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

Because we add componentwise in ${}\mathbb {R} ^{2}$ 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 ${}(w,z)\in \mathbb {R} ^{2}$ 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).

Zur beantworteten Studentenfrage