Math372 Fall2017 Lec16 ConformalMaps

[stijndhondt]

At the start of this class, the question is asked if the heuristic calculation done on the blackboard is enough to prove that g is holomorphic. I would think the answer is "no".

In order to do the calculation, the chain rule is needed. This assumes that g is holomorphic at z. So all the calculation tells us is: if g is holomorphic at z then f' cannot be zero at z.

[stijndhondt]

I don't know if you still teach this course but I have a nicer, geometric proof that holomorphic, injective functions cannot have a zero derivative anywhere.

It uses the concept of the winding number: basically take the formula to count the difference between the number of zeros and poles of a meromorphic function (the argument principle, page 90 of Stein & Shakarchi). You can prove this counts the number of times the image of the curve over which you integrate goes around the origin. So take a closed curve gamma. Then the integral 1/2pi*i * intagral_gamma f'(z)/f(z) dz = Z(f), with Z(f) the number of zeros of f inside gamma. This is also equal to the number of times f(gamma) goes around the origin. This is also easily seen in the following way. Move from the domain of f to the range of f. How many times does the curve f(gamma) go around the origin? This is then integral_f(gamma) 1/w dw = integral_gamma f'(z)/f(z)dz by the chain rule. To move from counting the number of time f(gamma) goes around the origin to the number of times it goes around a point w_0=f(z_0) we count the number of zeros of f-w_0 so it is integral_gamma f'(z)/(f(z)-w_0)dz since the derivative of w_0 is zero.

Now, to prove the proposition, assume f holomorphic and injective but there is a z_0 where f'(z_0)=0. Since f' is holomorphic too, its zero's are isolated (f is analytic & not locally zero), there is a disk D centered at z_0 such that f' is only zero at z_0 and non-zero everywhere else in D. Okay, call the boundary of D the circle C and call f(z_0)=w_0. Consider now the curve f(C) around w_0. Since f'(z_0)=0, in the power series expansion of f centered at z_0 in D you can easily see that for all z in D: f(z) = (z-z_0)^k*G(z) with G(z_0) not zero and G holomorphic in D, for some k >= 2. This means that the multiplicity of the zero z_0 of f has multiplicity k (at least 2). So f(C) goes at least twice (k times) around w_0. Now you can pick a point w close enough to w_0 such that it is still inside f(C). Then f(C) goes around w also k times. Therefore there are k points (possibly counts with multiplicity) inside the disk D for which f(z)=w.

These k (at least two) points are distinct since we took D small enough so that f' is only zero at z_0. So from the power series around any of the k w's you can see it has multiplicity one, since f'(w) is not zero!

This contradicts the injectivity of f, which proves the result!

This is a wordy proof, but if the groundwork of the argument principle in terms of winding numbers and counting the number of times the image of a curve goes around f(z_0)=w_0 is done, and it is established that if f maps z to w a number k times, f(C) goes around w exactly k times, this proof is very easy when you draw a picture with it!