Hi there. Our previous session of complex functions on a cycle by integrally We found a very interesting formula. This residual value theorem or Residue theorem. Here the importance of this the difficult integration process rather than just a value turning to the calculation. Of course, some of the finer points of being I hope that you understand their subjects. Now here our research with applications We will continue with the exercises. I hope that with this terminology, The concept will be further strengthened. We want to do the following integral. Denominator five plus four the cosine of theta integral, coordinates of the center of the team in the center of a on the radius of a circle in We want to calculate the integral. In these implementations encountered is an example of such an integration. It by conventional methods We can not calculate. The easiest way, as you can see the integral without the need for calculations, by calculating a value of a Find algebraic operation will result. Now before the complex integration Let's make the transformation of variables. We know that z equals x plus y i. x instead of r cosine theta, If we say that y r instead of sine of theta, plus i found on here because i cosine sine theta, Euler's formula accounted for. i theta is going to run over. We are going to do the process is equal to a on the unit circle so on. Therefore i z e to theta is happening. If we take the derivative of it, divided by d z d e to the i times theta theta would be good. If theta is the right place, we obtain differential form. But here again there was e to i theta e to the i theta here because it is the z z put, we reserve d theta, d theta d z i z will be divided. I will be in the denominator. If we multiply the numerator and denominator of the minus sign, minus one times the denominator, of course, i occurs but once in the denominator minus i for that i give it a. Thus, in the denominator of virtual We're getting rid of the term. Now, again with the term Euler Let's continue our preparation. Let's find five plus cosine theta We know the cosine of theta. Over one-half to i theta i theta plus to minus. i was sadder to our theta. i theta minus one-half to be z. Him the power minus one, let z. If we multiply the numerator and denominator with z denominator consists of two z. The denominator with z z z squared plus one gets hit by the opposite has occurred kosÃ¼nÃ¼s theta for the girls but in general terms, not This unit circle of radius We got it on a r. KosÃ¼nÃ¼s quartet of theta If we multiply four times because we We want to find kosÃ¼nÃ¼s four times theta, sadeleÅip share two twos coming. I take it's place. Five plus two times z squared plus a divide z, Get our common denominator as you can see There are two times as our frame of here. There are two. He last term. But five times z occurs. Hence we find we have the denominator, The above complex variables, in terms of direction. This is the value on the unit circle. If we take this place minus sign d z z d theta split times were coming from. We have calculated a little before. z z squared plus two times five z plus two denominator comes. As you can see girls are more simple. A minus sign is staying here. So we initial integral The integral over theta, on the unit circle, integrally on the unit circle that complex variables We brought in terms of this structure. Now that's the first step. This consisted of our preparations. In the second step the singular point We need to find. What is the singular point? This makes the denominator zero value. Because then the function value where the individual is going to have forever. But this is an easy job. Because our two z squared plus five plus two A second order power function. We know that the root of the known formula. Five plus or minus the square root of minus five in 25squared minus four times two and two. Four are already doing this. Try four times in 16. So 25minus nine out of 16 involved in root. He three course. Here the roots of minus one divided by two and cons involved. I really put in place. Let's try with that minus sign. Here is composed of eight. Here there are two eight plus ten. But the minus ten consists of five z. Thus, ten minus ten we see that zero. We see that provides. Minus one-half the gives the same procedure. Here only these two radically z is equal to minus one-half It is in two circles. Minus two is out. This residual value theorem On cycle and in the singular We are interested in point. Therefore, residual, residual value only minus one-half will come from. Now let's take this term and Let residues account. See here we have a minus here. Z z squared plus five is a two-plus-two. Minus one-half of that in it a z edecez account the value of the two. To do this, share and share that, i z plus one share split will multiply and z is a negative way to We'll find that limit. We had one a minus located. Here lopital rule when applying a share derivative, z derivative of the denominator four plus five. Z is also brought will account minus one. This minus sign since the beginning of our 're standing, move it. You will see this at the end how important it will be. Have a minus sign. Put a minus instead of z divided by two As you can see happening here is minus two. Gives two plus five minus three in the denominator. Now this is no longer worth Let the result theorem. The function of the integral, integration of complex functions of the two i pi times the sum of residues. But there was one residue in the cycle. Here, the residual value of I've also account here. There are two pi i see here. I once had a minus divided by three. How important this minus sign We also see that. i times the real value minus sign gives a plus. As you can see denominator has two pins, There are also three in the denominator. Already since the beginning of this integral Remove the surplus could be valuable. Because the cosine of the smallest t where there is a minus. Where the smallest Even minus four plus five. This means that all positive integral values ??of Because the sum of the function the real results will be released, as well as from here plus we can see there would be valuable. If you wish it or not have an minus sign that we made ??a mistake we could not have an, you would see the real-valued The value of the integral virtual output this time. Did you understand that you made ??a mistake. For him, this negative yi I need to move carefully. If the error would be moved. Mistakes would be. I'm giving you two assignments. Here is very similar integrals In order to get used to your hand. One of five plus three kosÃ¼nÃ¼s theta. We had just quartet. Yet here 25minus nine will be under the square root. From there will be four integers will be easy to value. There's something similar here. But here is the interesting thing, of course. In his statement sinus girls in terms of a good goes. So the number of virtual volumes i. Because i theta minus sine theta over to i theta divided by two to minus i. This is what is known from Euler's formula. That's a good show itself here again. Here you'll find the roots of You will see that i'l. But the result still needs to get real. Negative values ??involved because the same this integral and theta, pi If you split the difference between the two, would be cosine. The results of two of the same occurs integrally qualitative observations were also available. Now a slightly different kind of We want to do the integration. This integration is not on a circle, from minus infinity to plus infinity as above minus the absolute value of x, that means such values ??are decreasing at the ends. There are a plus x squared in the denominator. With this integral transform again or something is impossible to calculate. It does, however complex, with values We find that the residual value theorem. Where x and z instead of before obtained by writing complex integration Let's over and it's a closed loop, What we now see that cycle. Here, from minus infinity plus There are a trajectory forever but this is a clear trajectory, not a closed path. Directly in the previous problem closed loop on a circle, had closed curve. There was a cycle. There is a cycle here yet. Have an open area. Now let's do it like this: e to the minus z'l, z'l absolute value function from minus infinity plus forever on a straight line, also a part thereof, a half circle with radius R Let's make a loop by adding. See here for the function. We plus from minus infinity We want to go on forever. Let r is less than once before. Whereupon one, yet finite Let's add a half-circle. As you can see now was a closed loop. As a result, r forever 'll take it and we'll see on the semi-circular The value of the integral is zero. So we r forever As the journey, we want to x-axis from minus infinity to plus infinity on integration, we want this, plus a zero value, but it was very important to us that this is a closed loop curve, and it brings the opportunity to We will find points in the residue. Residual value points, We'll find singular points. Now let's start again with a preparation. We have just seen it again, this time to z, r than y times x plus i e im coming over theta. Here is not one. A previous problem was a r. r will be held until the end. We now integral Before we go on x. means on x, z means that x. There is not a problem. Here r e over two square You might say theta square. But we will calculate an x on it. And this half-circle will on account. First half circle Let's take a look on the button. Semicircle on the r, theta e to i. Let d z here. As we did in the previous example. We need to do one of z value, the absolute value of z r is happening, e to the i theta absolute value of will appear when you get one. Now it slowly We began to put in place. d z e to the i r i times theta. We are writing here. i, r, e to i theta d theta. because we can get out of r We are regularly out on the constant r. E minus the absolute value of z where r is coming to the minus. A plus r squared in the denominator is Coming over to two i theta. Now we r here forever if we take see something here each relevant R but it will be here in the denominator r goes to zero. E is minus extraordinary forces goes to zero. This is going to infinity, but e to is minus infinity more rapidly is gone, if more If you want to look at is divided over also when applying the rules to take is rapital a r e to remain above below. divided by infinity is a way to This goes to zero as infinity. So this integral r When finite finite value but zero contribution is gone forever. So do we we want the integral over z, z z x is equal to x is on. Here the absolute value of x, we are looking for was already integral. d x plus x squared. This is equal to the closed-loop comes because here we go. Plus this, but we have seen that the value is zero. So this is a real and on the right integral part of this contribution is zero We have completed a circle of circles. Now the rest is easy. Because this integral can be calculated immediately. Because it is now two pi i times will equal the sum of the values??. Now where would be the value? Makes the denominator zero values. z squared plus one is there, z, i and minus the value of i is zero in the denominator. Therefore, single spots are formed. But from these two poles, only singular point i is in the ring only The residual value will account here. If we account him, that's singular points and details, When we put our equals i See wherein the absolute value of i, an absolute value of i. The minus sign there, here, minus a negative going on, to minus one-half to one that is. We found this place. This residue, now to find the value z with minus sign, ie minus the singular point z bumped by the coordinates. If we take the derivative of this one. If we take the derivative of the denominator of two z. also i z at z in two Two good is happening when account. As you can see via this email minus one from our By combining E divide one-half to two times good times. This e Euler number. I will see that it's still important wherein both pairs of heat because there. Denominator as you can see i have these two pins. There are good times to two in the denominator. That's a good stretch of the simplification and will be a floating-point value again. As this change of variables etc. possible ways to calculate This is not integral. Therefore, this residual value theorem and integration of complex functions our ability to be able to account We have moved away is paramount. In the past we can not do, you know, coordinate transform, such as variable conversion ways we can not We are becoming integral does. One of the assignments I give the following integral: Gene from minus infinity to plus infinity. This time not to the minus z. Only d x divided by x squared plus. This integral can see the first analysis of course you know. This is an integral sub-tangent. So instead of x tangent theta If you make the conversion, d theta come to this, This would lower tangent. Old tangent at minus infinity value minus pi divided by two. The value of pi divided by two plus infinity, Remove the value of pi here. But I suggest you do with this vortex. Follow the same steps as you can see, e to the minus z is not here, for There will be here to see it, there is no future for E p. Watching these same steps again You'll find the value of pi. I'm giving such an assignment, You can examine it. If you are a pole in, The value of this cycle is zero. Here the given circle. The unit circle and the center of the third and radius of three given circles. Here you will see a little bit Thinking The pole pi divided by two in it, that is, a Buc, a comma comes in 57. However, this half of our unit outside diameter of the circle. Therefore therein There is no one single point. Therefore, these integrals are zero. If you look at those presented here pole You can find the value. Now we have finished a major part. Now you will say it again teacher says everything is important, but Green's theorem that the last part of the course also an important theorem will direct. Then in the plane of the integral curves on On closed curves, i.e., an on cycle We will encounter special situations. Produced from the orbital potential If a vector which independent of the orbit was involved The value of the integral and indoor environmental No matter, it gave zero value. Because you are starting from a wandering You come back to this point. However, as in all integral There may be potential to do. That's what we call Green's theorem theorems brings clarity to this issue.