The Real Part of a Power Series

We assume that the function

Express the coefficients $a_{1},a_{2},\ldots ,a_{n},\ldots$ in terms of the real and the imaginary part resp.of $f(z)$ on the circle $\left|z\right|=r$, $0<r<R$.

(continued). We set $\re f(re^{i\vartheta }) = U(r,\vartheta )$ and assume that $f(0)$ is real and $\left|z\right|<r$. Then

(continued). Suppose $f(z)$ does not vanish on $\left|z\right|=r$ and that its zeros in the disk $\left|z\right|<r$ are $c_{1},c_{2},\ldots ,c_{m}$. Then we have for $\left|z\right|<r$

where $\gamma$ is a real constant. [Consequence of problem 231 as problem 120 is a consequence of problem 119.]

The function $f(z)$ is regular, its real part is positive in the open disk $\left|z\right|<R$ and continuous in the closed disk $\left|z\right|\leq R$. If the real part becomes identically zero on an arc of the circle the imaginary part of $f(z)$ changes on this arc always in the same sense: it decreases as $\arg z$ increases.

Let $f(z) = a_{0}+a_{1}z+a_{2}z^{2}+\cdots +a_{n}z^{n}+\cdots$ be regular in the disk $\left|z\right|<R$, $f(re^{i\vartheta }) = U(r,\vartheta )+iV(r,\vartheta )$, $U(r,\vartheta ),V(r,\vartheta )$ are real. Then the equation

holds for $0<r<R$ provided that it holds for $r=0$.

The function

is assumed to be regular and to have positive real part in the open disk $\left|z\right|<1$. Then $\left|a_{n}\right|\leq 1$, $n=1,2,\ldots$ In none of these inequalities can we replace $1$ by a smaller number.

Suppose that the function

on the disk $\left|z\right|<R$. Then the inequality

holds for $0<r<R$. Example: $f(z)=\dfrac {z+1}{z-1}$, $R=1$, $A=0$.

We assume that the Laurent series

converges on the annulus $0<\left|z\right|<\infty$ (sphere from which two points have been removed) and that $z=0$ and $z=\infty$ are essential isolated singular points. The maximum of the real part of $\psi (z)$ on the circle $\left|z\right|=r$ is denoted by $A(r)$. $A(r)$ increases to $\infty$ faster than any power of $r$ as $r\to \infty$ and faster than any power of $\dfrac {1}{r}$ as $r\to 0$. Precisely

We assume that the function $f(z) = a_{0}+a_{1}z+\cdots +a_{n}z^{n}+\cdots$ is regular in the disk $\left|z\right|<R$ and we denote by $\Delta (f)$ the largest oscillation of the real part of $f(z)$ in $\left|z\right|<R$, i.e.$\Delta (f)$ is the least upper bound of $\left|\re f(z_{1})-\re f(z_{2})\right|$ for $\left|z_{1}\right|<R$, $\left|z_{2}\right|<R$. Then we have

The factor $\dfrac {2}{\pi }$ cannot be replaced by a smaller one.

Interpret this proposition geometrically.

Let the function $f(z) = a_{0}+a_{1}z+\cdots +a_{n}z^{n}+\cdots$ be regular in the disk $\left|z\right|<R$ and denote by $D(f)$ the largest oscillation of $f(z)$ for $\left|z\right|<R$, i.e.$D(f)$ is the least upper bound of $\left|f(z_{1})-f(z_{2})\right|$ for $\left|z_{1}\right|<R$, $\left|z_{2}\right|<R$. Then we have

The factor $\frac{1}{2}$ can not be replaced by a smaller one. Interpret this proposition geometrically.

Let the function $f(z)$ satisfy the following conditions:

1. $f(z)$ is regular, $\left|f(z)\right|\leq M$ in the disk $\left|z-s\right|\leq r$;

2. $f(z)$ does not vanish in the closed half-disk $\left|z-s\right|\leq r$, $\re (z-s)\geq 0$;

3. $f(z)$ has in the disk $\left|z-s\right|\leq \frac{2}{3}r$ the zeros $c_{1},c_{2},\ldots ,c_{l}$. Then

[We may assume $s=0$; problems 232, 120.]