Let $f$ be a Laurent series centred at 0 and convergent in $\mathbb{C}$\\{0}. Let $b$ be the residue of $f$ at $z=0$.

Question

Let $f$ be a Laurent series centred at 0 and convergent in $\mathbb{C}$\{0}. Let $b$ be the residue of $f$ at $z=0$. Prove that there exists $w\in \mathbb{C}$ with $|w|=1$ such that $|f(w)-w^{-1}|\ge|b-1|$

Answer 1

Consider the function $g(z) = f(z) - \frac{1}{z}$ and let us compute the integral

$$\oint_{|w| = 1} g(w) dw$$

using the Residue Theorem. From the problem statement we know that $f$'s only singularity is at $z=0$. On the other hand we know that $\frac1z$ also has a singularity at $z=0$ which is a first order pole, with residue $1$. (Can you back this up on your own?)

Therefore we have

$$Res(g, 0) = b-1$$

So, by the residue theorem we have

$$\oint_{|w| = 1} g(w)dw = 2\pi i(b-1)$$

On the other hand, we know that

$$\left|\oint_{|w| = 1} g(w)dw\right| \leq \oint_{|w|=1} |g(w)dw| \leq 2\pi M$$

where $M$ is the maximum of the function $g(z)$ on the unit circle, attained at some point $z_0$.

Can you show that $|M| \geq |b-1|$? Try supposing $|M| < |b-1|$ and get to a contradiction.