Theorom from Serge Lang's Complex Analysis - pages 89-90

I am a math hobbyist/amateur studying complex analysis from Serge Lang's book Complex Analysis.

I need some help regarding Theorem 1.1 on Page 89

The theorem and is proof as given by Lang are as follows:

"Theorem 1.1 Let U be a connected open set, and let f be a holomorphic function on U. If f ' = 0 then f is constant.

Proof: Let $\displaystyle \alpha \ , \beta $ be two points in U and suppose first that $\displaystyle \gamma $ is a curve joining $\displaystyle \alpha $ to $\displaystyle \beta $ so that

$\displaystyle \gamma $(a) = $\displaystyle \alpha $ and $\displaystyle \gamma $(b) = $\displaystyle \beta $

The function

t --> f($\displaystyle \gamma $(t))

is differentiable, and by the chain rule, its derivative is

f ' ($\displaystyle \gamma $(t))$\displaystyle \gamma $'(t) = 0 .... (1)

Hence this function is constant, and therefore

f($\displaystyle \alpha $) = f($\displaystyle \gamma $(a)) = f($\displaystyle \gamma $(b)) = f($\displaystyle \beta $)

Next suppose that $\displaystyle \gamma $ = {$\displaystyle \gamma_1, \gamma_2, ..... \gamma_n$} is a path joining $\displaystyle \alpha $ to $\displaystyle \beta $, and let $\displaystyle z_j $ be the end point of $\displaystyle \gamma_j $ , putting

$\displaystyle z_0 = \alpha , \ \ \ z_n \ = \ \beta$

By what we have just proved

f($\displaystyle \alpha$) = f($\displaystyle z_0$) = f($\displaystyle z_1$) = f($\displaystyle z_2$) = ..... = f($\displaystyle \beta$)

thereby proving the theorem

QUESTIONS

(1) Lang does not describe the nature of the elements of U - but I am assuming that he is taking them as complex numbers? Is that right?

(2) At first sight - to someone familiar with elementary real analysis, the conclusion of the theorem that f ' = 0 implies f is constant does not seem surprising? Any comments?

(3) The statements in the proof:

"

The function

t --> f($\displaystyle \gamma $(t))

is differentiable, and by the chain rule, its derivative is

f ' ($\displaystyle \gamma $(t))$\displaystyle \gamma $'(t) = 0 .... (1)

Hence this function is constant,"

seem to assume the proof since after assuming f ' = 0 Lang just states, "hence this function is constant".

I am lost at what is being proved - can someone please help?

Theorem in Complex Analysis - What is Proved?

Thanks.

Yes, it does seem a very basic theorem, but my Question 3 about what is actually "proved" is my real worry. We are supposed to show that f ' = 0 implies f is constant. Where in the proof is this actually demonstrated - it seems to be just stated as true in Lang's proof?

Examining the statements in the proof, we find the following:

"

The function

t --> f(http://www.mathhelpforum.com/math-he...65c1de79-1.gif(t))

is differentiable, and by the chain rule, its derivative is

f ' (http://www.mathhelpforum.com/math-he...65c1de79-1.gif(t))http://www.mathhelpforum.com/math-he...65c1de79-1.gif'(t) = 0 .... (1)

Hence this function is constant,"

__seem to assume the proof since after assuming f ' = 0 Lang just states, "hence this function is constant".__

More on the Theorem on page 89-90 of Serge Lang "Complex Analysis"

I think it is true that he is "proving" what you say - namely

"Either that or he's proving this: U is a complex domain (as described), and f is a function on U. So, he proves that if f'(z) = 0, then f is constant on all of U."

*My worry is that his proof seems to consist only of an assertion in the sentence -*

"The function t --> f(http://www.mathhelpforum.com/math-he...65c1de79-1.gif(t)) is differentiable, and by the chain rule, its derivative is f ' (http://www.mathhelpforum.com/math-he...65c1de79-1.gif(t))http://www.mathhelpforum.com/math-he...65c1de79-1.gif'(t) = 0 .... (1)

__Hence this function is constant__," <--- ISN't THIS JUST AN ASSERTION?