MATH2621 Higher Complex Analysis (1 Viewer)

leehuan

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This isn't actually a question I got from the course but I'm not sure if it can be done via 4U techniques so I decided to leave it here. I know intuitively why it works but I'm curious as to the proper proof of it







Feel free to change this question into an R2 question if that makes it easier. If the proof happens to be long just a description will be plenty.

(I actually don't know if this question is correct or not. It was a deduction I made based off a few CSSA papers.)
 
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seanieg89

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By invariance of the properties in question under rotation and dilation about the origin it suffices to consider disks of radius r < 1 centred at z0=1.

Then for any fixed angle w in (-pi,pi], we can consider the ray z=t*cis(w). Finding the distance of this ray from z0 is optimising a single variable quadratic (or remembering a formula). This ray has nonempy intersection with the disk iff this distance does not exceed r. This gives us an interval of w (|sin(w)| =< r) for which the ray meets the disk, so we can then consider the endpoints of this interval.

Then for these optimal w, the quadratic equation in x coming from y=tan(w)x and (x-1)^2+y^2=r^2 is easily shown to have 0 discriminant. i.e. the rays with optimal argument meet the disks tangentially.
 

leehuan

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Which of the following does the empty set satisfy:
- Open
- Closed
- Bounded
- Compact
- Connected
- Simply connected
- (is a) Region
- (is a) Domain

Is the answer all of them?
 

seanieg89

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The last two terms vary in meaning between different authors etc. One of them could very well have a nonempty requirement in your course ("region" sometimes has this in particular, "domains" are a little more standard and usually just refer to open connected subsets of C in this context.)
 

leehuan

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Oh I see.

For our course, "domains" are defined the same way you are, and this was how they defined "regions":

A set S is a region if it is an open set together with none, some, or all of its boundary points.
 

leehuan

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I can't assume anything about complex differentiability yet
 
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leehuan

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I remember doing a taylor series on that one, where e^z = 1 +z + z^2 /2 + z^3/3! + ...
At the time I wanted to avoid it because of the fact we hadn't done differentiation but turns out it was the next lecture lol. So I ended up just L'H smashing it.

Which is probably me cheating, because this is the definition of the derivative, but I'm in C so I'll cheat by using the Cauchy-Riemann equations instead
 

seanieg89

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At the time I wanted to avoid it because of the fact we hadn't done differentiation but turns out it was the next lecture lol. So I ended up just L'H smashing it.

Which is probably me cheating, because this is the definition of the derivative, but I'm in C so I'll cheat by using the Cauchy-Riemann equations instead
How does that make it any less cheating?

You should use the definition of the exponential explicitly (presumably it is the power series centred at 0, so the argument is much the same). Don't assume anything about the differentiability of the exponential if the limit you are asked to evaluate (if it exists) is the definition of the derivative.

Same with L'Hopital. Eg you could try to find the limit of sin(x)/x as x tends to zero by using L'Hopital and it "works" in that cos(x)/1 clearly does tend to 1. However how did we prove that sin(x) has derivative cos(x)? It turns out we need to essentially evaluate the limit of sin(x)/x in order to do this!
 

leehuan

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How does that make it any less cheating?

You should use the definition of the exponential explicitly (presumably it is the power series centred at 0, so the argument is much the same). Don't assume anything about the differentiability of the exponential if the limit you are asked to evaluate (if it exists) is the definition of the derivative.

Same with L'Hopital. Eg you could try to find the limit of sin(x)/x as x tends to zero by using L'Hopital and it "works" in that cos(x)/1 clearly does tend to 1. However how did we prove that sin(x) has derivative cos(x)? It turns out we need to essentially evaluate the limit of sin(x)/x in order to do this!
That being said,

When you first told me that the power series was the definition of the exponential I initially accepted it, but became reluctant to believe. Why is that the case? Or do they just teach things in the wrong order?

Because the power series falls out of differentiation as well.

I get your point, but I really don't know how I feel about the power series definition. I would've thought that the limit definition was more acceptable. (Hopefully I got that right because I didn't double check)
 
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seanieg89

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That being said,

When you first told me that the power series was the definition of the exponential I initially accepted it, but became reluctant to believe. Why is that the case? Or do they just teach things in the wrong order?

Because the power series falls out of differentiation as well.

I get your point, but I really don't know how I feel about the power series definition. I would've thought that the limit definition was more acceptable. (Hopefully I got that right because I didn't double check)
Because working with power series is pretty low level analysis, you don't need much theory development before you can start saying meaningful things about functions defined via power series. (Eg it is smooth and invariant under differentiation, which almost uniquely characterises the exponential.)

Same with trig functions. From power series definitions we can recover all of the familiar properties of these guys. This is more for efficient structuring of theory rather than mirroring historical progression (logs came first!) or being the best way to learn what a trig/exp function is (high school does a decent job here).

It is easy to show that your expression and the power series defn are equivalent, so it is a matter of choice. Again, power series are usually very pleasant to work with if we are doing calculus and thus preferable.
 

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