Integral q Watch

latentcorpse
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#1
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see q 5.11 in attachment.

in the solutions, why are we going in a clockwise direction on the inner boundary?
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Oh I Really Don't Care
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(Original post by latentcorpse)
???
latex - there is some kind of encoding for that attachment
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SimonM
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Open office doesn't like it. Could you post it up on TSR please?
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latentcorpse
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ok. suppose that f(z) is holomorphic on the annulus U:=\{\frac{1}{4}<|z|<4\}. Prove that if \frac{1}{2}<|a|<2, then

f(a)=\frac{1}{2 \pi i} \int_{|z|=2} \frac{f(z)}{a} dz - \int_{|z|=\frac{1}{2}} \frac{f(z)}{a} dz

the answer says

If R is the closed annulus \frac{1}{2} \leq |z| \leq 2 then R is entirely contained in U. Apply Cauchy's Integral Formula for a \in R to get the result and remember that the inner boundary gets a clockwise orientation (which explains the negative in the 2nd term)

my question is why is the inner boundary orientated clockwise???
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DFranklin
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Did you mean f(a)=\frac{1}{2 \pi i} \int_{|z|=2} \frac{f(z)}{z-a} dz - \frac{1}{2 \pi i} \int_{|z|=\frac{1}{2}} \frac{f(z)}{z-a} dz?

my question is why is the inner boundary orientated clockwise???
This is impossible to explain without a diagram.

Have a look at Figure 1.19 (page 33) of http://media.wiley.com/product_data/...3527406379.pdf. The idea is to use one closed contour to go around both your original contours, by using 'contour walls' (the bits A and B in the diagram) to connect the two contours. Because A, B are the same path, only in opposite directions, they cancel out. But if you trace the entire contour, you'll see you end up going clockwise around the inner contour.

Incidentally, I find it very hard to believe this hasn't been covered in your lecture notes.
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latentcorpse
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it has. i found it just after i posted this lol - although the diagram in those notes is much clearer than the one i copied down in lectures.

a simple question that i am unsure about.

Is a satisfactory definition of holomorphic:

"If U is an open subset of \mathbb{C} and f<img src="images/smilies/u.gif" border="0" alt="" title=":U" class="inlineimg" /> \rightarrow \mathbb{C} be a class C^1 function. f is holomorphic iff it's complex differentiable on U iff it satisfies the Cauchy Riemann equations on U."

The reason I ask is that my notes seem to be making use of the fact that holomorphic functions are continuous and this isn't mentioned in the definition we were given

cheers
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DFranklin
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I don't see that's a satisfactory definition. (Simply put, if you have more than one set of "iff" in your statement, it probably isn't a definition).

I'd go for something like:

"If U is an open subset of \mathbb{C}, then f<img src="images/smilies/u.gif" border="0" alt="" title=":U" class="inlineimg" /> \to \mathbb{C} is holomorphic on U iff f is complex differentiable on U.

Note that complex differentiable trivially implies continuity.
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