 Okay, good morning, yesterday we concluded the lesson with this important relation. If z is x plus i y, e z is e to the power x time cos y plus i sin y, okay? Which is also e i y, okay? This is Euler relation which is very important one. This tells us that the complex exponential is in fact periodic, something new compared to real exponential. In fact, if you take e to the z plus 2 pi i, you obtain that this is e x plus i y plus 2 pi and this is the same as e to the power z. This is easily seen directly using this relationship since cos and sinus are both functions which are periodic of period 2 pi. So that in some sense what happens here, this is the, imagine this is the gas plane and take this sub domain is repeated in strips, horizontal strips, okay? Furthermore, as I noticed last time, the exponential of z is never zero and actually this becomes evident when you consider the modulus of e to the power z which is e to the power x, all right? So if you use the polar coordinates, this is the modulus and this is, so e to the power x is raw, the modulus and y is the argument of them in the exponent, in the polar representation. Then from a plus b equal to e i times e b which was proved yesterday, we also have that we can obtain conditions for the sin and cosine sum and difference of angles, right, of complex numbers. So you actually are asked to prove the following, okay? Which clearly resemble the real case but they have to be somehow verified. I suggest you to use, to heavily use this relation without going into the definition of cos and sin. But also the relation with the cos, the complex cos and the complex sin with the exponential. So that we have found, we have already established this cos z is one half e i z plus e minus i z, correct? So use this and you conclude this. So this is another exercise for you, okay? Of course when you have sin and cosine it's natural to define by analogy with a real case the tangent can use this or tan of z to be the ratio of sin and cosine of z, good? And also for the tangent function, complex tangent function, there are similar expressions for the sum and for the difference of c w or a b whatever. So without wasting our time in these calculations, let us focus our attention now to possible vertebility of this value. So what we know from the real k is that the exponential is an inverted, the real exponential is a monotone increasing function or monotone decreasing function, this exponential we are considering. So the extension of the real case is monotone increasing because e is greater than 1, right? In any case it is invertible and the inverse function is called logarithm. So now if we try to do the same we want to solve an equation like this in the complex case. So given w, find z with the property that e to the power z is w. Of course w cannot be chosen arbitrarily because w is zero but we already know that there is no z with such a property. However if w is not zero we can use the expression we have found so far and maybe w can be also written in polar coordinate like this but theta is the arc of w which makes our research much easier because here we have on the left hand side that this expression is e x times goes to y plus i sin y so by comparison we obtain the following that modulus of w is e x where e x is the standard real exponential because x is a real number and here the exponential residue on the real numbers coincides with the real exponential. So we can solve this problem until we are considering w different from zero. So x is the logarithm, real logarithm I put R here of modulus of w and it is completing meaningful because w is not zero. The problem now is related to the determination of the arc or vice versa determination of w because this is not uniquely this is not uniquely determined because as we already observed this depends on the this is this depends on a periodicity of 2 pi. So if we want we can say well okay take w to be arc sorry take y to be the arc of w but this implies that the inverse function of this is what we are looking for so the possibility of defining an inverse function for the complex exponential depends on the choice of the determination of the angle okay so in general this is not an equation because this depends on the determination of arc correct. So there are infinitely many possibilities to define logarithm complex logarithm so in any case you solve this equation this different this difficulty is related to the definition of to the uniqueness of the representation in polar coordinates depends on infinitely many choices of the arc up to a constant times 2 pi. Similarly we have infinitely many functions it's a multivalent function in general unless you restrict your choice and in this case there is a determination of a branch of the logarithm okay. So the complex logarithm one of possibility is to find as follows so w is logarithm as I determine the arc so as for the arc there is no unique determination so for logarithm complex logarithm. However as I said if this is not a multivalent function sorry if this if a choice of the interval where the arc the range of the arc is assumed to be so we can also prove the following fact take the logarithm of a times b what is this this is should be and by to verify this also comes from the relationship found yesterday eab a plus b is equal to ea times eb okay. And finally oh yes sure let me observe observed that from cos z one half e i c plus e minus c we also have possibility to invert cosine complex cosine call this number w and look for z such such okay as before w is given you look for a possible z such that this is true cos z is equal to w if this is the case we normally define this function to be arc course okay like arc complex arc course like in the real case alright what do we have here notice that here we have the opposite of e i c sorry of the opposite of i z so that this is the inverse of e i c therefore on the left hand side so in this expression here we have a quadratic relation of e i c because we have we can also write this in this form e i c squared plus one over e i c correct and this is w so this is quadratic e i c so to have z we have to find e i c and then to look for the the complex logarithm in some sense right okay so from here we should obtain that e i c is w plus or minus square root of w square minus one I invite you to make this calculation these are very easy calculation because well you multiply here to e i c okay and you solve in e i c substitute e i c and well this is the solution which depends of course plus or minus well depends well of course there are two solutions in the complex there are always two solutions okay but look this number here so at this way w plus square root of w square plus minus one sorry and w minus turn out to be one the reciprocal of the other do you see this well multiply this number times this okay yeah w square minus this square okay so w square cancelled minus one okay so these two numbers satisfy this relation yes sure thank you yeah yeah okay so from this we have e i z as minus one and hence as we are looking for z these what i z is the logarithm complex logarithm of this number so times i you have if you multiply by time by by i have minus z is i log okay then minus one then I put minus i log then since this number this pair of numbers are in fact reciprocal the logarithm of the reciprocal is minus the the the logarithm of one over w one over k is minus logarithm of k right so at the very end of the story this becomes minus minus logarithm of the w plus and this defines the function arc cos of w to define arc sine of w we don't have to repeat the same procedure using the other relation with the exponential but remember that this is by over two minor arc cos of w here previous one okay here we have a minus in front and here we have oh sure sure thank you yeah sure i is here sure sorry yes what we do have and it is quite surprising that of course there's no counterpart in the real case is the following all say elementary transcendental function so these functions are the ones you start studying when you are in a course of calculus right and complex analysis are referred or to the inverse of x okay logarithm so the exponential is a very important function like in the in the real case even more I would say okay because in the real case the trigonometric functions are not related to the exponential in any in any in any way okay good and this is one fact so the next task is to continue exploring what is the plan of this part of the course in some sense we have the following say diagram we have defined complex differentiability or sorry or holomorphicity and we have seen that this is guaranteed for power series functions which are convergent of course right so if you define a function in terms of power series expansion convergent you know where it is defined then it is automatically complexity differentiable conflict differentiable or holomorphic and this is equivalent to cr equation satisfied for the real and imaginary part so for u and v now I want to use another another equivalent definition of conflict differentiability and then that then characterize geometric property associated to this characterization okay so this guarantees that the set of completeness rangeable function is not empty now that we have examples and here we have another equivalent condition to guarantee complex differentiability now I give you another one and then study the property and as I said this long somehow long list of equivalency will close up in a in a circle because we will prove that then any complex differential function is in fact complex analytic so it has power series expansion okay so to introduce something new okay let us come back to notation like this and remember that x is one half z plus z bar z bar being the conjugate of and similar y is 1 over 2 i z minus z bar z bar is x minus i y correct okay so yesterday we started by considering a function complex value function and in some sense focusing our attention to the real imaginary part of this function so the components and using the real derivation with respect to x and to y of u and v real and imaginary part so instead of f of z with the some abuse of notation but of course I consider this quite natural consider f depending on x and y and this is u of x y plus i v of x y and of course is also u of z plus i v of z with the same abuse of notation okay but now we have also an expression of x and y in terms of z and z bar what if we take that is to say x depends on z and z bar the derivative of f with respect to z bar of course we are only applying the chain rule right but what is the expression of this new operator it's differential operator right d over d z bar something new okay we apply the observation is correct the remark is correct we apply the chain rule how do we apply the chain rule well we use this we use that well this is the f over dx times dx over d z bar correct plus the f over dy dy over d z bar and we have the expression of this and this oh we can calculate actually this is one half right because from this we calculated the derivative with respect to z bar one half and here we have pardon me i over 2 right in other words we have one half the f x plus i dy the f over dy let me copy it in a new page so this is what we have obtained one half the f dx plus i the f dy if you want to use formalism of operators complex differential operators we can say that d over d z bar is in fact one half dx plus i dy right so you plug in the function f and you see what happens okay the sum of two real operators i okay plus i the second operator well this operator is known as Cauchy Riemann operator or von Neumann operator in some books why is it important well assume that f as a solution that is assume that f over d z bar is zero assume that f is a solution of the problem d z bar d so sometimes it's also indicated by d bar this annotation d bar f equal to zero what does imply why we come back to our standard notation and we observe that the following we have here one half u x times plus y plus i v x plus i v sorry u u y minus v y correct this is the d over d z bar f and since it is zero this has to be equal to this and this is to be equal to this so u x as v y and v and the u y is minus v x so Cauchy Riemann equations and vice versa if Cauchy Riemann equation are satisfied then obviously repeat the inverse path and we obtain that d f over d z bar is equal to zero so in some sense you can characterize or homophicity in terms of Cauchy Riemann equation or solution of the Cauchy Riemann operator okay now this important fact I will summarize it in a proposition f is holomorphic of course when I'm not very precise and I omit to say at z naught okay if I say holomorphic and I don't say where I mean either at one point which means in a neighborhood actually of this point so in that at that point if you restrict the Cauchy Riemann equations or you take the limit of the incremental ratio and so on and so forth everything is defined locally okay so f is holomorphic and well at z naught if and only if f satisfies and here I mean what is this I take I take I have taken the derivative with respect to z bar and found an expression for this complex differential operator which is related to Cauchy Riemann equation so I invite you to prove this is an exercise and just the exercise that formally this is the conjugate of the d bar operator so this is also indicated by df and the exercise is not this but here okay one half df over dx minus i df over dy that's why it's the use of the contrary so d bar is the other one because it reminds that you are differentiating with respect to z bar and here we are differentiating with respect to z okay but well you see that the imaginary part changes its sign so this is the exercise and another exercise is the following assume f is holomorphic because of course this can be defined for any function in any complex value function it has meaning okay so assume that f is holomorphic which means that d bar f is zero so if this is the case df of this is f primacy and this is another exercise for you so please I invite you to use the notation of f in terms of real imaginary part and use Cauchy Riemann equation this is one easy way to solve the exercise so first find this and then substitute f with u plus IV right expression using this operator use Cauchy Riemann equations and conclude that in fact this is ux plus IVx it is the derivative okay so the two operators d and d bar are very important complex analysis because they provide you well as solutions in one case d bar f equal to zero a solution which are the class of holomorphic complex differentiable functions and the other operator gives you the derivative of complex differential functions if you are restricting your operator to the solutions another interesting exercise is to calculate this you apply first the derivative with respect to z bar and then the derivative with respect to z or you compose these two operators so I think that you have enough exercise not for tomorrow not for Friday but for next Friday say no no not for the day after tomorrow but for Friday the first Friday of October okay so I will however collect the exercises and send you the list of exercises for okay just but in case you have time you want to think about and just if you well it's a good idea to repeat this stuff by making some calculation this is a very simple calculation I invite you to to well and by the way this is found in any textbook of the subject do you have just a vague idea of the answer to this question well it it is not sure to think of a second order of operator so up to a constant in fact this is the plush okay well enjoy your yourself with the calculation good now with this basic fact recalled let me continue by giving you some other interpretation of complex differential ability one is the following so again we consider this complex value function f to be to have components u and v okay and if we're if we use the our glasses in the in the real sense okay this is a function from an open set of r2 into r2 so typically what you do if you know some regularity of the components you look for the expression of the differential right and the differential is calculated like this the first component the first component is the derivative of first component with respect to x then the first component with y the second component so first component first first variable first component second variable so gives you exactly the position and then do by two matrix all right of course this then has to be evaluated for z equal to z naught or x naught plus i y naught the pair x naught y okay but in general we have this but assume that f is holomorphic all right at z naught so your colleague is suggesting me that well this is simply the modules of the derivative at z naught the module square right you said this what is okay what is the assumption we are taking f is holomorphic at z naught so that we can replace using the koshiriman equation instead of u x for instance you put y and u y minus v x so for instance i substitute u y with minus v x and v x here instead of v y i put u x okay and these two matrices are the same because we have this condition when you take the determinant the determinant is u x square plus v x squared so okay the determinant of the matrix u x minus v x v x u u x is u x squared plus v x square which is precise the modulus squared of f prime at z naught of course this is in sequence and this number turns out to be of course no negative it is zero if and only if the derivative of f is zero otherwise it is strictly positive and this has an interpretation and real analysis do you know the interpretation of the fact that the determinant no but i'm only talking about the the the the fact that the determinant is positive let's say not not not negative to be correct so when well i'll assume that the function the holomorphic function we are considering has derivative with never vanishes do you have a do you have an example do you have an example of holomorphic function with the derivative which never vanishes we do have we do have the exponential the exponential right well good so or restricts to your function to the set where the derivative is never vanishing in this case when consider as a function from our art from an open set over two into our two this function is orientation preserving because the derivative of the differential is positive sorry derivative the determinant sorry the determinant of the yes yes conformal maximum mappings are in fact somehow related to this but all this is not sufficient it's just orientation from serving conformal mark we mean nothing comes out from from something we are we are going to see in in a minute so conformality is a property of preserving angles here we are just talking about preserving orientation not angles okay not the amplitude of the angles okay so this is another geometry if you want this is the first geometric consequence of an analytic property of complex differential functions so complex differential functions turn out to be orientation preserving or well the rotation not determined because the derivative is zero right now I want to investigate in fact conform conformality and I'll do it making this very very simple considerations now let us make a kind of picture of the situation we have an open set u and the plane where f is defined and assumed that f is holomorphic so take z naught you take a curve passing through the knot so gamma is a curve for the sake of simplicity well okay take minus i a okay real number at the time zero well we can also take the open interval minus a a in r at the time zero the curve passes through z naught okay so and then I consider another curve call it w of t to be f of gamma of t this is well defined because gamma takes its values in u and f is defining u correct in order to prove conformality we have to talk about tangent vectors to curves tangent vectors or say vectors passing through a point applied to a point so the natural thing to think of is well I take this gamma prime of t naught and since I'm I want to have a vector here I note the zero vector I may assume that gamma prime of zero is not zero okay so I take a curve passing at time zero through the knot with velocity velocity different from zero okay this is the physical interpretation I'm assuming that derivative is different from zero in zero now with this assumption we have that this is another path is a curve of path okay w is a function from minus a a into c and w of zero is f of z naught what about the velocity of this curve at the time zero well we use a chain rule because well we can first I'm saying this is c one it has some derivative otherwise we are talking about nuts we assume that this is not zero because we want to have the tangent vector but we have to have we have to require that the function has a derivative first okay so it has a derivative and the derivative is not zero so if f is holomorphic then automatically we know that it has also it is complex differential also it has a derivative so w is complex differential and the derivative at zero is f prime of z naught times gamma prime zero right remember that gamma of zero is z naught so in fact this is gamma of zero and you are we are using the chain rule precisely the chain rule okay therefore we have gamma prime of zero and double prime zero being f prime of z naught once again we are interested in consideration about angles of vector so we are not interested in vectors which can be that zero vector so we already assumed that gamma prime of zero is different from zero but this does not guarantee that w prime of zero is different from zero because this is not zero but this can't be zero so we have to assume that the derivative is not zero for the function f at z naught all right so if if we have both conditions so function holomorphic at z naught the derivative at z naught is different from zero then we can say w w prime of zero is different from zero and to see conformality we have to measure the angle what is the tool we have to measure an angle the argument the argument of a complex number gives you the angle with respect to the real axis now what is the arc of w prime of zero well I cannot calculate I don't have any information but I can say that this is the arc of f prime at z naught times gamma prime of zero and remember that one of the property of the arc is that the arc of the product is the sum of the arc so this is the arc of f prime of z naught plus the arc of gamma prime of zero which is a very interesting interpretation so the difference of the two angles depends only on the value of the arc the two angles of the two vectors we have vector tangent to gamma to sorry tangent to z naught to f of z naught corresponding to the the velocity vector of the curve gamma and of the curve w the angles between them is the difference of the two angles depends only on this which is independent of the choice of gamma you say this well arc of w prime at zero minus arc of gamma prime at zero is arc of f prime of z naught arc of f prime of z naught is independent of the choice of gamma which means that if you take another curve passing with the same speed at the same time through z naught and then you consider the image using the function f holomorphic at z naught whose derivative z naught is not zero nothing changes about the arc of the the curve the image curve okay in other words we have the following so arc this difference is independent gamma now some consequences take two curves passing through z naught and having two vectors gamma prime of zero and alpha prime of zero okay the corresponding image curves both pass through f of z naught and if f prime of z naught is different from zero we have two vectors here one is as I said w prime of zero and here maybe I use another another letter is better prime of zero while you consider the differences and the differences between the arcs is the same which means that the function f is conform so they preserve angles not so only the amplitude of the angles not the they're not isometrics they are just preserving angles and conformity is a very important property in geometry and in the past it was extremely important for for sailors huh you know these famous cartography results well just let let us spend two minutes to say something it's not possible to represent the sphere or a portion of a sphere on a on a portion of a plane because of curvature reasons okay this is a result known and well it was um empirically it was known quite far away for quite a quite a long ago but it was proved definitely proved by Gauss if you want to keep an isometry between two surfaces well then curvature is preserved it's not possible well the only possibility is to use a sphere with the same if the if earth is considered as a sphere you cannot even represent isometrically a sphere with another sphere with a different radius so you have to use the same radius so it's not very useful okay as a cartography tool a sphere which is as large as the entire but you can have good conformal representation so representation of two surfaces preserving angles between points between so between vectors and this is the case for many many maps used by sailors the famous representation is the following so this is assume that this is earth this is the equator okay the idea was the following take a cylinder wrapping the equator and take a projection from the center of earth and at any point of of earth and project it on the cylinder so of course this is not an isometric representation but this angle preserving so it is conformal see any any point on the same with the same longitude is on the line and as you can see here the presentation is very good but when you move to the poles north or south poles actually become distorted terribly but for sailors it was not important because they couldn't even sail near the pole so they were interested in in a band around the equator this is the the so-called projection of Mercatore Kauffman it's very important and even today for for sailing it is important to have the angles instead of using any other indication angles with respect to some reference line can be the meridians or sorry meridians or the equator or latitudes right so knowing these angles you can make a you can you can say okay and using of course some GPS system as a system okay good so it turns out that conformal mappings are sorry that holomorphic mappings are conformal good now what about the the embers of this so let us consider the up the this this problem to characterize conformal mapping mappings uh complex value to conformal mappings okay i take f from u again into c and i assume some regularity assume f as components which are c1 i cannot make any consideration in terms of derivatives without this this assumption and then i notice that gamma of t can be also represented in this way and the plane or with some abuse of notation with a pair of real numbers x of t y of t which represent respectively the first and the second order in the plane so we are assuming that well well this function here is such that gamma of zero is x zero plus i zero is z naught as before and i also denote this to be x naught plus i y naught okay um and and moreover i'm taking f to have components so u and v who's derivative with respect to x and y are continuous now i take w of t which is f of gamma t so i'm restricting my function along the curve gamma and looking for the curve coming out from this composition when i have to consider w prime at t this is and this is as i said a function of t in this way right once again i use the chain rule you already observed this and i have that this is df over dx times at t at sorry at the point x y which i probably better omit to write times x prime plus the derivative of f with respect to y times one prime of t correct this is real analysis real calculus now i have x t y t and then i define sorry i use gamma prime of t is x prime of t plus i y prime of t and this is in fact what parameterize our variable in u it is like z of t right and gamma prime bar is x prime of t minus i y prime of t correct now keep in mind that we have this we have this w prime of t is i copy it the f over dx x prime of t plus tf over dy y prime of t and that gamma prime of t is sorry i have to copy it x prime of t plus i y prime of t and it's conjugate sorry gamma prime bar of t gamma prime bar of t is the conjugate then i make this one manipulation on this expression all right it's one half df over dx over t and then i put a one half in here and i use well minus i one half the i sorry i df over dy y prime of t plus one half df dx x prime of t and then i have minus i sorry plus i i'm sorry minus i one half i yes sure df dy y prime of t so instead of putting minus and minus i write minus and i squared and i use i a split i squared into i and i it's a trick effect but this becomes okay one half df over dx then i put plus i first minus i df over dy and then i have a gamma prime of t plus one half df dx plus i df dy gamma prime t bar do you see this okay one half and here and i df of dy is here but then i have x prime i y prime with a minus in front minus is here so i y prime and x prime and minus prime from here and the other part the part comes so i invite you if you if you're not convinced by me that prove this verify actually it just a matter of calculation so substitute this and this to this expression improve it okay last effort is a following so as i said this expression represents w prime of t which is one half df dx minus i df dy gamma prime of t plus one half d half dx plus i df dy gamma prime of t bar now you notice that i after these manipulations the operators the differential operator koshiriman operator is as appeared okay here and it's conjugate so i have d bar and d operator here i'm not assuming that the function is holomorphic i'm just making the consideration in general so i have some regularity for the the derivatives and this is the expression of the derivative of the image curve of the tangent of the image curve sorry in terms of the derivative of f of the function with respect to x to y and with and with respect to gamma prime and gamma prime bar what is this this is the difference over the two arcs of course because remember that now because because of this condition right as i said gamma prime of t is not zero so everything here is well defined on the other hand since w prime t as this expression w prime t over gamma prime t turns out to be one half df dx minus i df dy plus one half df dx plus i df dy of the portion gamma prime bar t of gamma prime of t and here comes the observation if angles are preserved by f so this arc the arc of the the portion has to be constant this difference to be constant okay but on the right hand side here i have something which is not constant and if t varies okay this vector here is modules one so it is a vector plus something and the vector of modules one so it is a circle so the arc varies with p but actually well this is not possible unless this number here so the radius of the circle is zero so the only possibility to have the arc preserved so sorry the angle preserved so the difference of the arc the arcs to be constant with respect to t as this number is zero are you with me so this expression here is one thing which does not depend on t plus something times a vector which has modules one remember that gamma prime is not zero and gamma prime bar is not zero as well but the the ratio gamma prime bar over gamma prime is back is a vector of modules one is a unitary vector so we have something plus something else and times a vector of modules one as t varies the only possibility to have this independent of t is to kill this part and if we kill this part then necessarily f okay i write it here so this is so with this in mind i write it on a new on a new slide which is the last slide for today also because i don't have any sheets of paper so in order to have the arc of its constant we have require one half the f over dx plus i the f over dy is zero because the difference of the two arcs is the arc of the quotient and the argument in the quotient as this expression so the arc varies okay unless this is zero so that we conclude that the function turns out to be conformal in the real sense if and only if f is satisfying the bar f equal to zero or Cauchy-Vaman equation or it is holomorphic so this is the important conclusion that's why in complex analysis sometimes to say well complex analysis is it's a topic which covers conformal mappings because in fact those functions which represent the main class studied in complex analysis holomorphic function turn out to be also conform in this in the sense in the classical real analysis sense all right so this gives you also another motivation to study complex analysis if i can because it provides geometric interpretation using analytic definition or calculus tools so it's a good mixture and a good training subject where you can mix your your attitudes in analysis and geometry and as you will see it it always requires some basic knowledge but some knowledge some some basic facts from topology because all sorts of positive will play an important role in complex analysis for instance we would say that a condition for existence of some branch of the lower is related to topological properties of the domain and this is also important to know okay now in the in the very last part of this lesson let me just okay give you kind of plan of the next two lecture lessons i want to introduce very basic fact about complex integration right so in some sense we will not make a general theory of integral functions and so on but we refer complex integration to real integration and use many many properties from the theory which i assume you know so what i will consider in particular will be the integral over a function of two variable real variables along a path in the plane so what is called typically line integral and some of the properties related to function and forms actually for line integrals in particular i assume that you are familiar with the fact that the line integral depends only on the end points of integration if and only if there exists a potential function or equivalently if the integral of a closed path is zero and we will characterize also a normal function using these integrals and then obtain cushy integral formula and with cushy integral formula we will show that any normal function is in fact complex analytic and this is what i consider somehow compulsory for a course in complex analysis to know then there are lots of other theorems of results concerning class of function properties general properties and in particular i want to show you properties of the zeros of homomorphic functions and some singularity for homomorphic functions okay at that stage of the course i believe it will be in two weeks we can decide if you are really bothered of this stuff and you already know everything and what and what follows we can make the big step and start talking about several complex variables okay otherwise we continue with some other topic which can be interesting or not but however more standard but definitely we need cushy integral formula and also in in the case you decide to to have some basic information about the several complex case okay so for the time being please those who already know everything wait for for two weeks okay all right in any case what i i i assure you now is that i'm not going to to stress too much the course about calculations so that in case you are worried about possible exercise in integration and so on well this is not the main part of the course right just i need to have the tools from complex integration to define a formula to define to have a theorem and define and obtain the results from the theory which is essential okay so i stop here and see you next Friday see you in two days right