 In this video we will consider some problems on Milne-Thompson method. Learning outcome at the end of this session students will be able to construct analytic function by using Milne-Thompson method. Pause this video and write down the answer for the question what is the formula for the derivative of an analytic function f of z. I hope all of you have written the answer. Now the answer is the derivative of any analytic function f of z is given as f dash of z is equal to ux plus i into vx and this derivative also has another form since f of z is an analytic therefore its real and imaginary part satisfies the CR equations ux equal to vy and uy equal to minus vx if we replace ux by vy and vx by minus uy we get the same derivative in the another form as f dash of z equal to vy minus i into uy. Let us consider one example find the analytic function f of z was imaginary part is e raised to x into x sin y plus y cos y solution here we have given the imaginary part therefore we will denote it by v step number one we are differentiating the given function v partially with respect to x and y so let us differentiate first v partially with respect to x treating y constant therefore we get vx equal to now here we can see that this is the product of two functions of x therefore we use the product rule to differentiate it that is first of all we will write e raised to x as it is and then differentiate the second function now here in the first term sin y is constant and the derivative of x is one plus in the second term x is not involved therefore its derivative is 0 plus write down the second term x into sin y plus y cos y as it is and the derivative of e raised to x is e raised to x after taking e raised to x common from these two terms we get e raised to x into sin y plus x sin y plus y cos y let us differentiate v partially with respect to y treating x constant we get v y equal to now the first term is of e raised to x which is constant let us write it as it is into inside the bracket while differentiating with respect to y x is constant and the derivative of sin y is cos y plus now in the second term we can see that this is what the product of two functions of y therefore in order to differentiate it we have to use the product rule writing y as it is derivative of cos y is minus sin y plus cos y as it is and the derivative of y is 1 now after simplifying it we get e raised to x into x cos y minus y sin y plus cos y step number two we know the formula for the derivative of an analytic function f of z that is f dash of z equal to u x plus i v x let us call it as equation number one now in the step number one we have obtained the expression for v x and v y and we can see that the derivative contains the term u x that is the partial derivative of u with respect to x which is not known to us now here now our aim is to replace this unknown function in terms of the known function in order to replace it in the step number three we are using cr equations the cr equations are given by u x equal to v y and u y equal to minus v x minus v x now here u x is v y replace in the equation number one u x by v y we get equation number one as f dash of z equal to v y plus i into v x the content involved in the right hand side is known to us let us replace v y and v x by its expression we get this function now to have the right hand side as a function of z let us put x equal to z and y equal to zero on the right hand side of equation two now we get f dash of z is equal to e raised to z into z cos zero minus zero into sin zero plus cos zero plus i into e raised to z inside the bracket sin zero plus z into sin zero plus zero into cos zero that is we get f dash of z equal to e raised to z as it is into we know that cos zero is one therefore z into one is z and here zero into sin zero is zero plus cos zero is one therefore it is plus one plus now here we can see that sin zero is zero again here sin zero it is also zero and zero into cos zero is zero it means this total bracket has the value zero therefore the second term is not written here now we have obtained the right hand side as a simple function of z let us integrate the above equation with respect to z to get f of z now integrating with respect to z we get integration of f dash of z is f of z is equal to integration of e raised to z into z plus one dz plus c integration constant now here we can see that this is the product of two functions of z therefore the generalized rule of integration is used to evaluate this integral we will consider z plus one as the first function and e raised to z as the second function now write down the first function as it is and the integration of e raised to z is e raised to z minus let us differentiate the first function z plus one with respect to z its derivative is one into one more time let us integrate e raised to z we get e raised to z plus c after simplifying it we get the final answer f of z as z into e raised to z plus c this is the required analytic function f of z let us consider one more example find the analytic function f of z was real part is sin 2x upon hyperbolic cos 2y minus cos 2x also find imaginary part solution as it is a real part we will denote it by u step number one let us differentiate u partially with respect to x now here we get here we can see that u is provided in terms of a rational function and the numerator as well as denominator contains x therefore to differentiate u partially with respect to x we have to use u by v rule now the rule is write denominator as it is and differentiate the numerator now the numerator is sin 2x its derivative is cos 2x into 2 minus write numerator sin 2x as it is into differentiating denominator with respect to x now it is a function of y therefore its derivative with respect to x is 0 minus now the derivative of cos 2x is minus sin 2x into 2 divided by denominator square now after simplification we get 2 hyperbolic cos 2y into cos 2x minus 2 cos square 2x minus 2 sin square 2x upon hyperbolic cos 2y minus cos 2x bracket square now in the second and third term we can see that it is possible to take minus 2 common and after taking minus 2 common the expression remained in the bracket is cos square 2x plus sin square 2x and its value is 1 therefore we get ux equal to 2 hyperbolic cos 2y into cos 2x minus 2 upon hyperbolic cos 2y minus cos 2x bracket square and similarly differentiating u partially with respect to y we get the derivative minus 2 sin 2x into hyperbolic sin 2y upon hyperbolic cos 2y minus cos 2x bracket square now let us use the formula for the derivative of f of z that is f dash of z equal to ux plus ivx let us call it as equation number one here vx is not known to us therefore let us use the cr equations the cr equations are ux equal to vy and uy equal to minus vx now here the second result is useful we can replace vx as minus uy therefore equation number one becomes f dash of z equal to ux minus i uy and we are knowing the expressions for ux and uy after substituting it we get this equation let us call it as equation number two now to get this right hand side as a function of z let us substitute x equal to z and y equal to zero on the right hand side of above equation two now we get f dash of z equal to two into if we put y zero it is hyperbolic cos of zero and hyperbolic cos of zero is one into cos of two z minus two divided by again this hyperbolic cos 2y is hyperbolic cos zero and its value is one minus cos of two z bracket square plus now here we can see that if I put y equal to zero here it is hyperbolic sin zero and its value zero therefore the imaginary part totally vanishes therefore we can write here plus i into zero now after taking minus two common from the first term in the numerator we get it as minus two into one minus cos of two z upon one minus cos two z bracket square let us remove one one minus cos two z from the numerator and denominator now we get f dash of z equal to minus two upon one minus cos of two z and again we know that one minus cos two z is nothing but two sin square z now we can remove this two and we get f dash of z as minus one upon sin square z and again we know that one upon sin is nothing but cos x therefore we can write minus one upon sin square z as minus cos x square z now the right hand side is obtained as a function of z integrating with respect to z we get now the integration of f dash is f of z is equal to minus is written outside the integral and the integration of cos x square z dz plus integration constant c and we know that integration of cos x square is minus cot z and this plus c as it is and finally we get it as cot z plus c now this is the required analytic function now to find the imaginary part from this analytic function we will write f of z as u plus iv and it is we have obtained as cot z and we know that cot z is nothing but cos z upon sin z again z is nothing but x plus i y therefore we get it as cos of x plus i y upon sin of x plus i y and it is equal to let us multiply and divide by sin of x minus i y now we will get this expression now we know the identity cos a into sin b is nothing but 1 by 2 sin of a plus b minus sin of a minus b now we will get this expression and in the denominator also we can use the identity that sin a into sin b that is equal to 1 by 2 cos of a minus b minus cos of a plus b now we will get this expression after applying this identity now this 1 by 2 will be removed and we get this numerator as sin 2 x minus sin 2 i y upon cos of 2 i y minus cos of 2 x that is equal to sin 2 x as it is and we know that sin 2 i y is nothing but i into hyperbolic sin of 2 y divided by and cos of 2 i y is nothing but hyperbolic cos of 2 y minus this cos 2 x as it is now if we separate it from the numerator we get u plus i v equal to sin 2 x upon hyperbolic cos 2 y minus cos 2 x plus i into minus hyperbolic sin 2 y upon hyperbolic cos 2 y minus cos 2 x now we can see that this is what the given real part in the problem and therefore definitely it is the required imaginary part therefore after comparing we get the imaginary part v as minus hyperbolic sin 2 y upon hyperbolic cos 2 y minus cos 2 x