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Complex line integrals

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OK I would like to start yeah was you at so the before I start I was told that the value of last class has been uploaded so please uh please check it and make it available no OK we are um last time there was a big Iran and a tree was that the computations so let me uh 0 what I have to do what I want to do today is um we mention a few consequences and when we start with something you namely was too complex genuine complex case of complex line integrals but 1st let me and expand on what we did last time so as you know all if X is conservative meaning it has a potential admits a potential then only on early on we had but this is equivalent to the curse of the line integral uh of vector feud x over them all loops see vanishes for all loops see in you'll and you'll was assumed to be a domain to be connected OK it was the class before last class and then then we also have that x is In rotation of so if a is a potential Vannevar rotation vanishes there was lots of the theorem and and moreover last class will we showed last class is that X is a rotational if and only if f all contractible loops on the line integral the line integral vanishes fall contractible loops and all this at all new see in you OK and the main method for doing this so this was last time theory 15 and as you observe has just 1 word difference on the right hand sides so in the case uh that the domain is a simply connected then these are equivalent to but in general they are aren't and I give you an example of the example that was given by is given by the vector field uh corresponding to UM what it corresponding 2 would have partial a this uh it's an entity you rotation and the potential would be the angle function of the angle function is not global however the uh gradient of the angle function is a vector field which is defined on all also I 2 except for the origin OK so indeed there is a difference between it can be can be a difference not each irrotational vector field is conservative enter main um ingredients for proving this was a formula which gave that's a general vectors used uh over a contractible loops see can be given by integrating this condition here 0 1 a B or vise versa and also the um Jacobian of a vector field minus the transpose of the jacobian applied to DHT t D. H ds ds dt what comes 1st DTDs where are the where h where h is a homotopy of the loop see so this is the new and a homotopy however or to the edge which contracts that you have given the loop see we want to contracted by this uh homotopy age so as far as I'm concerned this is a beautiful condition namely the irrotational condition that this year vanishes uh must be integrated over filling in well you take a look see you fill it in with the homotopy you integrate you condition here and right away you see that this is the outcome here himself and a very nice really very concrete reasoning from my point of view so let me draw all consequences snow in the coronary Our then so now so again perhaps I should write out the conditions we have this C 1 vector fields uh from you to there you you in and is a domain that is it's open and connected and then I have 2 consequences namely 1st is that dates if X is the rotational the rotational then and the way has what this is actually equivalent to them to the effect that um the potential but the potential exists locally not globally but locally so for instance I could say so X has a local has local put use purely local potentials and what local potentials there are various ways of phrasing this for instance I could say for all x in new exists a ball the are of X and you which is non empty so our the positive such methods an there exists a potential f on the ball the then meaning that the gradient of MSE insects the but this is only defined on the ball yet so we have a demand 0 point texts and we use uh the use of potential on a locally and perhaps be interesting case of this is that the the pretentious not global so in a case like this think again of the vector field of the example where 3 years global pretend that the potential would be the angle functional and angle function uh locally and angle function is no problem globally on the entire and it doesn't exist or case of this is 1 consequence and we have a mentioned before is simply that uh if you it is simply connected if you is
simply connected and it connected then OK there's no difference between a contractor blue and any loop and so in that case we have met and irrotational future is conservative here have selected out and language then an irrotational irritation of fields is uh conservative the so there exists so given given very a condition that the rotation vanishes they have the potential OK there's not much to prove this is really a corollary wiser so think about 2 so to their well as a study before any contractible loops there any loop is a contractible loops in a simply connected you so that means that uh that the atoms take in up there and so by the previous theorem uh we can we have that indeed there is a potential OK so tool this 4 just note that any any new is contractible yeah and so the claim follows by fear of 13 the main claim follows and for 1 to prove 1 knowledge that there a
ball which is entirely contained in u is contractible so you can see shrink any loop and to the midpoint of center of the ball and so uh
line integration the and then the problem you face is the past dependence the and the condition so so this is just general nonsense here himself um the problem problem may be past dependence the end if there is a condition which tells you that there be a there is no path dependence that the interval of all parts of Greece such as the irrotational condition was simply connected domains then this is called a integrability conditions in condition condition which guarantees guarantees for um past independence Prof independence of line integrals it is called and integrability condition the what but the and so we've actually discussed and integrability condition and for the problem of uh what flaws simple differential equation gradient of f equals x where x is given then f uh a seat and this problem arises in many different uh in many different contexts self which unfortunate can really tell you good example now but if you may face later on in his studies and so this is why all the entire sections any uh the intervals a a discussion of the integrability conditions world and so then I can come to the
next section of the class which is on the integral formulas into the formulas in the complex case follow morphic functions it will I so home for you OK in but the cell to simple simple enough so we wanted to all programmers now apply the general theory of real theory 2 weeks case this is what we want to do and um this something we have to do 1st which is to discuss the line in in the complex case services the 1st section complex the line integrals complexed line integrals there OK let me start I had a little earlier with integration about over and complex functions so recall from analysis the uh that if S is a Riemann integrable functions a continuous function to make everything stays but it has values in the complex numbers real members then we can still said uh the integral of f of t t the put if we only know the real integral and 1 to define the complex case and this is actually what we did in the class mainly separated from the real and the complex plot have class integral a b and as my missing the Portland I sorry plus on times the integral of the imaginary part the but OK so this was some of these functions the for real-valued functions defined really uh um Riemann integrability 1st and then for the complex case you simply have this uh setting to take care of the real and imaginary parts separately and of the main result the main result on the intervals is of course a fundamental theorem so the in couple of say it is a a C 1 function in the same interval uh but it's differential in is of primitive entered the river to some funds on all sensors of S is balance then so primitive means let's write it out as f prime equals and the then we have the fundamental theorem then integral from a to B of the derivatives of f of t completes a derivative is still given by the primitives evaluated at the into the interval boundaries and this is the fundamental theorem should be capitalized fundamental theorem theorem groups theorem of calculus had the so why is of fundamental have you realize that the new 1st year a tells you for instance how to evaluate integrals and try to see figure out what a primitive is and then it is by trial and error you found it useful formula it also tells you how to manipulate integrals by using the rules for differentiation OK so and there are many examples some of them I covered uh thanks to the Internet of this course so To make the longest frauds let's take a French like the to the i t and integrated from I don't know where what do we want it to be say you know what I need to do well the go out figure out the primitive what's a primitive well is 1 over the to be i t since about differentiated according to the chain rule that I had the I in front so this Council so this is in fact this is a minus sign uh and this has to be evaluated OK and this just a reminder on integrals and perhaps the most important functions to integrate besides the exponential is the power function so if you have f of z equals z to be in then heard you can in the primitive is uh 1 over N plus 1 uh set to be plus 1 and this works for any norm uh and on he could not equal to minus 1 cell I consider we God minus 1 and it works for uh that non-zero if
n is negative for uh in OK and the and in fact this is a simple I mean this is a simple difference quotient uh calculation to see that this is in fact the chromatin and um the OK and so by linearity and so forth you could you know what the a primitive of polynomials are and can integrate you can also uh integrate power series is scarcely a suitable rules and and and so forth so this is really the fundamental thing to know uh in order to into lots of functions OK so now let me come to remain definition here of this uh um subsection which is the complex line integral so what do we need a of c is a piece wise differentible curse from say a agreed to the complex plane audits perhaps more general and you can see came we given the curse and want to integrate a function this and to the road for the integral to exist really let just assume that f is continuous the complex valued OK then then the complex line into the complex lined integral I complexity coffin it is while it's written pairs F of f of Z deserved overseas and it's given by the by the interval of F may you plug in the curve C from and complex multiply with the crime of t t t yeah some of the main thing here is that this is a complex complex multiplication the no OK we simply define this kind of the line integral and in a way uh but all my my point of view is OK reason it's obvious that this is a unique definition but on the other hand we don't quite know what we've done OK if in fact uh I should uh erase this and to be careful I should say and and for seed in please see 1 take sums near the integral if the disease is equal to the sum of all of these and then many TI to t i plus 1 and I that I copied this the integral there the function is um is uh not differentible exactly at this very points and my convention will be then I will not take care of i will always work with piecewise continuous curve curves however I will always give the sons so a convention convention it's that whenever I use curves of qualities the piecewise to differential the and I skipped a sounds uh Rees suppress the this sounds and for simplicity the OK at the thank i then the slide those the any serious stuff from this uh in order to link it with a real integrals let me make a few remarks on properties of complex line integrals we all the good properties reviews so followed or and so if you the the the the the the time that I of care properties the of complex when integrals those in the so all we leverage the uh how explicit I want to be 2 of the equals Lavender indigo and from that's the uh deceased missing the the the if linearity we have path independence snow integral over see after a re parameterization phi of every the user
it is the same as the integral over C itself however must be careful for all give this holds if there is a fine who is uh a map is a diffeomorphism of the the respective intervals which are not explicit here and phi must be orientation preserving falsify orientation preserving diffeomorphisms of which is just a formal way to say that the derivative of phi is positive log so this is the parameterization independence p a and moreover we have the the facts Y K if I integrate OK if I take a AEA non orientation-preserving diffeomorphism so solemnly diffeomorphism with phi the website Prime negative so the standard example would be just take minus the sum every inverse curve then it it gives rise to sign change minus and steep using here for this is but this is for c in C is the inverse inverse curve of C meaning C goes false and C minus uh goes back of king In this comes in and I forgot to say waivers this holds well why is the same as in the real case well uh this is uh if you plug this in men you have to integrate f after c uh f of sealants like this f of C. after 5 8 times the of the fight crime In all use the chain will tell you that the derivative of C after 5 is the prime it's i times 5 prime and this holds the same in the complex setting as in real settings is uh and and and then performing follows by substitution so that means that this holds and also by the fact so here we use that this is the prime of times 5 prime and also in the in the in the 1st case I get to hear a sign change and sign change is responsible for the minus sign here quote came and what else do I have it they are something I would meet later on effect you can take some sort of curves so I have a curve C 1 and a curve C to say and I can take the sums so the initial point of C 2 must be endpoint of C 1 uh then this is also an additive in the line integral will some warm and the D. C plus and because the truth please look so and this is all what I would say this is not surprising the only the only thing is make sure that you understand that the year that the a chain and substitution work as well as for overcomplete complex case why is this so well since uh derivative rules are the same for complex differentiation OK and there's 1 more thing had to collect here which we need later uh the length of a curve to remind you is the length of C is given by integrated the velocity vector the scalar of the velocity so the In itself this is the and this is the uh uh definition we know from constant and the complex case in particular this means that this is the real part of c prime squared last plus the imaginary part of c prime squared square root uh t t OK and uh there is uh unimportant estimate for the for it for line integrals which we will need Austin when shares estimates I don't care if I have ask continuous and therefore in 2 or more generally integrable as and I have a curve C as usual piece wise uh piecewise differential then uh I can estimate the integral of the line integral of yes but they the maximum of of uh on CD times the length of the maximum of seen in traces see of uh if of officers OK this is also not so surprising right the with the line integral over like you his place see the line integral over my uh function and is given by the length of C times the maximum value of modulus of f I am way to the various and let me give
proof so this is uh a the integral of f of C. Ecclesiae is less or equal or is equal to me let's plug in the definition from a to B a F of of T times c prime of t uh due to and no I use uh a seemingly in uh innocent inequality
but I will say a word about this in a minute so this is less than them as a triangle inequality F of c of T times the prime of t the key so we can estimate this by the integral of the model marginals often have functional or in fact I can introduce uh and and models of the product is a model of the product of a model I so I can actually estimate this by the end of this year has a maximum over uh all t in a b namely this thing here so I can the remaining part is interval of c prime of t which is uh so it's light damage and trees sees and hears here is the integral of c prime over d t people and this is the length and solve this gives the estimated and let me say a word about this inequality here I used um which is from my point of view we only uh groups only and this was the only interesting er whole and here and we a the 1 of the the from OK so so here's OK we proved this inequality in uh where's quality here the inequality that models of an integral is less than integral of a modulus and analysis cause but let me remind you on how this goes since that's the only the only and property which is um someone non-obvious
caring uh still lower slightly and the the contains so here's use my In this I copied out of their loads of the analysis class analysis 1 OK what you see there um OK this is the the inequality we want to prove for complex-valued at for real valued if there's no problem right why is this so well if uh as uh at the number is always less than the number or its negative is always less than that it's monuments and so by monotonicity of the of the interval I can write this gives as plus minus interval is less than that of modernist and the left hand side here in 1 of 1 of the 2 left in size is equal to the modulus of into the left OK but in the complex case them the idea would be to use to decompose S into real and complex and imaginary part but there was would be the standard that idea out uh and receive it you will not arrive at this and uh this estimate but you will have an extra 2 in front coming from affected real part of s plus imaginary part of the modulus can only be uh estimate by 2 times models of this so there's a nice trick here which I want to remind you of uh on this you know it has died from who remembers remembers restricts said air so how does this go well the the hit the look of a formula live displayed line 1st we um 1 to estimate the integral of F dx think of as complex values I write an E to the I T in front of and the idea is to use an the value for T which makes this entire uh real so effective value to choose but he is the argument of the integral actually minus the argument of the integral yet is written down here nutrient yes so if you choose this value for T then and it will rotate this complex number integral of FTX back into the real line In minus the argument means minus the angle of minus the angle so much indication was minus angle will make it real will place number realize this is then the real number infect once you do this yeah but this works in general also solicited 1st look at the general calculation if you if you use this OK this is the same by the definition of integration and uh integrating the real part of the effect of times and ends and know the real part of the number a is always less than its models also so this applies in particular to the to the number e to the i t and the models of the to the i TF is infected f itself this is written here in fact yes of the real part of the to be ITS is as usual is less than its modulus and the modulus of a number with models 1 times as well you can forget about the united right he it's an essay right so we arrive at a and if you choose T as the argument then this is a real number so can I can forget the real part here and so that makes makes infect them this uh to be integral itself the 2 to the modulus of the integral itself right since unchanged so the modulus here is uh estimated by into of the models and this is 1 of the very few things in their analysis 1 where sort of the complex case is not immediate from taking the real and of a complex part and since this is because of complex and I was there for that 4 point your attention to this vector now it's good moment for the vacant afterwards we continue with vector fields so I would like to continue and the so we have define complex line integrals and now here is the fundamental example of the line integral the son of solid line integrals cell Chanda our mental of and since it's fundamental I would call I call it a theorem the the although hold that doesn't look more so great on the 1st side OK so we take just circle of radius R I so the simply take the circle of radius R O and Collis C. T. maps to the to the onto the EIT and we want to integrate the most so our of course our positive then on how I want to integrate this to the end on and my curve C I and depends on what in this uh for n equals minus 1 it's to type so this is really vicarious interval 1 over Z dessert the of more OK on 1st sight it appears just as a number well this number is distinguished since otherwise for all other values well of end and uh minus 1 for all of the various it's 0 so this is an somewhat surprising him why is it exactly that for if I integrate 1 over Z 2 I get a non-zero number but 1 of his it's grant gets 0 1 over said something that's where we get 0 z squared itself gets 0 and so for only 1 over it gets the value in which is non-zero what the proof um this simple and straightforward but doesn't explain you why this is the case so far for the case n equals minus 1 but we do falling so what do we have to integrate 1 over Z uh is a for short and so what do we need to do just like in the definition take S at the circle and uh multiply with the derivatives so this means I if I need to integrate uh 1 the words my curves are to be E I T or to really hard to and see what the derivative of C are with respect to t is while I get the I in front of that's everything by the channel so this is tied to the R E I T T T and now it's not that hard to calculate the integral the art of be ities consul so have to integrate just a number on a from 0 2 2 pi i and the number I integrated from 0 to pi is 2 time would to clear yeah so this is see this is the CEO of the
CIR and this is the archive the of now has all context so just by definition nothing uh nothing particularly end for n money could not equal to minus 1 I do the same thing the I with that but I write it in the form said to the end and from over OK what do I need to do well plugin Michael onto the i and integrated on its domain which is from 0 to 2 pi so what is this this is ought to be an it told me I n t it by exponentiating and now I need the derivative of this the primal farmer uh as before this is our times on a times e to the t and this is integrated with respect to dt so what do I get well as are and where I anyway I constants so there's this artery implants 1 times I terms integral of uh what's remaining he to v i a i n t e to the IT so I to the n plus 1 have I n plus 1 TD TE be I'm from 0 to 2 pi and what is reduced well uh this is uh given well OK half half of your physicists or why don't you you can save calculation I don't know if you're interested this but it's clear what we do here right we have the we have uh what this in the wilderness we we are on the unit circle and you go around altogether and has 1 times in so we do differences for n equals 1 we do this right so this is very center of gravity of going and times larger units so that it will be is it it has to vanish quite clearly yet but we can also uh calculators and so what is well it's uh I need where n plus 1 in front of n plus 1 I need the a uh what is it on here and this is the priority and and 1 and the so the primitive is current III to VI N plus one T era is it's to high since if I differentiate this the denominator uh the Council would be I n plus 1 interest I copied whatever visitors if you plug in there and 2 pi here and just by the fact the 3rd and in n plus 1 is integer there is always a plus 1 right itself is 1 minus 1 so it's 0 OK so by calculation we convinced ourselves that the claim here is true and much of what follows in complex analysis give area about understanding and the understanding the example here and in fact what we do here if you think a little about it and what we do we have over 1 over Z is as the well what is the yeah what what should the primitive people of the logarithm while the logarithm and hence uh the real part is a is of the logarithm of says the prime but the imaginary part the imaginary part uh is not a defined entire complex plane it has exactly the behavior of the angle function as a 2 pi get and this is the reason for the 2 pi here we will see this more clearly later but it it's exactly the example of the river grape every year potential uh is the angle function which is appropriate here for this example the OK so after the after event and and forms the around to the other bowlers let me know come only to the interpretation of the of the complex line integral in real terms where the location so all use use the next subsection and nature it's about co she's had that tentacles Tehran a more more you know OK in order to use in order to use the information we derived in the last section about irritation of fuse having a potential on simply connected domains and let me let me try to write down right out of the complex line integral in real terms so this is this is the test shown for the that's perhaps less and uh complex line integral Brian integer where real integrals that's a task at home and the following lemma achieves this by and what in that so but given the curve C all right uh really tool you won't see if and a function f which we want to ride out what areas you plus on the decomposing it into real part and complex parts so this is a function which just continues in our head defined on you and uh has range of c so in this representation just as z equals x plus I Y without saying it explicitly I mean with you and be of real value so I V are these imaginary part of why don't we write it here there we apply sorry real parts and his imagination this hockey the so and then if we have the following formula telling us that bound the complex line integral in real terms so I want to write this book this is a complex number let me ride it and perform real something plus the manager is something with a something being integrals and line integrals line integrals over vector fields namely the effective real vector fears in R 2 equal C given by you minus these DS and the other 1 the Maginot for the imaginary part this is the yield of times Ts OK so this is by the standard then to identification like this is a vector field in the plane the value the value of the integral is a real number the real number plus i times a real number is a complex number now this is what I mean but now I have a value like this represents a complex line integral in terms of real line integrals and the proof of this is a calculation straightforward
calculation just use the definition of the line integral integral F of said said is equal to 0 is equal to 0 uh say intervals 5 plug in my curves at sea I have f of CEA uh well perhaps I don't want to use the argument T so I'm I'm just writing f after c times c prime the dt sigh skip the Ostsee off here to make it a little shorter and OK so this is what came if I want to write it in real form the real part of ESA's you imaginary part is the wall so let's do it any to be solved now I write a complex number as a real to vector you after seeing the after and this is multiplied it still complex multiplication of all I write a right in a real vector here the times OK what's what's this vector written in real terms as new part times imaginary part and I skipped the off he throughout so this is still a complex multiplication and this and all I mean here is i I use my the identification of C with R 2 he is still a complex modification of not anything real at the moment so what is this complex multiplication well if you multiply 2 numbers the real part is the Pugmire merit we're part times real parts minus ImageNet part imaginary part and imaginary part is given by the cross-terms right so yet so remember that this is still a complex multiplication it yeah so I can write this in the form of solidity years to write it out but it's obvious what I do have to do this times this the real part of C try and my nose this times this times the imaginary part of C prime and now I have the imaginary part which is given by the cross terms so all I have taps are right you integral the the yeah so this is you you times imagine at time the last z times the real part and it should be D teams he also sigh on OK so whatever done it too complex numbers a b times you v as a product a U plus and minus T V and so forth practices what used here OK now I want to collect my forms I want to OK this is known as to all real numbers which is a real number plus I times a real number no I want to write this in the form of a real line go so all the way to do this is that let's use a scalar product who currently to and let's make it is scalar product and the while we want to we want to get to this form so this is the scalar product of this vector fields at the curve C times the prime so why don't we the prime here and see if you can arrange for this form well if you write the prime then it's real part of the prime imaginary part of supply this is this right in front is you after seeing and minus you have to see as a vector you after C minus the after this and now i do the same thing for the 2nd term again I want to write it s a real as a real line integral what if the prime so this and that hovered terms for c prime so I have to start interpreting as this as the scalar product this is the top component and this is the bottom all the or OK I you get what I'm doing here it's in a way it's just that it's just computation and rearranging OK the so this interim smothered right so this is exactly what we desire this is the vector field you all you minus the integrated over C fact of their you ruled minus the the s integrated overseas and similarly so here the this is the you what integrated overseas and this is my very claim here OK found this is the proof so that means I have represented around as presented a minor complex line integral in real terms and it's somewhat of a and ideally um ideas stuff we have to integrate yet but it's not too bad it's the while this is F and the real part of f this is minus imagine a part of 5th so I can use this as as power now as the complex conjugate of s and this cycle could uh interpreted in terms of i its conjugate I will write this down in a minute OK so so this formula know good makes it possible to deal with complex intervals by using the real theory and just the next thing I wanted to lemma grow up to be who is right care deviate a little from what's the course notes contain but it's the same content it the OK so let f be a C 1 function complex function and no uh I returned to of the complex uh theory I've covered time I want to relate to micro of intervals my line integrals uh was the home of the city of s and the claim areas and very beautiful so red and then by F equals you VII is home or the course if and only if the this on the payoffs these 2 vector fields here our irrotational it if it on if the school vector fields but the and let me write amount in
various forms of Daisy I switch I switch false and back between complex notation and identifying this with vector fields here this is real part this is minus imaginary part so 1 vector field so I call it makes this is the 1st uh relevant vector field and in if I 1 to write it in terms of so I could write it and so real part of f minus imaginary part of and this is nothing about the complex conjugate of F using all these identifications and hope this uh you're not bothered by all the I didn't identifications use that is a vector is equal to complex number we interpret the complex number as a number not to the and why status fields which is the you and and the you OK is imaginary part of the answer real part of the essay and if I want a one-line complex way of writing it out I would say it's I pierced by here if I take this and complex multiply with on the what do I do well I take the and I take the real part of you and write it out in the uh imaginary part so you goes here and I take the imaginary part had take the so the real part goes to the imaginary but and if I multiplied imaginary part by a by I then I end up with minus the real part so uh the imaginary part minus the multiplied by i gives a a gives a minus so double minus gives the inferior part OK so this is indeed if you there's no it there's no need to write it out like this but uh it's a neat form OK if these 2 vector fields are irrotational if the 2 vectors is are is a rotational neural uh OK I got criticized for leaving obvious and comes ordered almost automatically so not spelled correctly prologise for the 1st half of this class OK so this is an interesting is this is really interesting home the of of the function f means that if I make you from F vector by writing really n minus imaginary part is this or as this then uh I have no rotation in and as I have no rotation all the fury we've whoops sorry we on all the theory we um covered uh in the previous section applies so we know something about the integral of loops and so forth yeah so we we know that the line integrals depend only on the real depend on the on initial and terminal point and all that OK let me give improved the which is quite simple so so how can I relate and what how can I related and being irrotational would follow mosaicity well both all conditions on derivatives right irrotational means uh that while derivatives of you in the eye equal certain derivatives and uh homotheticity means that the course the Raman equations that also relate derivatives of the real and imaginary part and so in fact the student editions of a very sick so the proof is there is little more freak Cleveland tools is equivalent to f satisfies the satisfies the course you remind equations course remember the creations and said and this is a covalent tool well let's just compute but the rotation of n x and pretation of why it is pretation effects is take the 1st partial of x 2 minus the 2nd partial of X 1 and what is this wall this is really the 1st partial of minus the a so in complex mentation physicists here so I look at here him that in the 2nd component is minus the minus the differentiated in the 1st direction so in the x direction minus the differentiated with respect to the x direction minus and now I need x 1 which is you differentiated with respect to Y and I get so I get the eggs so this is the x plus uh you why well by the Shiromani creations Baeza equal and similarly so for to retain its rotation of Y which is the again 1 8 Y so I wanted to minus D 2 and y 1 so let's check right so uh now I have to look at and why which is the como use so this is you will differentiated with respect to x and I need uh why 1 is the differentiated with respect to the segment why and so this is the 1st course remind equations u x equals y a wire the why and the 2nd 1 is this so by the caution remain occasions these finish so that's it that's
straight forward so just look and and so my point of view is just stare at the complex line integral integral F of the d z and the see what you really calculate with this line integral and you see that and for holomorphic s full romorphic yeah you integrate irritation of the of OK that makes a series of a very simple OK now let me see still time left so but put so let me now drawn the consequences of real theory which I can do everything fear and 20 so it's just a little reorganized after now had in comparison to the course notes that's the same thing so we have for a function f in C 1 you see is a equivalent the the following is a coolant should read uh for the the equivalent so now I collect various properties which come actually out of this uh observation that polymer ficity yes which I write out 1st is equivalent to rotational so irrotational fields uh s will more on you use domain I should uh should write this somewhere I think this is my general assumption that uses domain so connected and all use wall so 2nd for the each contractible will for mean there differential same defined on the interval uh going into you a rule is contractible loop 1st the contractible contract to build blue um the we have before each contractible but which should be for each for each contractible loops uh we have so that the integral over the loop of s renditions for no OK and this is 1 to 2 is usually what is called co she's very now we have another version of this which is so here we had loops now we can also formulate the same thing with 2 curves which are homotopic to 1 another um solve this statement would be led to see knowledge the day piecewise differential curves uh sale chemically ride initial and terminal point out uh lived in a kind of from I see no of a people's pleased to see a lot of B equals Q then so the then um and then the line integral over C man in the the line integral the integral over C of as the of CdSe um the actually OK paths a right out as I have demand modes so let's make this the end point let's call it so z and let's use a different variable zeta is a common choice is a term uh so this is the line integral over another curve C but I wanted to have the same initial and terminal point the end for each kind of see in which has the same importance scientists same points and is homotopic you to see if not and it is you what what topic that to see within you and increase With this land and to go ahead and agrees with the line integral of c the right now that iteratively right then the line integral reaches the Chair agrees with this the integral over signal the so if I have the same initial and terminal point and has homotopic cos I get the same value and this is interesting since it tells me what a potential is the the potential in this case is a primitive so that's why we have the next statement uh as local primitives the Mount potentials but in this case primitives and 1 way to rival South orders on the board you for each uh as they had in you exist b r of C A set contained in you Our positive as usual a such the the and of the
and exists uh and primitives for Goldman and as the and for exists say this is a primitive the primitives a dreams capital S and C 1 all of you at the our be out of the to see off it office the OK so this is a long list of occurrences but since we've done a similar things for real it's not so surprising right so the real real version of this theorem avoidance stated in this form is the and here is a vector field which is a rotational then the the V integral over this vector field over loops vanishes for each contractible loops men was what we proved last time then OK this is equivalent to the line integral from 1 point to some other point to depend only on the endpoints in the case of uh the homotopy curves and last uh the that if I have such a vector field then there is a potential so this last thing at the and actually I stayed there is a local potential to the right result as local primitives here a restricted to be off right so this is if it's the rotational has local it has a has a local and potential but the all of all what stated there is uh stated in complex form so now we can perhaps proved um the I 1 of 2 implications that not everything it was the new of she was located so 1 and tools the 1 in the columns of 1 and tool is basically uh what we did Internet will be um that comes from the fact stated in the lemma and the last lemon uh together with real theory so s is careful Mosaik the then by the last lemma lemma 19 and we have x 1 s in the line of lemma x and y are irrotational you should know and this at a rotational then we proved last time that this was and 15 that the real line integrals the integral x over C uh 2 years and into a y overseas 2 years then mention so both marriage but for um contractible loops for or or contractible predictable loops the c and so that means that's equivalent to taking this plus i times that uh managing the so integral careful of Z d z equals integral xt if plus I times 2 years C equals 0 complex number vanishes if and only if its real and imaginary part vanishes for contractible loops see so the this is really this is only since the move proved what uh what we uh how FIL amorphous at can be expressed in real terms namely for the spectra-physics and why there were a rotational and then it's obvious from what was done before OK tool and free although I think I've I'm a little sloppy here but but before I showed you that um it's equivalent for of 4 curves and loops so the fact that the curve integral of the line integral is independent a only depends on the end points is equivalent to the fact that been to go over the OPS measures and or you need to do is take the take the true curves and define a loop like that from it and then you see that uh this equivalent of the loop integral and the um and to to uh um line integrals vanish and in fact and of this relation here it's equivalent that the uh the loops are
contractible and the curves are homotopic although I didn't Clyde stated the citizens here so this is follows form FIL and 14 I would say yeah and noting that really V noting that under review and of this relation which I explained before uh a contractible loop goes to honor to homotopic past and vise versa yeah noting the it mid term homotopic loops correspond to uh the contract of uh homotopic curves correspond to contractible loops this way for book loops you know if you can so in this you come from and this OK and the next is there a little longer saw perhaps have to split proof and continue at this very point uh next time colleges of this comes in middle of but it'll is more there are more or less directly a consequence of of all real fury what we do know and that's 1 thing to say I forgot to say this is and that wanted to each implications 1 to 2 the score is called co she's integral theory you OK so I thank you very much for attention and will see 1 of and we can
Subtraktion
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Derivation <Algebra>
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