3

This is a question from NBHM $2022$ exam. It asks to find the area of the region $\{z+\frac{z^2}{2} \mid z\in \mathbb{C},|z| \leq 1\}$

Now $z+\frac{z^2}{2}$ = $\frac{(z+1)^2}{2}-\frac{1}{2}$.

The $-\frac{1}{2}$ part is just a translation, so it does not change the area. But I couldn't understand what is happening in the $\frac{(z+1)^2}{2}$ part. The resulting region is not a circular shape as the images of $i,-1,-i$ are $i,0,-i$ which lies on the same line. My question is how to understand what's happening in the $\frac{(z+1)^2}{2}$ part. Edit: the answer is given $\frac{3\pi}{2}$

4 Answers4

5

$\textrm{Area Theorem}$:Let $f:\mathbb{D}\to f(\mathbb{D})$ is a conformal map with $f(0) =0$ and $f'(0) >0$ so that $f$ has a Taylor expansion $$f(z) =\sum_{n=1}^{\infty} a_nz^n$$ $|z|<1 $ and $a_1>0$.

Then $$\textrm{Area}(f(\mathbb{D}))=\pi\sum_{n=1}^{\infty}n|a_n|^2$$

Given $f(z) =z+\frac{z^2}{2}=\sum_{n\ge 0}{a_nz^n}$

$\begin{align}\textrm{Area}(f(\mathbb{D}))&=\pi\sum_{n=1}^{\infty}n|a_n|^2\\&=\pi(1+\frac{1}{2})\\&=\frac{3\pi}{2}\end{align}$

SoG
  • 13,327
  • 1
    Can you give any link about this theorem? i couldn't find anything upon searching. –  Mar 19 '23 at 14:00
  • The proof of the above theorem follows from the @achille hui's proof combining with cauchy product of power series. – SoG Apr 09 '23 at 08:08
  • 1
    It can also be seen as an application of Parseval's identity (compare https://math.stackexchange.com/q/1764827/42969). The conditions $f(0) = 0$ and $f'(0) > 0$ are not really needed, only that $f$ is injective. – Martin R Apr 12 '23 at 04:54
3

Hint: the disk $\{z + 1 : \lvert z \rvert \le 1\}$ can be reparameterized as $$ \{ r e^{it} : -\tfrac{\pi}{2} < t < \tfrac{\pi}{2}, 0 \le r \le 2 \cos t\}, $$ so its image under the map $z \mapsto z^2$ is $$ \{ r e^{it} : -\pi < t < \pi, 0 \le r \le 4 (\cos \tfrac{t}{2})^2\}, $$ whose area follows from polar integration.

L. F.
  • 2,703
1

Following is a way to evaluate the area using complex coordinates $z = x+iy$ and $\bar{z} = x-iy$. The key is in terms of $z, \bar{z}$, the area element has the form:

$$dx \wedge dy = \frac{1}{2i} d\bar{z} \wedge dz$$

Let $\omega(z) = z + \frac{z^2}{2}$ and $D$ be the unit disk and $C = \partial D$ be the unit circle.

Notice the map $C \in z \mapsto z + \frac{z^2}{2} \in \partial \omega(D)$ is 1-1, the area of $\omega(D)$ equals to

$$\begin{align} \verb/Area/ &= \overbrace{\frac1{2i}\int_{\omega(D)}d\bar{\omega} \wedge d\omega = \frac1{2i}\int_{\partial \omega(D)}\bar{\omega} d\omega} ^{\text{complex version of Stokes' theorem}}\\ &= \frac1{2i}\int_C \left(\bar{z} + \frac{\bar{z}^2}{2}\right)(1 + z)dz\end{align} $$ Since $\bar{z} = \frac1z$ on $C$, we have

$$\begin{align}\verb/Area/ &=\frac1{2i}\int_C\left(\frac1z + \frac{2}{z^2}\right)(1+z)dz\\ &=\frac{2\pi i}{2i} \verb/Res/_{z=0}\left[1 + \frac3{2z} + \frac1{2z^2}\right]\\ &= \frac{3\pi}{2} \end{align} $$

achille hui
  • 125,323
0

Let $f$ be a holomorphic function bounded on an open connected bounded set $D$. The area of the region $f(D)$ is $\iint_{f(D)}dudv$. With the change of coordinates $u(x,y)\to x$ and $v(x,y)\to y$, the area is $\iint_{D}|J|dxdy$. $J$ is the change of coordinate matrix and because real and imaginary parts of a holomorphic function satisfy the CR equations, $|J|=|f'(z)|^2$.

Thus the area is $\iint_{D}|f'(z)|^2dxdy$.

$f=z+\frac{z^2}{2}$ is holomorphic and bounded on the unit disc. $f'(z)=1+z$. So, $|f'(z)|^2=(1+x)^2+y^2$. The area is $\iint_{|z|<1}(1+x^2+2x+y^2)dxdy$. Transforming to polar coordinates so that the integration can be done easily, the jacobian is $r$. So $dxdy=rdrd\theta$. The area is $\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}(r^2+1+2rcos\theta)rdrd\theta =\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}(r^3+r+2r^2cos\theta)drd\theta=\int_{\theta=0}^{\theta=2\pi}(1/4+1/2+\frac{2}{3}cos\theta)d\theta=\frac{3\pi}{2}$.

subrat verma
  • 69