# How to Find Analytic Function When Imaginary Part Is Given – Complex Variables – Engineering Maths 3

Hello friends so in this video we are going to learn a concept based on analytic function where we are going to find out the value of analytic function when the imaginary part of that function is given so it means we want to find out the analytic function when only imaginary part is there but real part is not given so for that what we’ll do is we’ll consider 1 analytic function so let us consider analytic function W that is function of Z and we know that it is defined by U of X comma Y plus I V of X comma Y so let F of Z be the analytic function now we have to find out the analytic function that is f of Z when this V is given since V is the imaginary part so there are certain cases in the numericals where we’ll come across the imaginary part that is V will be given and they will ask you to find out the value of analytic function or analytic function f of Z so in such cases if u is not given then how to get the value of f of Z so let’s see these steps so in step number 1 we will find out the value of dou v by dou X and dou V by dou Y since V is given and we get this value of dou v by dou X and dou V by dou Y by just partially differentiating V with respect to X and y after that we are going to use the identity or the property of analytic function so according to property of analytic function if F of Z is analytic function then its derivative F dash of Z is given by the formula dou V by dou y plus I do B by dou X so this is the formula of F dash of Z if f of Z is analytic function now since it is said that imaginary part of analytic function is given it means the given function or the function that we are finding is analytic and hence we can use this property of analytic function now whatever value that we got in step number one we are going to substitute these two values in step number two and we are going to get the value of M dash of Z but let me tell you my friends that since V is function of x and y and you are differentiating V with respect to x and y partially you will get value of F dash of Z in terms of X and why because dou V by dou n dou V by dou X both are in terms of x and y now the question is to get the value of analytic function f of Z and for that we need to integrate this F dash of Z with respect to Z and if I will integrate that F dash of Z with respect to Z then I should have that F dash of Z in terms of Z now the question is how to convert this function f dash of Z which is now in terms of x and y into terms of Z so to convert this F dash of Z in terms of Z we have to use something and that method is called as milk thompson’s method so friends I am going to apply the milk constants method in which we will convert this F dash of Z in terms of Z by substituting or by replacing X with Z and Y by zero in this F dash of Z so here I will say apply Mill functions method by substituting X as Z and Y as zero and once we get the value of F dash of Z in terms of Z we are going to integrate that F dash of Z with respect to X with respect to Z so here I will say integrating F dash of Z with respect to Z and we will get the value of F of Z so friends we are going to apply this method which I have just explained you in one numerical so we have a numerical where the imaginary part of the analytic function is given which is V equal to X square minus y square plus X upon X square plus y square and we have to show that the real part U is minus 2xy plus y upon X square plus y square plus C now friends if we want to show that the real part is equal to given value then first of all we need to find out the value of f of Z that is the analytic function and from that analytic function since V is given F of Z we know we can easily find out D u that is the missing term now how to get the value of f of Z from V and for that we are gonna use this steps so in step number one we will get the value of dou V by dou X and dou V by dou Y by differentiating the V term with respect to X and y partially so let’s start so here I will say V is equal to X square minus y square plus X upon X square plus y square so therefore here we’ll get dou V by dou X is equal to 2x this the derivative of Y with respect to X partially will be 0 next year X we have a numerator and denominator so we are going to use u by V root and according to it we will get in denominator X square plus y square the whole square next X square plus y square the derivative of X that is 1 minus X as it is and derivative of X square is 2 X and derivative of Y square is 0 next so here we get the value as 2 X any of you will get minus 2 X square and here X square so we will get minus X square Y square is positive so y square as it is upon X square plus y square the whole square similarly we will get the value of dou V by dou Y so for that we’ll differentiate this V with respect to Y partially so derivative of X square is 0 Y square is minus 2y next plus X is constant outside and the derivative of this term is minus 1 upon X square plus y square the bracket square into 2 y so here we will get minus 2y minus 2xy upon X square plus y square the whole square now after that we will find out the value of M dash of Z which is dou V by dou y plus I do V by dou X so for that we’ll put these two values in step number 2 so we will get so now we got this value of M s of Z and we got the value of F dash of Z in terms of X and mine now we are going to apply this milk Thompson’s method to get the value of F dash of Z in terms of Z and for that we will substitute X equal to Z and y equal to zero in F dash of Z so here I will just say buy milk Thompson method X is Z and y 0 so this would give me F dash of Z is equal to Y is 0 so first term is 0 y 0 this term is also 0 to Z I Y is 0 so this term is 0 here I will get minus Z square into I so minus i Z square upon y 0 z square the whole square that is e raise to 4 so you have be able to get to Z I – I upon Z square so now we got the value of F dash of Z so to get the value of F of Z we will follow the last step that is integrating the FF dash of Z with respect to Z and for that here I will say integrating both sides with respect to Z so so here we will get F of Z equal to 2 as it is I as it is the integration of Z is Z square by 2 next – I as it is and the integration of 1 upon Z square is minus 1 by Z plus C so here by cancelling this 2 and we can take this I common so we will get Z square minus – plus 1 by Z plus C so here we got the value of f of Z but my friends if you will see the question then here it is given that we have to show that the real part is this now here I got the value of F of Z but I should find out the value of real part so to find out the value of real part what we’ll do is we will again resub stitute Z as X plus iy because there is a complex number and then we will take real parts together imaginary parts together and we’ll see whether we are getting the real part that is U as the given question or not so for that I will put Z as X plus iy so here I will say put Z as X plus iy in f of Z so therefore the value of F of Z would be I in bracket we had Z square plus 1 by Z so this will become X plus iy the whole square plus 1 upon X plus iy plus C so by solving this here we will get I the square of the bracket is X square plus 2i XY plus I square Y Square where I square is minus 1 but for simplicity I will keep I square as it is here next this 1 upon X plus iy we know that it is X minus iy so here I again use the formula that is 1 upon X plus iy is equal to X minus iy so here we will get I X square is the real term or real part here I squared is minus 1 so we will get minus y square which is real so X square minus y square and here we will get plus X next in these two terms I can take I common and here we will get 2xy minus y plus C so we will put Z as X plus iy in f of Z so therefore F of Z will become I into X plus iy the whole square plus 1 upon X plus iy so I into X plus iy the whole square plus 1 upon X plus iy plus C now we will simplify this terms so here we will get I as it is this will be X square plus 2i XY plus I square Y square so I am applying the formula a plus B the whole Square and here since we have the complex number in the denominator we will rationalize it so here we will get 1 upon X plus iy into X minus iy and at the same time I’ll write the same term in the numerator plus C so I as it is this will be X square and here I square is minus 1 so we will get minus y square so X square minus y square plus 2i XY here we will get X minus iy upon X plus iy into X minus iy is X square plus y square plus C now if will separately divide each term in the numerator by X square plus y square then we would get eye X square minus y square plus 2i XY plus X upon X square plus y square minus iy upon X square plus y square plus C now if we’ll take all the real terms together then we will get X square minus y square plus X upon X square plus y square and from these two imaginary terms we can take I common and we will get to X Y minus y upon X square plus y square plus C now we have one more item outside the bracket so I will take this I inside so here we will get this real part as imaginary because we are going to multiply this first term with I and the second term which has I will become I into I I square and that will be converted into real point so here we will get is equal to I into X square minus y square plus X upon X square plus y square next this is I into I I square that is minus 1 so we will get minus 2xy minus y upon X square plus y square plus C now if I will write the real part first then this will become minus into 2xy that is minus 2xy this will become minus minus plus y upon X square plus y square so this is the real part and the imaginary part would be X square minus y square plus X upon X square plus y square plus C so here we got the real part and imaginary part of F of Z where the real part is called as u and the imaginary part is called as B and if you will see the question then the question was that show that the real part U is minus 2xy plus y upon X plus y-squared plus C if you see here then we already got the real part as minus 2xy plus y upon X square plus y square and this C so here we can say that therefore real part that is U is equal to minus 2xy plus y upon X square plus y square plus C so this is the required proof thank you

I was dis appointed when you simplifey the complex conjugate but latter great

hahahahha

awesome bro . you know what students need .

well explained bro

Sir Sara question again upload kariye plz

I want to show cauchy theorem by ekeeda

sir short trick btao solve krne ki

and sir civil geo tech. start krdo and fluid mechanics and hydraulic machines

how to find analytic function when real part U is given ……in Cartesian coordinates ……..plsss upload the sums of the given statement

how to find analytic function when real part U is given ……in Cartesian coordinates ……..plsss upload the sums of the given statement

how to find analytic function when real part U is given ……in Cartesian coordinates ……..plsss upload the sums of the given statement

Didn't know about the Milne-Thompson method. Thanks for uploading!

Nice job

1/(x+iy) = x-iy ??? 😂😂😂😂😂 Very good teaching stuffs

Sir please upload some examples with proof the existence of CR equation doesn't imply the analicity of the function .

bro please solve milne thompson method

Trick tell

GREAT SIR

THE EXAMPLE U HV SOLVED IS ASKED IN ESE 2017

Samiksha message. Me

Is there any way to do this without MT method? we havent covered that yet

Please put less advertising….this is an educational institution…not for the sake of money

How -ve sign has come in dv/dy=-2y-2xy/(x^2+y^2)^2

Hello Friends,

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can anyone please help solving this.?

prove that lim x-0(1+x)1/x=e.

and

if f(z)=u+iv and u-v=ex(cosy-siny) find f(z)

Got exam tomorrow this was only left which I decided to leave. But gave a try to your video and surely I'll attempt it tomorrow and will clear it 😊.

Tysm sir

Abe chutiye, galat kyu batar 1/x+iy= x-iy , agar galati hui ho toh vo part delete kar deta , upload kyu kiya