Soit ${\displaystyle (a,b)\in \mathbb {C} }$, soit ${\displaystyle n\in \mathbb {N} }$. ${\displaystyle (a+b)^{n}=\sum _{k=0}^{n}{n \choose k}a^{k}b^{n-k}}$

Les éléments a et b commutent car ${\displaystyle \mathbb {C} }$ est un corps commutatif. La démonstration est alors la même que dans un anneau commutatif.

Soit ${\displaystyle (a,b)\in \mathbb {C} ^{2}}$, soit ${\displaystyle n\in \mathbb {N} ^{*}}$. ${\displaystyle a^{n}-b^{n}=(a-b)\cdot \sum _{k=0}^{n-1}a^{k}b^{n-1-k}}$

Soit ${\displaystyle (a,b)\in \mathbb {C} ^{2}}$ :

${\displaystyle {\begin{aligned}(a-b)\cdot \sum _{k=0}^{n-1}a^{k}b^{n-1-k}&=\sum _{k=0}^{n-1}a^{k+1}b^{n-1-k}-\sum _{k=0}^{n-1}a^{k}b^{n-k}\\&=\sum _{k=1}^{n}a^{k}b^{n-k}-\sum _{k=0}^{n-1}a^{k}b^{n-k}\\&=a^{n}-b^{n}\end{aligned}}}$

Soit ${\displaystyle z\in \mathbb {C} }$, soit ${\displaystyle (m,n)\in \mathbb {N} ^{2}}$ tels que ${\displaystyle m\leq n}$.

• si ${\displaystyle z=1}$, ${\displaystyle \sum _{k=m}^{n}z^{k}=n-m+1}$
• si ${\displaystyle z\not =1}$, ${\displaystyle \sum _{k=m}^{n}z^{k}={\frac {z^{n+1}-z^{m}}{z-1}}}$

Soit ${\displaystyle z\in \mathbb {C} \backslash \{1\}}$, soit ${\displaystyle (m,n)\in \mathbb {N} ^{2}}$ tels que ${\displaystyle m\leq n}$.

${\displaystyle {\begin{aligned}\sum _{k=m}^{n}z^{k}&=z^{m}\cdot \sum _{k=0}^{n-m}z^{k}\\&={\frac {z^{m}}{z-1}}\cdot (z-1)\sum _{k=0}^{n-m}z^{k}1^{n-m-k}\\&={\frac {z^{m}}{z-1}}(z^{n-m+1}-1)\\&={\frac {z^{n+1}-z^{m}}{z-1}}\\\end{aligned}}}$