Here the complex functions under a coordinate transformation and the same two variables in two variables will consider the scope of the transformation. But here's the the fact that two variables, x and y are going to a variable. Now that we have done our the following:z i y x plus call. This is a function of z. Of course often complex will be a function. That u and v are the real and imaginary parts. But now we've made them x where a coordinate transformation Let's call it but as bi coordinate transformation to be two variables we have to go to two variables. We have simple here. We are a new z for him line, and so let us two variables x and y and z from z line lends itself to conversion. That's our line of z i in the negative i define by replacing and it is called the conjugate of z. Now here we calculate x See I have collected them y and z plus z line but would go. But two of x is formed We split for two. If we remove them from each other If we change our line of z, im going here for two years if we subtract time. This means that we find y We need to divide by two in. A means wherein We provide coordinate transformation. x and y and z and z lines the relationship between. Now we say inasmuch as it usually x turned to the line y z z, z and z will be a function of the line. But we would like, your job, we insist z lines are not in this transformation. E z line that you're not z lines of the derivative function, means that the partial derivative is zero. That'll do it. So we did earlier not much different from the V.. Where u from a to z. have the same thing in our line. But we insist that we are already putting, of complex functions structure and strength here as well. Secondly independently. Second, the means to ensure independence from accordingly means that there is a partial derivative. So do not change accordingly. Now, the following preliminary do:we saw earlier. There will be a lot of longitudinal partial derivatives. x y z in the z-line by Let's take the derivative with respect to the line. Because the chain rule for derivatives We will apply these to come. Here is one-half as you can see and one-half times i occurs. We saw earlier bivariate in the chain of partial functions derivatives as follows:f z the ratio of the line with respect to x derivatives of the function x According to our line derivative. Plus the partial derivative of the function to y According to the partial derivative of y with z czar, we have seen this all before as u and v are on duty. This line and this z We want it to be zero. Because the complex function z line can not be addictive. Here, let's open the f. plus i have fun because usually complex will be a valuable function. d * d z and d y d line Let's put in our line. We put in place for plus i V. Now we can put d.times.d our line. Similarly, instead of f u plus i put the d y d z V. good times for the line immediately We put one-half. Let's take them. We need to allocate as follows. See terms here i'l There are terms and i'siz. Here the term but outside i'siz i'l buras for being bi, This term will be i'l b. i once i minus the product of his We will obtain such a value. Now here's the real and complex values If we divide the virtual parts I would get it. See u here is divided d * can discard two. d u d v divide divide d y d x minus coming. Because of these good times i by multiplying the number of real happening. If racial conflict if i'l customers divided by d v d * here. Here, too, the outer i d u d y are thus divided. Now this is a theorem of these here we are proven Any of the function a complex function, x and y are functions of the function just enough to be attached to z and the conditions thereof individually is zero. Because of f z, z are independent of line sufficient and necessary condition of having this. So it is divided by d * d v d y and d have to divide evenly divided d x to y of the minus According to the partial derivatives are equal. This is of paramount importance to a theorem. Now, a new question may come to mind. iii when they made the conditions As if that w is no longer fun As the only one with the function derived Can we make the process as variable? How do we define? As this definition. Inasmuch as in one variable went down in z z We will calculate the change in function. Here we will calculate the negative changes. d z will divide. Functions of one variable x If you were going to do it the same. But here we remember the zoom is not just a x. x plus y i. Delta z, of course, the delta x I will be delta plus years. There are special cases where b We can ask whether. This univariate functions we know. Yes, there is. This is double what you univariate such as variable functions. See where x is because we are increasing. y are increasing as well. We do not do it in this partial derivative. We BI artÄ±rÄ±yo one. We BI artÄ±rÄ±yo one. In univariate just a There was one for him consists of a special structure here. It does not hurt. So you know it has been After we put in place. Below the x does not change. Partial derivatives yo x Thank y ranged varied. Where by definition of z both are changing. Now here's the tangent plane We can use the approach. We have already seen in the Taylor series. Taylor series is also the first to tangent plane, The term is considered. When we opened this series, see There are changes in x, y have the change. Thus, the first term before change with respect to x plus delta x. delta change y to y. We're doing the same thing to have. Put them into place When we edit here See here, here u will go to the u'yl. Wherein wherein there is going to v'yl these terms and staying back. Now I wonder where this delta x and delta y is different from going to zero? See, even though only delta x delta correctly goes to zero for x b. But we also have the delta y, There is also the delta x. Up to a point and the delta x We are far from the delta y so. Here the two variable function as endless as the center so this delta x and delta y to the point where we can reach zero. Here we reach in different ways here may be as delta y, here may also, where possible. Delta x could be here, where possible. Therefore, we can come in endless ways. But these two variable function While there was always this limit. Now it y delta and 'll take the delta x to zero. One particular form of Suppose go. So in centers such m slope Suppose that come with lines. This account simplifies a bit. As you can see we're getting with this statement. You can follow sophistication but simple algebraic their installations. Now the question is:I wonder. M Is dependent from these results, m independent? In this function of two variables in general we have seen. Here the Cauchy-Riemann conditions If we have to use x to y, u, there is the use of x minus y, we we see here is the following is organized as:Here d u d y divide here. But there is a minus rather than divide d x d y d u divide Let's write instead of d v d y d x d u divide. So here Cauchy-Riemann We use conditions. See, there are longer. This is when we collect terms here are of a divided by d *, d x, d u divide has also been good times. M going on here plus i times. When we look at here, is to divide d y'l terms, see here The term is divided by d y'l out negative, Let's take out the minus sign. There's one here, where di is divided by y plus there. A plus. So if we multiply by i m I m plus comes a minus. As you can see i plus i m The common denominator in terms. Do I have a surplus in the denominator and so a plus sign Does the more simple and backward results of the remaining independent of m. We are proven wrong? Delta x and delta y in what way Let's get to zero if, while along the predetermined directions, Cauchy-Riemann conditions, if provided, We assume that d w d z definition of a valid identification is going. Because the limit of arrival our way of going independent. Here in this theorem We say:Cauchy-Riemann as well as derivative terms only identified as valuable to sufficient and necessary conditions are happening. , The Cauchy-Riemann conditions using different Expressions can also be obtained. So we are d w d z appropriate changes derivatives with respect to x or y by i times We can show that derived by equal. I do not want to go into these details. Here we're going to keep in mind When conditions Cauchy-Riemann derivatives of single variable functions such as derivatives are able to define. However, both have x, and y are. But these are very special conditions together with the In a very simple manner because it depends of course connected to the x plus y Because Cauchy-Riemann If conditions ded, f d z can be calculated as a single. If the derivative of complex functions and there is a finite analytic call. Or here for a weekend getaway There are point call. One exception, these singularities point with an analytical function minus z z z is zero point zero n'yinc force given. We call this the pole. Now here's a few examples I'm giving:u given in the following way. there is provided in this way. Now in its u'yl V. By combining an analytical Does Does not have functions, We need to figure it out. The only condition to be analytical Was to ensure the Cauchy-Riemann conditions. Now see here is divided by d * d have got to be divided is equal to y. Indeed, in respect to x When we take the derivative e to the x cos y involved. v is the derivative with respect to y we get When going out the same thing again. And similarly to y u When we take the derivative of minus hence the minus cosine The term sinus is coming. This divide minus d v d must be equal to x. Indeed, this negative Comes from the Cauchy-Riemann conditions. It came from the negative derivative. Minus cosine derivative came here because it is negative. But here with respect to x We will take the derivative of v is. He gives the same thing. So the Cauchy-Riemann We provide conditions. Thus, in general, plus i have the x and y is a function of two variables connected to. Functions of one variable can be written as. Indeed, if we combine plus i V, e to see the x partner. In the first are the cosine, I have sinus in the second year. Plus i y e cosine of Euler's formula so i y e above. When you combine them, e to e to the x and y i, e to be x plus iy. E that z. So to have this function over z, this complex function. The real part and the imaginary part there is also the functions. We are one more example. We wer u plus i have function Whether we want it to be analytical. But just given. V. How we choose this function as an analytic function that Allow the one written in terms of z. In this case, for our base, what happens? Now Cauchy-Riemann We know from the conditions again. is divided by d * are account here, two x's. Other than that there has to be divided is equal to y. It integrates a time If we come both x and y. But it is also an integral fixed income. Bura, the gun can be x, in general, because we take the derivative with respect to y When g x will be reduced. Now this is the first Cauchy-Riemann condition I've obtained from the condition of this structure. Now the second Cauchy-Riemann Let's use condition. This means that v is xa We'll take derivatives. This derivative of u to y should be equal to minus. Here we make this calculation, v is the derivative with respect to x is coming two years. g and g base comes from derivatives with respect to x. These two must be equal to y. Because di divide d The minus sign comes to y. Here is the account if we divide d y, minus two years. But circumstances have a negative here, say that two years. As you can see both sides can balance each other. This is also what g demonstrates the need. g of the derivative must have been zero. So it can be fixed more than one gun. E it is uncertain at constant We can choose to zero. So there was also b. Indeed x squared minus y frame X, we combine two good times, i see x plus y squared is happening. Let's open it:x squared minus y squared. Because of i y squared minus plus two times the product of the two sounds, i once both x and y. E z squared it happens. So we Cauchy-Riemann conditions using a analytical portion provided the other part of the function and here we found the analytical We can build function. Derivative thereof, so that no single We have come to multivariate functions. At its derivative, we know univariate function z becomes twice as derivatives. w x seÃ§sek see, this is a very seemingly simple example. Ph.D. exam to a friend non-analytic give an example, I was asked because as everyone analytic functions He wants to know, because they are useful, Does the narrative, but subject he had asked for it. Now, we're looking at things, is to divide the d x, Cauchy-Riemann conditions, d d * u where x is divided by the plus i have to type it w as, we would only x u x, v is zero. One can see that. When we look at d v d y We see that it is zero. So the first Cauchy-Riemann the condition is not provided. If you look at the second one, it is not provided. But the provision of any one, Failure to provide is sufficient. So x function w is an analytic function is complex valued in these functions. Here I have given some homework. They show that the writing. Here, given w, ui x squared, v's ie the imaginary part y. As for that z Do not post indicates. So Cauchy-Riemann conditions Show that you will provide. If it does not, We found previously between X, y, in the x and y'yl, transformation between z and z lines. That only in this way can be written I'm just saying show. If this function is real portion provided herein. x squared minus y squared minus x. The imaginary part. This Cauchy-Riemann conditions that provides and hence f z can be written as we would like. I replied how to seems to be something. Others in the same structure. So now we related to derivatives we finish our work. We learned quite a few issues. The next section, multi-storey integrals. Three kinds of multi-layer integration. A:Cartesian coordinates. But in Cartesian coordinates We saw a two-storey integral. But we will see a three-storey coordinates. These cylindrical and spherical coordinates. Have you been to this triple integrals. But let's not forget one two There storey integral type. We have not yet processed it. We have two floors of integral integrals in the plane rectangular and circular as coordinates have seen. But there are also surface in space. For example, consider the surface of a sphere, Think of a dome, consider a cone. E their three-dimensional and surfaces in space. A plane thereof We can not be obtained within. But they are two-dimensional space. So are the curved two-dimensional space. On account of the surface thereof We will be an important issue. The next portion We will continue working together. But the parts related to derivatives We finish the application. In contrast with integral We'll see about new topics. Please goodbye until we meet again.