Kurs:Mathematik für Anwender (Osnabrück 2019-2020)/Teil I/Repetitorium/5/complex inverse/Studentenfrage/Antwort

Because we add componentwise in ${ℝ}^2$ the negative of $(x,y)$ is indeed the componentwise negative $(-x,-y)$, i.e. the pair for which $1$.

But the inverse is the multiplicative inverse and not the negative. As you correctly say the inverse of $a$ has to fulfill $a\cdot b=1$. So to find the inverse of $(x,y)$ you have to find a pair $(w,z)\in {ℝ}^2$ such that $(x,y)\cdot (w,z)=1=(1,0)$. Plug this into the multiplication rule for complex numbers and solve the resulting system of equations for the variables $w$ and $z$ and you find the inverse of $(x,y)$.

Alternatively, try to prove the formula in Satz 5.19  (5).

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