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I am trying to understand the above theorem. I don't understand how we get the first line of the proof.

1- Why is the area of $E_r$ defined in this way?

2- How is $E_r$ two-dimentional?

3- How does a set have an "area"?

Source: P. L. Duren Univalent Functions

gbd
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    Your questions are not entirely clear to me. $E_r$ is the interior domain of the simple closed curve $C_r$, what are your doubts about it being two-dimensional? And the area is not defined that way but can be computed that way, see for example https://en.wikipedia.org/wiki/Green%27s_theorem#Area_calculation. – Martin R Feb 27 '22 at 08:41
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    See also https://math.stackexchange.com/q/2561793/42969 for some derivations of the area formula. – Martin R Feb 27 '22 at 08:52
  • @MartinR, I mean for example where did $\bar{w} dw$ come from? Why is $w=g(z)$? – gbd Feb 27 '22 at 09:01
  • $C_r$ is the image of $|z|=r$ under the function $g$, that is parameterized as $C_r(t) = g(re^{it})$, $0 \le t \le 2 \pi$. Therefore $$ \int_{C_r} \overline w dw = \int_0^{2\pi}\overline{g(re^{it})}g'(re^{it})ire^{it} dt = \int_{|z|=r} \overline{g'(z)}g(z) dz. $$ – Martin R Feb 27 '22 at 09:09
  • @MartinR, But why $\bar{w}dw$ and not "$wdw$"? Also, is $E_r$ a region containing the origin? – gbd Feb 27 '22 at 09:14
  • $E_r$ is the interior domain of $C_r$, that is a bounded domain. It may contain the origin or not, that should not be relevant. – Martin R Feb 27 '22 at 09:50
  • See also this: https://math.stackexchange.com/q/257743/42969 – Martin R Feb 27 '22 at 10:07
  • @MartinR, In the last link it is shown that $$\mathrm{Area}(D)=\frac{1}{2i} \oint_\gamma \overline{z} ,dz.$$ and I understand why that is from the proof in the answer. But the proof uses $z=x+iy$, however in the area theorem we have $z$ or rather $w$ equal to a function $g$ and not $x+iy$. So how does this still give the same $\mathrm{Area}(D)$? I hope it is clear what I mean. – gbd Feb 27 '22 at 10:12
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    You can name the integration variable whatever you want. $\oint_\gamma \overline{z} ,dz = \oint_\gamma \overline{w} ,dw$. Here, informally speaking, $g$ maps from the z-plane to the w-plane, and we want to calculate an area in the image, therefore the letter $w$ is chosen as integration variable. – Martin R Feb 27 '22 at 10:17

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