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We learned about conformal mappings and various properties of locally one-to-one, analytic mappings of the unit disk. I am having trouble with the following problem, can anyone help?

Let $f(z) =\sum_{n=0}^\infty a_n z^n$ be analytic and locally one-to-one in the unit disk, $|z| \leq 1$, and suppose $f$ maps the unit disk onto a domain $D$ whose area is $A$. Show that $A = \pi \sum^{\infty}_{n=1} n |a_{n}|^2.$

Also, if it is known that $f'(0)=1$, can we determine whether $A \leq \pi$ or $A \geq \pi$?

EDIT: I have attempted to show this using Green's Theorem since we talked about this theorem in class and proved several important results related to area. Here is what I was able to come up with. Are there any mistakes?

Green's Theorem can be used to show that if $C$ is a simple closed contour and $S$ is a region bounded by $C$, then $$ \mbox{Area of }S = \frac{1}{2i}\int_C \overline{z} dz.$$

Since $f$ is locally one-to-one, $D$ is a simple closed curve. Therefore, $$ A = \frac{1}{2i}\int_{\partial D} \overline{w} dw.$$

By changing variables, we obtain

$$ A = \frac{1}{2i}\int_{\partial D} \overline{f(z)}f'(z) dz.$$

Changing to polar coordinates and using the fact that $\partial D = \{e^{it} : 0 \leq t \leq 2\pi \}$ implies:

$$A = \frac{1}{2i}\int^{2\pi}_{0} \overline{ f(e^{it})} f'(e^{it}) dt = \frac{1}{2i}\int^{2\pi}_{0} \left(\sum^{\infty}_{n=0} \overline{a}_n e^{-int}\right)\left(\sum^{\infty}_{m=1} m{a_m} e^{-i(m-1)t}\right)ie^{it}dt = \frac{1}{2} \int^{2\pi}_{0} \left(\sum^{\infty}_{n=0} \overline{a}_n e^{-int}\right)\left(\sum^{\infty}_{m=1} m{a_m} e^{-imt}\right)dt.$$ By expanding the product of sums and using the fact that $e^{ikt}$ integrates to 0 as $0 \leq k \leq 2\pi$, we have:

$$A = \frac{1}{2}\int^{2\pi}_{0} \sum^{\infty}_{k=1}k\overline{a}_k a_k dt = \pi \sum^{\infty}_{k=1} k |a_k|^2.$$

Also, if $f'(0)=1$, then $a_1 = 1$. Thus, $A = \pi + \pi \sum^{\infty}_{n = 2} n|a_n|^2 > \pi$. Equality holds when $a_k = 0$ for $k > 1$, that is, when $f(z) = a_0 + z$ for some constant $a_0$.

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The answer to the second question is immediate from the formula. What is $a_1$? –  Jonas Meyer Feb 19 '12 at 18:00
    
I think you want to use Green's theorem somewhere. –  ShawnD Feb 19 '12 at 18:15
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This looks very similar to the question Georges links to. Another related question: math.stackexchange.com/q/47052 –  Jonas Meyer Feb 19 '12 at 18:25

1 Answer 1

up vote 2 down vote accepted

$D$ is a simple closed curve.

A domain cannot be a simple closed curve. It may be bounded by a simple closed curve. More importantly, $f$ being locally one-to-one does not imply that $\partial D$ is a simple closed curve, or even that $D$ is simply-connected. For example, this is the image of the unit circle under $f(z)=(z+2)^{8}$ (which is locally one-to-one due to having nonzero derivative in the closed unit disk):

8th power

The plot shows that image of the unit disk under $f$ wraps around the origin and overlaps itself. At the same time, $|f|\ge 1$ in the closed unit disk, which means the image has a tiny hole (not seen on the picture). Thus, the boundary of the image has at least two components.

Let $f(z)=\sum_{n=0}^\infty a_nz^n$ be analytic and locally one-to-one in the unit disk, $|z|\le 1$, and suppose $f$ maps the unit disk onto a domain $D$ whose area is $A$. Show that $A=\pi \sum_{n=1}^\infty n|a_n|^2$.

This is false as stated. The computations in your post lead to $\iint_{|z|<1}|f'|^2 = \pi \sum_{n=1}^\infty n|a_n|^2$. The integral on the left measures the area of image, counting with multiplicity. For example, the small loop on the picture shown above will be counted twice. Therefore, $\iint_{|z|<1}|f'|^2>A$ whenever $f$ fails to be globally injective in the open unit disk $|z|<1$.

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