Question

Obtain the analytic function f(z)=u+iv, whose real part u is ex (x cos y-y sin y).

Answer 1

Answer:

Analytic function F(z)= u + iv becomes

$f(z)= (x cos y - y sin y) + i(\frac{x^2}{2} sin y + x sin y + xy cos y) + C$

Step-by-step explanation:

Here, we have analytic function f(z)= u+iv

and given the real part value,

u = (x cos y - y sin y) ..........(1)

so, by the Cauchy Equations

$\frac{du}{dx} =\frac{dv}{dy} \\and,\\\frac{du}{dy} =-\frac{dv}{dx}$

Partially Differentiating Eq. (1) with respect to x, we get

$\frac{du}{dx} =cos y$ .........(2)

Partially Differentiating Eq. (1) with respect to y, we get

$\frac{du}{dy} =-xsin y - sin y - y cos y$ ........(3)

Eq. (3) equals to the $-\frac{dv}{dx}$

or

$\frac{dv}{dx}= xsin y + sin y + y cos y$ ..........(4)

Now,

integrated equation (4) with respect to x, we get

$v(x,y)=\frac{x^2}{2} sin y+ x sin y + xy cos y + C$

so, analytic function F(z)= u + iv becomes

$f(z)= (x cos y - y sin y) + i(\frac{x^2}{2} sin y + x sin y + xy cos y) + C$

Where, C is the airbitrary constant of integration.