 Hello, everyone. In this video, we will discuss the Millet-Thompson method to construct an analytic function. Learning outcome, at the end of this session, students will be able to construct an analytic function by using Millet-Thompson method. This method is actually used to construct an analytic function f of z when one of the real or imaginary part is given. Now here we will consider an working rule depending upon whether the real or imaginary part is given. We will discuss this situation in two different cases. Now consider the working rule to construct an analytic function by Millet-Thompson method. Now here in the case number one we will consider that real part u is provided and we want to find out an analytic function f of z. Now the step number one is to find out the partial derivative of the real part with respect to x and y. That is to find dou u by dou x and dou u by dou y and immediately let us use the definition of derivative of an analytic function. The derivative of an analytic function f of z is given as f dash of z equal to u x plus i v x. Let us call it as equation number one and we know that u x is nothing but dou u by dou x and v x is nothing but dou v by dou x. Now in the equation number one on the right hand side the expression involved we can see that this u x is known to us just we have determined. Now this v x is the partial derivative of imaginary part but the imaginary part is not known to us. As we are going to construct an analytic function f of z its real and imaginary part definitely satisfies cr equations. So here in the step number two we are using cr equation to replace this unknown partial derivative in terms of the known partial derivative that is now the cr equations for u and v are u x equal to v y and u y equal to minus v x. Now as we want to replace v x second equality is useful in this case. We can replace this v x as minus of u y. After putting v x as minus u y in the equation number one we get it as f dash of z equal to u x minus i u y. Now all the quantities present on the right hand side are known to us. Now to convert the above right hand side to the function of z only we will put x equal to z and y equal to zero after putting this we will get that right hand side as the function of z and if I get the right hand side the function of z by simple integration we will get the required analytic function f of z. Similarly in the case number two we will consider that imaginary part v is given and in this case also we want to find out an analytic function f of z. The step number one is same we have to take the partial derivatives of the imaginary part that is to find dou v by dou x and dou v by dou y and immediately we have to apply or use the formula of derivative of an analytic function and that is f dash of z equal to u x plus i v x. Let us call it as equation number one and if I observe this expression as a real part u is not known to us so obviously u x is also not known to us as f is analytic its real and imaginary part satisfies c r equations now by using c r equations we are having u x as v y and u y equal to minus v x. Now here we want to replace this u x and here we have obtained u x as v y where v y is known to us which is determined in the step number one. Now putting u x equal to v y in equation number one we get f dash of z equal to v y plus i v x. Now the expression present on the right hand side is known to us. Now our aim is to convert this right hand side to function of z only to convert it we are making the substitution x equal to z and y equal to 0 in the right hand side of f dash of z to get f of z as the function of z after making these the substitution we are getting f of f dash of z as a function of z and if we take a simple integration with respect to z we get our required analytic function f of z. Now pause this video and write down the partial derivative of this function phi equal to log of f of x comma y. I hope all of you have written an answer now the problem is to find out the partial derivatives of this function. Now the partial derivative of phi with respect to x is given as 1 upon f of x comma y into dou by dou x of f of x comma y and similarly the partial derivative of this phi with respect to y that is a dou phi by dou y is given as 1 upon f of x comma y into dou by dou y of f of x comma y. This is the formula is used to find out the partial derivative of the function phi equal to log of a function. Let us consider one example construct the analytic function whose real part is log of root of x square plus y square. Now here we have given the real part so let us denote the given function by u and we want to find out this f of z. Step number one is to find out its partial derivatives. Let us differentiate u partially with respect to x we get. Now just we have discussed how to find out the partial derivative of log of a function. This is the formula 1 upon root of x square plus y square into dou by dou x of this root of x square plus y square. It is equal to 1 upon root of x square plus y square as it is into now the derivative of root of a function is 1 upon 2 times root x square plus y square into derivative of x square plus y square with respect to x is 2x and after simplification we get this derivative as x upon x square plus y square again differentiating u partially with respect to y treating x constant that is dou u by dou y equal to again using the same formula we get 1 upon root of x square plus y square into dou by dou y of root of x square plus y square that is equal to 1 upon root of x square plus y square as it is into now the derivative of this root of x square plus y square is 1 upon 2 times root of x square plus y square into derivative of x square plus y square with respect to y is 2y and after simplification we get it as y upon x square plus y square and let us write down the definition of or the formula for the derivative of f of z that is f dash of z equal to ux plus ivx here u is provided therefore ux is known to us vx is not known to us so let us write down the cr equations for u and v these are the cr equations now the second from now from the second equality we can replace vx as minus uy after putting vx equal to minus uy in equation number one we get f dash of z equal to ux minus i uy now here ux we have obtained as x upon x square plus y square minus i as it is and uy as y upon x square plus y square call it as equation number two now our aim is to convert this right hand side to a function of z to convert it we are making the substitution x equal to z and y equal to zero in the right hand side of equation number two now we get f dash of z equal to z upon z square plus zero square minus i into zero upon z square plus zero square after simplification we get it as one upon z now we have obtained the right hand side as a function of z by simple integration we will get our required analytic function f of z let us integrate with respect to z integration of f dash of z is f of z is equal to integration of one upon z dz and we know that integration of one upon z is log z plus integration constant c this is the required analytic function whose real part is provided as log of root of x square plus y square