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In my script a complex number is defined like that:

$\mathbb C := \mathbb R [X]/(x^2+1)$ and the imaginary entity: $i := [X] \in \mathbb C$

$\mathbb C = \{a+bi | a, b \in \mathbb R\}$

Question: How can I get the inverse $(2+i)^{-1}$ in $\mathbb C$ using these definition rules?

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    In $\mathbb{R}[x]/(x^2+1)$ : $\frac{1}{a+bx} =\frac{a-bx}{(a+bx)(a-bx)} =\frac{a-bx}{a^2-b^2 x^2} = \frac{a}{a^2+b^2}-\frac{bx}{a^2+b^2}$ (using that $a^2-b^2 x^2 = a^2-b^2 x^2+b^2(x^2+1)=a^2+b^2$) – 2017-02-07

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$(2+i)^{-1}$ is, by definition of $^{-1}$, the element $a + bi\in \Bbb C$ such that $$ (a+bi)(2+i) = 1 $$ (if it even exists and is unique, that is). Now expand those brackets as with normal polynomials, but remember that $i^2 = -1$. This gives you an equation with the unknowns $a$ and $b$ that can hopefully be solved.

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    is $2a + (2b+a)i-b=1$ correct? Can you explain how I should solve that if I have only one equation and two variables? – 2017-02-07
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    You can solve it because this is a polynomial equation with $i$ being the variable. That means that the constant terms ($2a-b$ and $1$) have to be equal, and simultaneously, the first-degree terms ($2b+a$ and $0$) have to be equal. – 2017-02-07
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    I am sorry, I still don't get it – 2017-02-07
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    In $\Bbb R[X]$, do you understand how the equation $(2a-b) + (2b+a)X = 1$ can be solved, with $a, b\in \Bbb R$? This is _exactly_ the same thing going on. – 2017-02-07
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    Ok, so I have two equations: $2a-b=1$ and $2b+a=0$. After that I have inserted the first one into the second. So $a= \frac{1}{2,5}$ and $b= -\frac{1}{2}a= -\frac{1}{5}$, right? – 2017-02-08
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    @jublikon That looks good. But don't trust me. Test it out instead! Is it really true that $(\frac25 - \frac i5)\cdot (2+i) = 1$? – 2017-02-08