Cauchy's integral formula
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Cauchy's integral formula

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7

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15

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CC Attribution  ShareAlike 3.0 Germany:
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2014

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English

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00:00
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03:49
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11:35
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15:52
Rotation
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18:04
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22:37
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31:49
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46:23
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1:03:10
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1:19:57
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00:06
OK I would like to start the hello I 7th class on convex analysis and the In fact we have the middle of the proof so let me um Rekall a few things uh from last class them so I hope it's not too bad this is readable OK nobody says no how OK great so also you should know everything on on the slide right so please come down OK which defines complex line into grows by just a formula which looks the same as in the real case however this multiplication . EIS stands for complex multiplication OK with uh also computed some examples in the 1 example which is really important in which you uh have to memorize is that 1 over the um integrated over a circle um gives of any radius r the gives 2 pi way as any other powwow uh any of the negatives pose or it's a positive power over and the same circle gives 0 look at this as an example and we just computed by plugging in the tools of art R times e to the IT into this formula now In order to use the real theory for a line integrals we computed that the complex line integral in real terms it turned out just by um just by figuring out complex multiplication gives and rewriting it in this form it turned out that have to integrate to real effectively use 1 for real part of interval it's you minus we we have a for the imaginary part it's the you so if a tool real vector fields you minus the and the EU which we could also interpret as if by the end the 90 degree rotation of as far solids i times as this tool real vector fields and uh we also showed that these 2 vector fields um a rotational on the rotational indicates that F is or more and uh to figure that out all you need to do is write out what the rotation is for X and Y and see that these this is the very core she Rhemann equations so uh this is very obvious and now let me do something for visualization purposes you can already see it here I'm interested in the um invade uh visualization of these fields X and Y for particular functions before I do that let me tell you something about the rotation which I realized only and once thought about these examples um
03:50
the don't care so they may do let me tell you something or on the board which is is the
03:59
following so now I want to tell you what rotation really is and I realize I should have done this a lot earlier but I must admit that only um when I looked at these visualization examples uh I realize what they have to do so here I just had some bad words about intuition so it's about understanding rotation or and uh what I realized is that all you need to do is look at the formula cell considered and consider the say i j playing In our and so the plane spanned by the EIA and the jvectors you like so there are insect there are n times n minus warm such claims in Rn the and or let's look at the definition of the rotation of the vector field X which is TI which for around as a T i x j minus the j x i and for the moment simply call this the i j component of the rotation vector which is actually what physicists to except for already avoid uh double indices enough 3 all came mn and now I want to look at what this means what does this formula here me uh and in order to do that I am only consider the i j plane who OK so in the i j plane I have a vector if you say X and I want to look at its components said this is the i direction and this is changed uh then I have components maybe I use colors uh I have components say x i and x j the and I consider these as spectral components for a moment here OK so now all M I want to study 1st order effects of and um in order to do that so study 1st order effects have restored effects at some point that say P. and you end up in order to uh so 1st order means I want to would discard for of a constant uh for over uh I want to subtract the constant vector field X at this very point so uh that means I consider I would like to consider the consider the subtraction x till the equals x minus x over this point which is a vector field there are no this as a vector field defined on some domain you I didn't write em all the details of out there for all of these tasks is I want to subtract the constant a constant part it's a peak so that I have x till the offices of p equal to 0 which makes the makes it easier to see yeah and also known so so x till the of P is 0 and the rotation is unchanged at any point actually the OK and now let me let me just draw a picture at to see what's going on air just what does this formula mean and will so it's easy to see for it's still there so let's look for uh for the case that save 1st term is positive the positive energy and I want to look at the i j playing once again size vector you that at the point p vanishes and which has positive the what does the j component dual while if this derivative is 0 and the value x 0 0 well it's it's positive here it's negative here and this would be the linearization uh in the eye direction then the linearization on can't either right she in on the direction and actual student EI direction made whatever it whatever it may be here may be different but I don't care and since I look at 1st order effects so if I I have a positive rotations say I would like to consider the effects a but this time is also all this term including the sign is also positive so this term on its own is negative so let's look at tj X or Y is negative what does that mean well now I have to look at the XI component if it's negative that means that some of it is decreasing so that it he it used to be positive if it's 0 at 0 and it becomes negative OK I hope you get the picture there perhaps you see in the very rotation this is something I should have done right of the beginning of introducing when I introduced this the but I didn't have this picture in mind now so in this case this means the rotation of X at this point is positive and in fact it's the sum so you look at the resulting effect but if you have a test body uh which may be say it the disk on the ball or just and say tool bars uh in so then once this is positive views and you really see but it rotates in the poor taste only in the i j plane for each search planes there is a rotation for instance in our you can have 2 perpendicular planes with 2 independent rotations this OK so once you've understood this it's a little harder once you ate the x of piggyback ride Ben if you add a constant back then you have to add to these uh rose uh a constant vector field and it's a little harder to see so I have some pictures but I'm and not tool and the illuminating having said this having this being said let me now go to agree um let me now go to view pictures of X and Y right around case some new sh to um the not OK
11:37
OK so the here we have very simple functions there's a race as a choice of functions implemented but let's just look at the 2 simplest choices which i z n to end and I think the and he said over more said the 1st 2 ones OK so here you see the vector fields uh and the vector fields or and In other words this you should be a large you know grow strategist go all that some some layer the error rate in this uh OK it's probably
12:22
the well can we get most of it is while OK it's the number of hits a screen resolution problem what we've case here whatever bad OK does get better but it's good enough I believe the OK so what do we see here well if the vector fields if a family is f of the if standard is when let's look what our X and well these are you n minus the for instance cT then is real part x and my the imaginary part with a minus its X minus Y and that's precisely what you see here the if you have if you have a vector the position vector like this when the vector field is just a reflection the reflection of the real line so 3 y components negative this is the blue piece of the blue arrows OK then I have a little test body here and I should point with the mouse I have a little test body here OK which you can move in and think of this as a fluid in fact it's an incompressible fluid and if you look at what the divergence of these 2 vectors of these 2 vector fields are uh it's again a cushy you remember thing that prizora divergent so this corresponds to a flow and to flow of an incompressible fluid uh I will not write out any details for this but for physicists it should be clear and also since there the department is good at fluid dynamics and the university to this is perhaps worth pointing out OK so now an let's look at this test for the so we can let it flow end OK you have to believe me that this is all correct but it is this this as a test body doesn't rotate since well this result here it's the vector field X for any holomorphic F OK this is you the simplest 1 is irrotational and by this interpretation really various no rotation of a test body so it will simply at translate along the small landslips go here translated along the flow lines like this OK now the vector field why is which is going to show next vector fields and why is a 90 degree rotation of perspective it so the Y vector field should the um the predicts that to what we see here and I'll hope this picture will convince you that's perpendicular to the blue spheres and once again if I as follow along flowline world periods of followers low line then the test body doesn't get rotated just by the reasoning here and you can also see the wasn't of again that's OK this is wrong side but if we go to is part OK OK the so how so far about this very simple function next Sloke ed a function
15:53
which is not homeomorphic right which is f of as of said it is or what it's just the models at z of minus offset so it's just the unit vector uh corresponding to this here's all we do is we divide by the length OK so we should be comparing the 2 vector fuse we should see the same directions go to complete picture that it's and it's only a part so the arrows should point in the same direction no all what's
16:29
different is that these are you unit vectors yeah since this is the unit vector field considered as not to vector here and now what we should see is um since this is this actually has rotation is is obviously a non holomorphic function so and in fact it has the rotation so if you moves our test body around then we see it's starting rotate it stars rotating yeah so this is from and the same thing happens for the 90 degree rotation which is in fact this is X and uh we 1st look at the 90 the rotation so again here um as you go around it curves and you do it in a way that there rotation yep at the mocking rotates meaning that this test body uh experiences rotation and think of it um the boy a minor boy you're in water this is what it us OK so I hope this gives you some feeling for what we're doing there are also many more functions implemented but it gets more complicated now if you look at those so you can do that for yourself I decided it's better not to uh to to many different cases but this is the intuition for what they're doing the what is but what what you all and Karen I
18:12
the OK so the let me now the come back to the theorem I started with last time he hates so obsessed it's a good idea to write out the sphere and fear head of the board last time has the following and we have a function for 1 of the main and it's differential bill so a the and we say the following is equivalent to as it is cruel and the following is a kernel and the furthest f is holomorphic when you on all of you M. segment all 2nd integrals over loops uh vanish in I manage in case uh that the loop for on honesty and for all come in case the loops are contractible I use of groups and you tractable loops see in you know OK 3rd statement the and we actually proved uh but these are equivalent and this is called uh this implication here is called the co she's integral the the river cautious integral theory man what can refer property is almost like this on the that we don't consider loops but we consider curves with different endpoints so all I want is an efficient way to write this integral of C 1 then F of at CdSe equals F you will cell given 2 different curves between them for the C 1 and C tool with the same uh to the initial and terminal point at the end but because must be homotopic then this is you can also quality so C 1 C 2 0 on the topic in you don't want to pick yeah can you energy by definition homotopic means as same and same endpoints and I no the and the 4th property a state there are potentials or in in complex data analysis language back our primitives of a function so f there's local primitives and primitives and meaning that uh for all points in you there exists a open boils with positive a it contained in you just this whole it's uh by effect we work with the main so n VECSYS primitive f from uh it's slide with and C 1 the also the r of C knowledge to see brown hair with then f prime equals on the the and so this is what I mean by local parameters once and for and later purposes let me underline some things which there Our not necessary the case I work in a simply connected domain so the can mango and
22:39
then this is a little more complicated I need I need a little more to say that OK the so last time when we prove that these 2 I equivalent and now all of them live and uh not only these 2 but these the 1st 3 are equivalent the emphasis this it's very simple just by splitting and open to 2 curves all by uh joining 2 curve so uh to form a loop the but what about the primitive year this is something I have to prove this you to prove namely uh let me do this now if I go from and I won't show it semblance but I will show for now on implications so if I have this property benefits well the essential idea is very easy here what do I need to do well just integrate out my vector field integrate up my effort to obtain my uh primitive if the capitalist right and 5 is property focus with the same endpoints homotopic curves uh this uh agrees and if I work with in the ball B R then all curves are satisfied the uh homotopy uh assumptions so if I work with curves in the then all curves with the same endpoints uh will give me back the same value of the integral so this is all I need to do but OK it's a proof so I have to give the details and saw how does it work and so Greece said a on OK we said interval as of C or I write zeta in order to have a C. for the left hand side full the see going from so the know what to say that within the board B R within the so what I'm doing here is I have large perhaps not simply connected domain and what I do is I use this ball so this is my domain you you and I use this this ball with respect to the knowledge and I will consider curves from seized tool the knowledge to see the also this not a pretty picture OK so now I want to use my real fury by want to use the real integrals OK so um since this integral here has the real and imaginary parts as this and these are um are real line integrals I can use real theory separately on the real part and on the imaginary part so that gives me back some some some potentials which I need a name for so I say in this um since this is well defined and it capture you the the next few the of DS where X is chosen as on this slide and this and I let me also use the notation it's implementations for this and it will have to go from said knowledge to z and it's welldefined whatever curve I choose to carry and why a sorry the equal to the integral similar integral over the imaginary vector field y did this 1 can so these are the real versions you an imaginary part of this integral and now I know that these are actually potentials him of potentials all X and Y X and Y respectively OK so what does that mean well it means that that Grant u equals and grant the gradient of the equal to Y on a on B R schools on the that was just by the real theory if I have a well defined but find uh line integral Pennec and uh and use it to define potentials and these potential satisfied that the gradient is the given vector field OK and now let's come back to the complex version who cares how do we do this we no capel areas to be the vector function you the capital you have to use its series of this goes from this goes from b r of C in order to do well on Tuesday in a manner that would be see of course what is the Jacobian then the the the Jacobian of this of this uh of this vector valued function is given by law take the 1st component differentiate with respect to x and y and the 2nd component differentiate with respect to x and y will look back at it uh this the gradient of you this is the gradient of the so this is by very by this I see the gradient uh of fuel here so this is X transpose X written as a line vectors and Y transpose I need more space in the little more narrow but this is lying just to indicate this yet just by the fact the the these 2 hold and what is X transpose and and why transposed well this is this is by the fact that we have uh x equals you might the i can simply write it out you mind as we and the 1st line and of the 2nd line i have to copy the you these are not minus signs that this was really good so this is the Jacobian OK look at the jacobian when good shape if you think of this map as going to see then it tells us that the Jacobian corresponds to complex multiplication so this is a holomorphic map emphasis will trick OK so as see value to so F equals you plus I really going from the ah of these terms see it is holomorphic uh and actually help here a set of so 1st I should I should say satisfies scholarship green yep since this these 2 elements of the same and so it's a Fig is end for holomorphic maps we know variety at some time ago so corporate on the slide so we must think man so that this is really poorly readable sorry for that 1 let's just thank you the change lines OK perhaps a bit
31:24
that little better now so far OK if I have a uh holomorphic map benefical she women equations ends the derivative is just given by lakes because I VAX all of the y minus I Q 1 OK let's use this in order to see that um my function capital f indeed is a primitive
31:52
is a primitive of is clay won't change who the OK so we have S 4 was fake investors holomorphic when we know that S prime can be expressed as you x you X plus the and uh by the fact that the OK by the fact that I know what you X and the X I look at the matrix and over trees are little you and 1 little thing services Q + IV just telling from the Jacobian and was also we know that for every this is again this is a this is a really um just a decomposition of fest in real and imaginary parts OK so altogether if I'm is that content so as it is a is primitive as clicked primitive gain P of of seen locally why did I we need to restrict to be our signal not too little ball well any simply connected to a subset of open subset of my domain uh would have done that all I wanted to have is that any 2 curves within this set of homotopic so um I think I actually in the version of the course notes which are provided to initially uh I did this therefore is for any simply connected subdomain which is also fight OK so we've seen that the uh weeks that the well that it if the a line integral uh happens to be the same for any 2 uh folktale from a topic for any Capello for model because then we have a primitive of a function now and how do we go back well this this little problem that here we work in the ball was uh while uh the other properties are globally stated however with it to the logical assumption so turns out it's a little easier to go from 4 to 1 to show Home ficity edit and save the midpoint the North and in a minute you will see why this is so so let's take any point in and then essential point here is that for each could have the c in B R well if uh we have a primitive here so that means that we can ride F S we can write this as a capitalist Prime OK how do we make use of it all in terms of line integrals we ride out uh f of C D is these which is by assumption s time but to and all this can be written as uh what is sailor line integrals so this is therefore of U of T times the prime of t detail OK and by the chain rule this sisters f that if after uh after CD differentiated the the good uh t t right left out the argument and solve this is by the rules on complex integrals this is just F of the terminal point minus F of the initial point that was done the same calculation of a real case once you have a potential it's very clear that uh the the line integral only depends on the end points of the curve chosen is only little so we will be finished in the air were allowed to work a globally however and this is in B R so how do we get around this little problem well I can say the following a look at the other statements here they are certainly true in BI and without any restrictions on the topology of a curve since B on the ball is current is is simply connected so any 2 curves in the ball element topic any any loop in the ball is contractible so I can simply say I use the proven parts of the theorem prover case that u equals B R and then what I get is so this shows that freely holes for u equals B and since this is equivalent statement is a column on the simply connected to a man to tool and 1 I can say the tools quotes for u equals so the special case that u equals this ball and actually 1 holds uh 1 holds for u equals the R him by so these arrows of by what we stand before you already proved I put already show yeah and here this exactly the same so that means I know all that edges homomorphic on beyond our of C not well in particular it's a holomorphic at them my much points you know and this is an arbitrary point so that it's automorphic throughout I In scene or arbitrary meaning that uh um yeah that's what I want to show if system will can it's in order to for complex differential well 1st thing be complex differential this complex differential at sea order and so on uh but think that the notice arbitrary it's still better like this that f is holomorphic the and we have done yeah so this part is only a little uneasy due to the fact that my
39:07
assumptions as I stated glued on the cursor stated globally and under topology assumption that I have to to go way around it the OK so far so good this proves the theorem and as as I actually used here in the pool of um difierent simplifies indicates that you interconnected and this is what I want to state next uh so stepsisters let's do it directly Corollary uh was the number of 21 who can I suppose you simply connected set post you simply connected it then then the theorem holds with fear hold while with 1 as about to without word contractible tree without word homotopic and fall without the word local so it is actually a primitive kept left on all of you the here without what words underlined in blue you are all we yeah so in particular the uh you don't need this assumption will boil at all you need to stateful floor instead S prime because s on the on all of you so I don't think I really need to write this out explicitly but as OK and this is just by the fact that any token in a simply connected to make any 2 curves with same endpoints are homotopic and any any loop is contractible end we can make we can actually apply this proof of 4 to all of you without any problem so false floor to make for to prove the coral reef off all we h run from this proof and all we need is what's written on this part of board here yeah so prove for all is only this part the yeah this here if you understand what's going on you understand what I've written on the book OK so as know it's a ball and we do 1 example perhaps before going to the break that um 1 example the which is worth uh looking at is the example of this slide let's look at a a bit a on the other slide infected it is let's look at 1 of his the integrated you OK Siam at this line here the integral of 1 of his these to pin 2 pi i vi you once around the region and also all other uh powers it 0 so integral uh 1 over Z T is then it on these curve C I R a equals 2 pi i uh that shows that uh shows meant c without 0 cannot be simply connected to something I couldn't do that cannot the simply connected the so that's effect I couldn't prove directly remember yeah it's it's pretty obvious that if you go to the entire plane or to disk and remove 1 point and to try to you have a loop and you could want to contractors who yeah it will is it it will not go beyond that at this point however in order to prove it you need a twodimensional version of the intermediate value theorem and this this topology and not so easy and as cell this a uh a dire I get a diet proof of this is is uh beyond the scope of what we've covered so far however the fact that this comes out non 0 so a rule shows the for loop within this domain the function as a war figure shows that this dual here see without the written cannot be simply connected here a words look at the very for any the corollary says for any loop is arm and a tool for any loop in and this see I must have the Indigo equal to 0 yet in case I'm simply connected so here I cannot be simply connected OK this is the 1 the other example now what about to be into the other into goes all the same curve so and not 0 they're not equal to minus 1 well this example shows that although all the assumption of a corollary is not true yeah if I look at you again I look at u equals C without the origin but this is a nice holomorphic tune so Mr. cell or the assumption is not true it's not a simply connected domain however the statements all true yeah it it's in fact true that any for for this case and not be equals is I'm not integrating 1 of then it's true that any interval of a loop that gives me back her retransmit value 0 any for any other tool homotopic curves not for any 2 curves no matter if they are more topical not I get the same value depending only on M points so uh of course this can happen you have the assumption of the theorem is not true that the conclusion is true made assumption of coronary I should say so here is essentially the ascension of Corollary In the faults but statement is true this can happen justice statement it's true however OK so the perhaps this is a good moment to stop the for a brick mn so let me do the to diplomacy now
46:30
now and that and now we come to something which sounds almost the same but it's different it's co she's integral formula co she's integral formula and on the holomorphic function has loop integral 0 of 4 uh uh contractible loops is co she's theorem the formula that is what we will do next OK we need in order to do that uh we need in that notation to simplify life so normalized of 1st 2 pieces of notation 1st for domain where it no you in this CDB a demand how and then we use the following um the in order to have a shorthand for in solid line integrals we want to ride in the interval over the boundary of the domain which could say is not a welldefined so what we do is we define that as something namely we define it as the integral over some curves the book was the digs in plants integral over OK of sign and number and and all of the desired here the situation before I write out all the details the situation i've I have in mind is the following uh it's a domain with many holes say so this is my say this is my domain you the OK now DU has several connected components here in this for my example here 3 of them so what I want to do is I want to take choose curves C 1 and C 2 and see free which parameterize the boundary of this domain in a good way and notice that the arrows point for instance in different directions OK so when I do this I want this subject to following subject to came 1st 1st thing is that uh the boundary OK anything more interesting than it has where isn't a OK DU is the that there OK more you it's not so hard to start and that's easy to start is of where is it somebody must be very busy right now OK the so 1st I will I want that the boundary of my domain you is just the disjoint union of traces of all of the curves involved please see yeah the starts meaning uh that I talk about destroying careers but this is not to end these curves also they don't that they don't go around to Austin so I want to add that these curves are injective where uh C. K. from say a k or in piecewise let me say this in piece continuous from a k k 2 were you more effective I a few uh Our injective Our injective all so since these are closed curves yeah these are loops I want them to be injective on this halfopen interval so what this really prevents is that I go around twice for of and many times you have this I want to disallow but OK and there was never um assumption I make is CDK um leaves uh 107 k is such bread you is on the left this is a common I left off ck and this means sorry long enough space was means so this it's now about the direction the orientation of these I want them such that if I take a while you and store never paycheck if I have a tendency Prime then the 92 you rotation of attention is being on normal to the domain the OK so this is the of the nite to the rotation pointing inwards this is what I want to write out so uh his election II i e that what is a 90 degree rotation of a tangent line in the complex case I can simply write it as i times the prime the of t it is the inner normal in a normal to the you and that means that if I go on the outside I go this way from the inside I need to go this way yet question is um they should go into you know um they should go into the you you you are what you correct that their enemies of the Iast so in this case you with not a domain which is uh this is not good what I'm doing um but but but but but but but but the I'm sorry and solid you the be a close and what I wanted connected OK so long the closest connected so that on but to sit at home OK thank you sulfurous OK OK as is the in a moment to d you I hope this this actually defines what I and I'm doing so example will be I scan right now I can write for instance things like DBI at the midpoint B of F with the d z where in fact I mean that I choose a parameterization C R all of the or the art the clothes uh B plus the 2 R times e to the I know I need a t OK the yeah of FCPs right I want I want to do this kind of shorthand notation here here and not that just by writing this isn't considering this is the set it's not clear which way I uh only in my boundary curves Nolte that I don't go a around to all segments of forth broken never notation I need on
54:48
his her that if I have sets and content and be in our analyses whatever you prefer men um I want to write when I say that they is compactly contained compactly contained indeed and a right so this a double inclusion as and a contain contained in the in the case that good indicates that a is compact In is compact and I a has compare closure i want i bars compact so has complete closure so I want to these sets I don't want them this little bit there will be open and all the applications but i don't require this right no and but aid is contained in the in a minute you will see an example of this notation namely the next lemma I the the slowly users movies of the back the uh and it should be OK so I 1 we stated that the integral of 1 over z over a circle centered at the origin finish is equal to 2 pi i I want the same for for any silicone not necessarily centered at the origin this will follow from the next lemma the so OK b r of C b r of C compactly contained in the B R of the so the situation that I'm in is the following i have a larger ball centered at the the radius r and I have any smaller I don't care if contains B little be or not any smaller ball centered at z of radius R came and now I consider a domain you and domain with OK actually what I wanted to contain is the the set in between these 2 circles so the annulus a solid say with b r i b without be part of Z is contained compactly in you so you may be a set like this where I can also harm perhaps a userdefined a different color I could also there has the whole here so this maybe this may be you for instance OK so I have a litter annulus have some lose contained in you and now I want I want to use I want to look at holomorphic functions on you which may be may have singularities in here OK then I went and for each of each horrible freak or more fixed function uh and let's call it since the 1 in the lemma this lemma will give a fearing were rich will apply to f so now we talk about GE and we say events er an integral over d b r of B of GE is the same we is just the same as the integral of a little t d are of the she of the time DCT OK so no change uh send Missy so if my function is holomorphic on this annulus I really don't care how large uh the circle I'm integrating over really is so goes 1 over the uh centered here at the origin say than any non non concentric circle will give the same interval the 2 pi i him there will be an example of no it's an interesting example in areas that used to be our office said uh contains and in the uh I have the integral of 1 over Satan distinct it's 2 p 2 pi over DBI the uh TB 5 of state cell and known concentric version of what we have the sampling had before OK and here's and approved is somewhat it's pretty in the sense that it mainly uses uh users of the concept of homotopic curves and the formula there's not much to calculate so what we do is the following if we look at the array from z to be In case be coincides with the With said there is no problem we can take any way home OK we take such array and then we have this annulus into 2 pieces in these 2 pieces of well into 1 piece I should say and this 1 piece is simply connected OK I made a picture in a minute of communities so I have side to of once again items so here's the and here it is be here's my the who cares and the chair 1 I want to use the part of array which is inside the annulus to define curves so what I want to do is the following and 1 to you look at some kind of gamma which way around is it going belief ghost of 1 that is that's like this 1 and the same could have Gamma going uh . B is no gone gamma and gamma minus going inside and outside exactly on the Bluray OK and I want to connect this with them by bounding circle so I want to go on like still the and then I followed gamma and I want to follow this little circle inside and then a followed them gamma minus and they get a closed number so constructed close to the the closed curve uh say called it's c equal to OK let me
1:03:14
perhaps I start here I go 1st round the last circle DB on all of so do you this is correct yes of being and then I follow so I followed gamma then I follow this little circle in the opposite direction B all of it is the and need the opposite direction minus the and then I move out plus gamma minus came so this is is the rope it's piecewise continues it please of French a book and that I claim it's contractible claims loops see is contractible contractible within you the you remember that you may have some uh that there may be gaps in you within this little so all I need to do is uh a show that this is a contract of a cold curve within you and the end what I want to do is H I want to slide is art gamma minus once round till it comes back to again so prove so let me let me just draw it entry into formulas it will be harder so this is the initial the initial version then where need arrows the then what I do is say so basically the open here now I want to cut something away they're like so and I can keep on going so like this caring and arrives at a situation that all the interior is gone the so the so I can I arrives at a gamma plus at the loop gamma minus plus gamma the emphasis the loop gamma plus gamma minus I start going with minor gamma minus and return with gamma and now I can certainly contracted this by making it shorter say to the initial point and so I can make this shorter until it finally collapses to the initial point which was used to be here OK self uh this is reproof for of the curve being contractible and if you want you can figure out of the formula so far but city I don't want to do that and solve this curve this loop is contractible within you so and there various other ways of doing this here but this is sort of I find a neat way of doing this now if it's contractible I can apply the cushy theorem so what I guarantee is that the loop and will oversee off my holomorphic function G vanishes the she and oversee vanishes and by quotients coaches intervals fair enough book the holomorphic function integrated over closely managers and now using the properties of line integration uh I do the following this is equal to the sum of 4 Kurds written out here so so the db are of being also she these the plants in the role of gamma over gamma of G is the groups and the I skipped of c plus and all of the little ball minus B r of C I have reverse orientation of GDC class integral finally all over gamma minus of cheese now this is the property of line integrations of the additive line integrals are additive on the sums of curves it's additive uh with respect to some sort sums of curves and is another property which is that if you have to integrate over curves and over the curve in reverse orientation then you get a minus sign so these 2 here Council of the OK why said yeah so remember into gamma is equal to interval gamma minus with the minus groups with a minus sign there was just by direct calculation OK so these to cancel out what you remain with of these 2 end result is 0 while this is the very claim yeah that these 2 uh 2 0 0 would only out it yeah OK up to sign that the sign is also great this miners so I can make a minus here and so I get a difference so I write this out I cross this out and a writer a minus here and then I actually get uh the integral as it's written here and I take this to the other side and typically so this is not a hard just playing around line integrals the what should the no OK in now them I can use his results to prove because she integral formula the if the the the yeah it the the we in so that's name 23 and its exactly co she's co she's integral formula so what does it say well it's about integrating a function which is not holomorphic on the domain so I start with automorphic functions let f from you to C World War the whole almost think back the but yeah what's going on there so the
1:11:47
OK springtime men so please stop and and I 1 a ball uh mn say the honor of the uh contained in you or I could also write I could also like this whatever you prefer dimension following formula in and F sorry F of so all of this will work for all points in with him this sport the I can express f of Z by an interval and the interval is 1 over 2 pi times what I want the line integral of f of Satan over Zadar minus C D zeta over the boundary of this ball the on all that's coaching formula so what does it do well the here's the ball B his his being this has radius R. I. then I take any point z in the open ball any point said to him and it expresses the value of f the value f of Z is a boundary integral office functions where I do not integrate s alone but I divide by this dangerous theta minus it's minus Z which has the pole uh and said but since there are so at this point but since I integrate over over the boundary alone there's no problem with the denominator the so 1 way to Sun I to decide explaining or proving the formula the sins and she I would like to prove it for but let me OK let me nevertheless do some some something uh and say regarded as a way to to if the way mean value of formula meanwhile you formula would I mean well if I plug in V a particular value for that will be B. the what I get F of can particular I get F of BI can apply this to uh efforts to uh z equals to the you would get well 1 over 2 pi integral over T B R of of graph of the data over C minus B OK he is not it's not a direct to see that but let's plug in let's plug in a parameterization go from 0 2 to play and plug in a circle so what is the circle well it's center that B has radius R and the to I t so do the same here then I have the plus the + the I T minus B all be concerts the OK and so what do I get well I get and I forgot to write out sorry the but now that I want to come to something I realize that and what I'm what uh what is forgotten the rate the derivatives of C the derivative of this curve is i times are to be i t OK this just plugging in a circle a circle centered at the of radios uh OK little are and capital are the should be OK I use little are here since OK it should be capitalized OK yeah but it OK this is the task of thank you for anything else here not but had no it's on the boundary we integrate over the boundary of a circle this is really what we into it about him so that's why complied in my clothes my of a B plus this is like a of the B + R to the EIT from 0 to 2 pi is centered at the has radius capital R and parameterize historical such that the domain is on the left OK and um OK now everything constants nicely b r to be E I T goes away and the i councils also also I get 1 over 2 pi uh 1 over 2 pi interval from 0 to 2 pi integral of s was be while f of the 1st part of the I T D T so what is this what the sum values of f scanned over the boundary of the circle and what's what did I do here so time time for this class to be finished but not yet OK um so this is this means here it's the average the the yeah mostly integrals often and it was I use for averages so I'm ever F over the over circle so the value at the midpoint of the circle is the average value of uh over the boundary so to in distance off so this which is why I call it a mean value formula now if you if you move your point so that was the case of A Z equal to B. the midpoint if you start moving z and to the boundary then this uh works like a weight and I don't notice I have something prepared OK I better do it next time uh to show you how this formula really works then so what is a good choice now would have done all of this uh um the best choice would actually be to show you what's going on uh 2 means the in I'm sure yeah here we go goods in slowly show it is very the then slowly uh but the location it the hi OK so I want to show you how this find out how this formula works indicates that uh and in OK In the case that I very my points within the
1:19:59
ball what happens this
1:20:02
functions in a it's not constant like here but it takes different values OK it's not quite what I want you to you OK and the so let me do the following the uh I show you 1st so
1:20:28
OK yeah good I show you 1st the weight function so I can sorry this slope of what I want to do I want to show you 1st the function of the denominator OK can you see that yeah OK so I want to understand this term and all I can do here in the graphics is I take the modulus so it's not the correct a complex picture I don't want to show you separately for real and complex parts instead I just take like here so the function you see there is just 1 over 1 over the distance 1 over and perhaps form our consider the case p equals to 0 so the function I'm considering here over is just 1 over zeta 1 over the MPs 1 over as data the distance from the boundary so this has a pole it's 0 and the pull is cut off the clearly and it's the radially symmetric function OK so far so good no and when I use a point and I can't do that we CAN BE because there was not a good choice sins now this now the center see I want to I want to move the center that means that the pole of this function goes to the boundary so let me move this point said to the boundary well important point here is the boundary values of a function since he here I only integrate oval boundaries so it's a red values of this function shown a modulus here the red values of a function which matter yeah and these are multiplied Eichel convoluted with and on the boundary so if I moves around the obvious thing here yeah ever go close to the boundary this function gets singular and ultimately would be delta function on the boundary OK now I have a given function minutes take this the way in and he's just some arbitrary choice of function f and it's only shown on the boundary local yet it's it should be defined all over the place but only the boundary values matter so only the boundary values are drawn so what does this formula do well is a weighted I consider this 1 over Z 2 minus the a z the weight on this function which I've integrated so let's do both things and let's function with my the weight function so this is a weird French perhaps I 1st used the midpoint version which perform number Reichert as you what if had weighted with a constant with a constant as function here this function is the same goes all of the boundary nothing changes so it's the same function but now if i take off an office and of erosion so I move my point around it means that at certain points I take a larger weight you know so this as this goes to delta function as the boundary it will it will actually increase the importance of certain boundary values of F so this case the boundary use here in front of the and our weighted heavily and the more I moved to the boundary the law of the type function which is the weighted function the purple function will go out and so it is actually not a surprise that I mean if you take the limit of this formula the z goes to a certain binary point then this formula should actually approach the value of a boundary point while since this goes to a delta function this should be the case and this is what this what the graphics here can show you yeah so all I can do now is move around and show how weighted function a weighted function works yeah yeah I don't knows that tells you something about this formula but I think of it as a way to as an average of F weighted by this factor which depends of the boundary distance and next time I would give you were approved and so we've arrived the end of the class thank you for listening the