Hi there. In our preceding session both along the trajectory, orbital perpendicular to both of the currents We were interested in examining. Especially along the trajectory of flux We've tried quite a few to be independent of the way. But also that the flux perpendicular to the orbit I have seen can be transformed and the Finally we asked the following question: How could these two types of integration also ie the projection of one of t a received vector of a certain integrally along the trajectory or When the projection on n, while i.e. from one side to one side of the line which calculates the switch on the other side the sections are both integrally independent of the orbit to the condition that We examined these two conditions are met, and We saw that with the arrival can be achieved. Addition of the components of the vector We have seen that provides a Laplace equation. Also common to both the potential itself We have seen that the Laplace equation provides. As you can see Laplace processor something which can occur frequently. But you also have immediate recall. 187previously. page and in the following sections As we will see a sign difference Cauchy-Riemann far beyond them, conditions is not something else. So with complex functions had an affair. Thus complex functions of this issue immediately and in practice the relationship is also very type of a function that works. Now we are going. You know only a function can be written in terms of z was supposed to certain conditions. These were the Cauchy-Riemann conditions. To the real part of this function, If you also have other virtual section, so 're getting a complex function. The complex variables x and i than y times is achieved. I'll remember, of course, the square root of minus one, i.e., real A number of non-unit imaginary number. Of course a complex number plane in order to show both x and y components is needed. So the vectors have a benzeÅiklik. Similarly, a complex function vectors u and v as components can interpreted as a also able to show with the function. Now we have this complex function d Let's hit the z'yl integral. This is an integral from A to B. It is up to an orbit in the plane. Because the x and y-z plane component is given by. This means that a curve in the plane says an orbit. Plus i have u as f Now Let's place value. z from the d z d x plus i Let's place as d y. Comes to the following structure. Just things that come to mind here, it is to separate the real and imaginary parts. Real part of x u times is coming. We are writing here. In addition, the product of I and II minus one for giving, v d y is coming from, there is a minus y There are also two terms plus i'l. One of them is to have this x comes from the product. There is a good beginning. Secondly u'yl d y the product comes from. There are a good. So the integral of the function of this complex consists of a real and an imaginary part. Just see here, Is there a difference in the v mark. If you change the negative V with v, See here vek, components u and v are negative the projection of the vector t, Once we see that u t d s of integration is nothing more than an expression. Where V is the gene When we replace the ex exactly perpendicular to the orbit becomes an integral component. Therefore, this complex function with two integral, It is possible to handle both. If we say that the real part j, there are also cons with this V When we change, we find the same thing. Here, too, there is less change with V, the k to say it, perpendicular to the orbit of this flux gives the integral. So these complex variables kind of overlap directly with integral and the application will find an important place. Now that a significant We will see the application. Now we have brought this building integration. Want to call the Cauchy-Riemann, If this function analytical Provides the Cauchy-Riemann conditions. E, Cauchy-Riemann conditions, this independent of the integral trajectories of from the requirement of having It is not something else. Wherein the second component v is negative with respect to x derivative of the first component to y will be equal to the derivative with respect. Wherein the first y is derived based on the component, the second component with respect to x b will be equal to the derivative. Therefore, a little confused here Although it may seem like separate them If we consider a separate one is multiplied by u'yl x, a y an d are multiplied by minus v'yl. Here also a bumped to v'yl d *, y an d are multiplied by u'yl. Therefore, u and v to a Instead of staying stuck in the first if we consider the second component of the component, This is the first component you have. To y of the first component, i.e. by the second variable derivative is equal to the second component derivative according to the first variable. If there is a minus here. The first component to the second being equivalent to the derivative with respect to the variable, second component, the second component but V'la minus the derivative thereof according to x. On the one hand they Cauchy-Riemann conditions on the one hand a little while ago, we found both along the trajectory of the integral The way to go is different We see the conditions. As a result of here involved:a little astounded at the transmitter. If this is a complex analytic function, its line integral is independent of the orbit. This supremely powerful thing. So when you go from A to B. Analytical If your function, to deal with complex functions Returning to the specific software intensive summary, all these integrals As long as the analytical orbit independent on the orbit. Off we go on a trajectory also b from A to B. When combined with the a'yl This integration also involved always zero. But here are a key requirement. I always got to be analytical. What does it mean if you analytical? There is a singular point. There are many types of singular points. We shall not go into the general but most importantly the most frequently encountered, a singular point of the month, a small Hoops can be separated species. Here you get a single point. We are around it We're passing around a small circle. We're cutting it artificially. Our mainly external in our orbit. Here, as an artificial We make a cut. Inside a small circle, radius of an epsilon We are creating a circle trajectory. We combine these two. Now get in now, we see this region does not contain singular points now. Single point was excluded. If you like it a lot, likewise each can be excluded. See the limits on the way to the end of this After such while in orbit around a person standing inside in the trigonometric sense plus valuable accept wander We accept that. This section comes along like this before. Here you can see the interior point According to a person standing in the opposite direction turns. So it comes in the negative direction. As we walk out of here in We're the opposite of the way we came floors. Go to the limits of these two lines When we combine destroy each other is this integrals. This is more than a single point, though This singular decomposed separable spots, all these in separating the be analytical, analytical functions a structure that would bring. Now let's start to one. Called simple singularities, As the so-called pole singularity. F's a function everywhere analytical, but at one point, If you are creating a zero in the denominator of this f g divided by the negative z z is divided by zero. Let's call this point z is zero. Now it's that when we exclude As the total integral around we do:We're going here. Now let's write the integral once. We say f z d z. f, g, We are writing negative z z to zero again. Because of this function singular There is a point at z is zero. Where z epsilon times instead of e to the i theta say. More precisely, the negative z z to zero. Came to see it. We have d. Whereby z z is zero minus epsilon E is over once i theta. d z also via E epsilon constant i e to the i times theta theta. d theta income. See above, you're writing it here. I have also here because of the d z. Epsilon times in the denominator I got over to theta. As you can see in this epsilon is more simple, e, i theta, they are more simple. When we go to the limit so this infinitesimal And when we bring the circle to zero an increased value of the remaining leaves. Not all of this is a function f, singular point of f but that helps to distinguish itself The gun remains analytic function. As you can see here, we go to the limit When we go to zero epsilon fall time for the epsilon where z in the z goes to zero and here comes just a two pin This is the integral of theta. I went Epsilonl theta. There are also a particularly good. Two pi times g i z remains zero. Now we here forever made on a small circle. c epsilon that we, in the sense that the circle of radius epsilon. c so that the outermost regions, we See the outermost plus the integral of the Enter verifies along destroyed each other a while just stay here for the c epsilon. But we are wandering in the direction c epsilon minus. So it over all The integral gives zero. But on the plus value of c that c epsilon minus the value on Because it was then moves to the right side. But now in this thing, We calculate the second integral. Just have calculated here. Two p i, g z was zero. This g z is zero, see this from f konksiyon growing back, in the sense of remaining. This residue in English, or by Fransz using the same words, In many western languages the same value is used. This residual value to Turkish, the remaining value is rendered. This residual value of f to g z is zero, If you want to use the term in English, Turkish is also used, It's called residue. Now g of light in many cases It is not certain. In some cases is confidential. For example, if you receive a slash our sinuses, value of zero in the denominator of which This function is singular. E sinus z values ??that zero is zero, pi, the two pins, minus pi to pi minus two singular points, has infinitely many singular points. Any function To find the residues, inasmuch as this singularity setbacks, sounded negative z z in the denominator zero. We share a z z is zero minus If we take the limit and puts it, z is zero remainder is happening. Let's see sine residues account. Let residues account for a share of the two pie. We're putting our share minus two pi. We take this limit. Limit above zero when you get there, below zero. We know that in earlier courses, There is zero lopital rules To remove the split zero. We take the derivative of the share, here. We take the derivative of the denominator. He is also the sine cosine derivative. Z also still equals two reckon with the value of pi. We find that a residue. This residual value, the remaining value. The remaining value of the pie the same function, now we'd like to find the value of this We will multiply our time with negative p. Gene lopital in accordance with the rules take derivatives denominator in the take derivatives, we calculate then zero divided by zero is eliminated. Minus pi is happening here. If we had looked at zero, Let's see where p is not. Gene would be zero divided by zero. Yet when we receive derivatives, z is a derivative of the sine cosine derivative. This time, not two pita the account will be zero. Direct cosine is zero, giving it a. So find the definition of these residues as non-analytic singularity In the point, singularity at the point z minus We find our hit reset. If at this point the function is not infinite, is finite, the residue is already here are zero. One thing remains. However, in the singularity of a the rest is worth going to. Now worth going. If you are going to f z forever, that means There is a term in the denominator goes to zero. The zero divided by zero uncertainty disappears in this limit. Now to say that we've brought the following results. On a cycle, zoom function on the closed curve for if the integral over all these regions and in f z is zero Analitiks. But, for example, if an analytical If z is zero at the point of singularity, This would be for residues of z. The residue, the residual value We are in the account description: z negative z hit the reset We're taking the limit of the account. If you have more than one, of course, each Once we obtain the contribution of the two pins. We won a very important thing here. Only an integral derivative able to reduce the import process. So you were able to reduce algebraic manipulation. Taking a limit, no limit, We have a value calculations download. This is a very important achievement. Now here I am taking a break. Meanwhile, after making this sample We will try to understand the subject thoroughly. Bye for now.