# How to find the bound for this function?

According to journal entitled Certain subclass of starlike function by Gao and Zhou (2007), it was proven that $-\frac{r}{1+tr} \leq Re \{\frac{z}{1-tz} \}\leq \frac{r}{1-tr}$ where $|z|\leq r<1$ and $0 < t \leq 1$. It is possible to find the bound for $|\frac{z}{1-tz}|$ using that information?

• As written, it can be unbounded. Just set $z = \frac{1}{2}$, say, and $u = \frac{1}{2}$. – Ray Yang Jul 10 '13 at 9:02
• Maybe it should be fixed $u,\alpha$. The question does make it clear that $z$ is the variable, and ranges in some disk around $0$ which lies strictly inside the unit disk. As that is compact and $g(z)$ is continuous in there, an upper bound exists, and a lower bound. Of course the OP should include info about $u,\alpha$ being fixed reals. – coffeemath Jul 10 '13 at 9:17
• @RayYang: I already edited the question.. it is clear? – DRN Jul 15 '13 at 4:20
• @coffeemath: I already gave the references, it is helpful to make it more clearer? – DRN Jul 15 '13 at 4:22
• What are the assumptions about $r$ and $t$? I'd guess they are fixed real numbers, and perhaps each is positive and less than $1$. This would make the assumed inequality on the real part of $z/(1-tz)$ make sense. – coffeemath Jul 15 '13 at 10:47

Edited. A first version of this question asked for a proof of the estimate given in Gao and Zhou's paper. Here is this proof:

I assume that $0<r<1$ and that $0< t\leq1$.

Since $$z\mapsto{\rm Re}{z\over1-tz}$$ is a harmonic function it takes its extrema on the boundary of the allowed domain, i.e., when $z=re^{i\phi}$. Therefore it is sufficient to consider the function \eqalign{g(\phi)&:={\rm Re}{re^{i\phi}\over1-tre^{i\phi}}={r\cos\phi-tr^2\over 1-2rt\cos\phi+t^2r^2}\cr &={1-t^2 r^2\over 2t(1-2rt\cos\phi+t^2r^2)}-{1\over 2t}\qquad\qquad(0\leq\phi\leq\pi)\ .\cr} On the right hand side the first denominator is increasing with increasing $\phi$. It follows that $g$ takes its maximum at $\phi=0$ and its minimum at $\phi=\pi$, and one easily obtains $$g_\min=-{r\over 1+rt},\quad g_\max={r\over 1-rt}\ .\qquad\qquad\square$$

When only an estimate of $$\left|{z\over 1-tz}\right|$$ is needed things are much simpler. Again the maximum of this expression is taken when $|z|=r$. The points $tz$ then lie on a circle with radius $tr$, and among them the point nearest to $1$ is the point $tr$. It follows that $$0\leq\left|{z\over 1-tz}\right|\leq{r\over 1-tr}\ .$$

• @Landscape: You are right. I have changed that; the result remains the same. – Christian Blatter Jul 18 '13 at 11:21
• @Landscape: is it possible to find $|\frac{z}{1-tz}|$ with the information of the bound for this real part? – DRN Jul 18 '13 at 14:34
• This answer just establishes the bound on the real part, already mentioned in the OP's question. If a similar approach obtained a bound on the imaginary part, the two might be combined somehow to get a bound on the magnitude of the function. – coffeemath Jul 18 '13 at 15:00
• @coffeemath: means that I have to find the bound on the imaginary part and then combined lower and upper bound for both of them? Am I right? – DRN Jul 18 '13 at 15:32
• @ChristianBlatter: tqsm.. it is really helpful. – DRN Jul 18 '13 at 15:38