \equiv 1 \mod 3$ where $5^{-1}$ is any multiplicative inverse of $5$.
