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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!

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*You are on the right track. Did you try any specific $x$'s? –  AD. Apr 1 '11 at 5:08

3 Answers 3

up vote 1 down vote accepted

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 $5$.

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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.

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thanks for the response! but I do have a question. like you said, I ended up with two equations: 2a -b + 3m = 1 and 2b + a = 0. Note that I have a "m" there, which makes 3 unknowns. am I missing something? –  user8969 Apr 1 '11 at 5:32
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@mathcat: Yes, I think you're missing something quite essential. If I understand correctly, you introduced $m$ because the equation was mod $3$ and you wanted to convert it into a "normal" equation? First, note that in that approach you would also have had to write $2b+a+3n=0$. But more importantly, you don't need to do this. The nice thing about equations mod a prime is that you can treat them just like "normal" equations, because the congruence classes mod a prime form a field. So e.g. you can multiply $2a-b\equiv1\pmod{3}$ by $2$ to get $a-2b\equiv2\pmod{3}$, since $2\cdot2\equiv1\pmod{3}$. –  joriki Apr 1 '11 at 5:43
    
yes, that was exactly what I was thinking. thanks a lot for the clarification :) –  user8969 Apr 1 '11 at 5:59

HINT $\ $ Rationalize the denominator of $\displaystyle \rm\ x\ =\ \frac{1}{2+i}$

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