hello and welcome back to complex analysis a video series where we are still talking about Contour inter quotes in the complex plane and indeed in today's part 26 we will talk about a very special content integral and this will help us to prove cauchi's integral formula which we will discuss in the next video however you already know before we start I really want to thank all the nice people who support this channel study via PayPal or by other means all supporters can download the quiz or the pdf version of this video and with that let's start so first please recall in the last video we have generalized koshi's theorem for such domains here and now today we will talk about curves that have exactly the shape here on the boundary and now because it looks a little bit like a keyhole it's often simply called a keyhole Contour for this reason I also keep this name and now today let's discuss what we can do with such Contour intervals so First We Take a function G that is defined on a disk there the radius we call R and the middle point zero moreover we want that this function is holomorphic with only one exception point at zero of course there we already know some functions that fulfill that for example 1 over Z however here the function G can be anything it just needs to be holomorphic with this domain hence we have our disk here with the exception point zero in the middle and then we consider a counter integral in this domain that looks like such a keyhole so for example it could look like this so it's a closed curve inside the domain of G therefore as we have learned in the last video koshy's theorem tells us that the contour interval along this curve is zero and please note this holds no matter how this Keyhole is formed or orientated in particular we see we can introduce two parameters here and the first one should be the radius of the Inner Circle so let's call this one simply Epsilon and the second parameter should be simply the width of the coin door here and maybe let's call this one simply Delta hence you see we assume that the outer radius here for this circle is fixed for our discussion here okay then the name we give the curve is gamma with index Epsilon Delta and then our result here is no matter how small the positive number is absolute and Delta are we always get that the Contour integral is zero okay that's not new but now we want to use that and split the integral up into four parts and I think it's clear how we do it we take the Outer Circle the two lines here and the Inner Circle okay then I would say let's give them names where we use upper indices hence the Outer Circle should be gamma 1. then comes the first line which is gamma 2. then the Inner Circle which is gamma 3 and then the last line gamma 4. okay now I can tell you what I want to do in the end is to send Delta to zero and then the question is what can we say about the counter integrals of all these curves here obviously the first thing we can conclude from above is that the sum of all these counter integrals along these curves here is zero now of course this fact holds no matter how small data is chosen therefore in The Next Step let's go with Delta to 0 and let's see what happens to all interquilts separately in other words let's start with the first one gamma 1. there you should immediately see if we set data to zero we have to full circle therefore let's call this curve Capital gamma with index one okay and then let's look what is the difference between both Contour inter quotes so we consider the absolute value between the 1 and the other one and then we simply know this is the absolute value of a new Contour integral namely the Contour intercool along this small line there so maybe I don't give it a name I just draw a small line there so you see we don't need a name because we can immediately do the standard estimate which is given as the maximum of the absolute value of G of Z where that goes through the Curve times the length of the Curve and now it should be obvious that this length goes to 0 when Delta goes to 0. so we can simply conclude the whole thing goes to zero so you see this is a very important fact because it means for the limit Delta to zero this integral here goes to this integral so in other words in the limit we are able to use the full circle okay then let's go to the next PATH which could be our other Circle this small circle in the middle we first consider this one now because you see the whole argument from above works here as well this means that in the limit Delta to 0 here we can also use the full circle however the only thing you should note is that the orientation is different from before and moreover here we also use the index Epsilon because it's the radius of this circle okay so you see half of the discussion is done so let's go to the corridor this means both lines here we should consider together you see this because if we send Delta to zero they will get closer and closer together hence you see the idea here would be that we close the curve such that we have a closed curve so for example what we could do is to use such a rectangle and then again as always by koshi's theorem we can conclude that the closed curve integral is just zero and now you also know what one can do we split it up into four parts again and then we use this estimate from a buff for the two lines we added and because they get smaller and smaller when Delta goes to zero they integral along the small lines gets also to zero therefore we can conclude that for Delta to zero only the two intercourts along the two lines remain moreover the result is then the sum of both inter quotes vanishes therefore we can conclude in the limit Delta to 0 the corridor is not important at all so in summary only the two circuits remain in the limit process to see this please recall this was our equation from above where all the four paths were included now the last thing we showed above is that this middle part here vanishes and the other two parts converge to full circles in addition please also recall there we used Capital gammas okay then we can bring this on the other side and then we have our result namely the integral along the Pix circle is the same as the integral along the small circle for this please recall that the orientation of the small circle was different from the orientation of the big circle and this is where we have the minus sign okay I think the best way is to remember that with a picture so if we have a counter integral with a very big circle you can just substitute with a small circle with the same orientation hence this is a very nice result because it means we can use a circle as small as we want and please note the only assumption we used here is that we have a holomorphic function with one exception point zero here okay and now this nice result we can use in the next video where we want to prove koshy's integral formula therefore I would say let's meet there again and have a nice day bye [Music]