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!

Douglas's answer is good and efficient. In case you wouldn't have come up with that idea, there's also a "pedestrian" way to do this: Write $x$ as $a+\mathrm{i}b$ and then consider $(2+\mathrm{i})(a+\mathrm{i}b)\equiv 1+0\mathrm{i}\pmod{3}$ as two real equations, one for the real part and one for the imaginary part. That gives you two linear equations for two unknowns, which you can solve like any old system of linear equations since the congruence classes mod $3$ form a field.
