2

Let $f(z)$ be a complex polynomial of degree at least $2$ and $R$ be a positive number such that $f(z) \neq 0$ for all $|z| \geq R$. Show that $\int_{|z|=R} \frac{dz}{f(z)}=0$ and deduce that $\sum_{k=1}^n \dfrac{1}{\prod_{j\neq k}(a_k-a_j)}=0$ where $a_i$ is $n$ distinct roots of $f$

I can indicate $\sum_{k=1}^n \dfrac{1}{\prod_{j\neq k}(a_k-a_j)}=0$ by Residue but I wonder why $\int_{|z|=R} \frac{dz}{f(z)}=0$?

Desunkid
  • 1,291
  • See https://math.stackexchange.com/questions/2804888/showing-that-sum-i-1m-frac1-prod-j-1-j-neq-im-a-j-a-i-is?rq=1 – Robert Z Oct 09 '18 at 15:59
  • Since $f(z) \ne 0$ for all $|z| \ge R$, $\int_{|z| = R}\frac{dz}{f(z)} = \int_{|z| = R'} \frac{dz}{f(z)}$ for any $R' > R$. Now send $R'$ to $\infty$... – achille hui Oct 09 '18 at 16:01
  • send R' to $\infty$ and then? – Desunkid Oct 09 '18 at 16:04
  • 1
    for large $R'$, $|f(z)|$ grows like $|C|R'^n$ where $C$ and $n$ is the leading coefficient and degree of $f$, so $|\int_{|z|=R'} \frac{dz}{f(z)}| \le \int_{|z|=R'} \frac{|dz|}{|f(z)|} \sim \frac{2\pi R'}{|C| R'^n} \to 0$ as $R' \to \infty$. – achille hui Oct 09 '18 at 16:08

1 Answers1

1

Hint. Note that if $f(z)=\prod_{j=1}^n (z-a_j)$ where the complex numbers $a_j$ with $j=1,\dots,n$ are all distinct then $$\frac{1}{f(z)}=\sum_{k=1}^n\frac{A_k}{z-a_k}\quad \text{ where } A_k=\dfrac{1}{\prod_{j\neq k}(a_k-a_j)}.$$ Then for $R>\max_{j=1,\dots,n}{|a_j|}$, $$\int_{|z|=R} \frac{dz}{f(z)}=\sum_{k=1}^nA_k\int_{|z|=R} \frac{dz}{z-a_k}.$$ P.S. Do you know what the Residue at infinity is? Well in our case such residue is zero: $$\text{Res}\left(1/f;\infty\right)=-\text{Res}\left(\dfrac{1}{z^2f\left(\dfrac 1z\right)},0\right) =-\text{Res}\left(\dfrac{z^{n-2}}{\prod_{j=1}^n (1-a_jz)},0\right)=0$$ because $n\geq 2$.

Robert Z
  • 147,345