Let $f:\mathbb{D}\rightarrow\mathbb{C}$ is an univalent function , and $f(0)=0,f'(0)=1$ , find the minimum of the area.

Question

My idea: I want to use the surface element as we know $dzd\bar{z}=d(x+iy)d(x-iy)=-2idxdy$ and use the mean value principle，but I can't put it all together and I don't know if it's true.

Any help will be appreciated!

My personal strong suspicion (with no actual argument to back it up): $f(x) = x$ gives the minimum area. Just to be clear: $\Bbb D$ is the unit disc, right? — Arthur, Aug 22 '20 at 08:31
@Arthur: Your suspicion is correct (only that I never would use $x$ for a complex variable :) — Martin R, Aug 22 '20 at 08:44

score 1 · Answer 1 · answered Aug 22 '20 at 08:32

You can express the area of the image in terms of the Taylor coefficients of the function, compare Let $f$ be an analytic isomorphism on the unit disc $D$, find the area of $f(D)$. Therefore

$$ \text{area}\, f(D) = \pi \sum_{n=1}^\infty n \lvert a_{n} \rvert^2 \ge \pi |a_1| = \pi |f'(0)| = \pi $$

and equality holds if $f$ is linear, i.e. for $f(z) = z$.

Let $f:\mathbb{D}\rightarrow\mathbb{C}$ is an univalent function , and $f(0)=0,f'(0)=1$ , find the minimum of the area.

1 Answers1

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