How Do You Find an Analytic Function Where the Argument is xy?

[DianaSagita]

1. This is something from complex analysis: Find the analytic function f(z)= f(x+iy) such that arg f(z)= xy.


2. $$w=f(z)=f(x+iy)=u(x,y)+iv(x,y) (*), w=\rho e^{i\theta} (**)$$

Here are the Cauchy-Riemann conditions...
$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\,\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$$

And polar Cauchy-Riemann conditions...
$$\rho\frac{\partial u}{\partial \rho}=\frac{\partial v}{\partial \theta},\,\rho\frac{\partial v}{\partial \rho}=-\frac{\partial u}{\partial \theta}$$

Also what might be helpful - Re and Im part of analytic function are harmonic functions, so:
$$\frac{\partial^2u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$$
(Laplace's differential equation)


3. Next what I tried to do (if it's correct)...

I transformed (*) to (**) so,

$$w=\sqrt{u^2+v^2} e^{i atan\frac{v}{u}}$$

$$atan\frac{v}{u}=xy$$

$$v=u\,tan(xy)$$

$$w=u\sqrt{1+tan^2(xy)} e^{i xy}= \frac{u(x,y)}{cos(xy)} e^{i xy}=u(x,y)+i[u(x,y) tan(xy)]$$

$${u_x^'}={u_y^'}tan(xy)+\frac{x u}{cos^2(xy)}$$

Now, I assume I got to get system of partial differential equations (by using second condition)...
Is this ok, and is there any easier way of getting analytical function?
Main thing that confuses me is the given argument which is included in real and imaginary part...
Pls help!

 

[DianaSagita]

Well, I must find modulo...
and this one you entered is not analytical function... :(

 

[Unco]

$$atan\frac{v}{u}=xy$$

$$v=u\,tan(xy)$$[/b]

From here and sticking with Cartesian coordinates, writing out v_y and equating with u_x, and writing out v_x and equating with -u_y. gives two equations; then substituting (for instance) u_y from the second equation into the first gives a PDE for u_x (notice there is some cancellation) which can then be solved for u in terms of an arbitrary function of y. Proceeding this way gives the desired analytic function. This is not to say that using previous results from solving Laplace's equation you may be familiar with are not helpful, but they are not necessary.

 

[DianaSagita]

Thanks, and that's what I did... now I got...

$$u=e^{-{\frac{y^2}{2}}}cos(xy)C_1$$
solving by $$u_y^'$$

$$u=e^{\frac{x^2}{2}}cos(xy)C_2$$
solving by $$u_x^'$$
$$C_1,C_2$$ constants
$$C_1=e^{D_1},C_2=e^{D_2}$$
so I have functions...

$$f(z)=e^{D_1-\frac{y^2}{2}+ixy}$$

$$f(z)=e^{D_2+\frac{x^2}{2}+ixy}$$

How two? Now I really... :(

 

[DianaSagita]

I've got it!

$$f(z)=e^{D_1(x)+D_{1num}-\frac{y^2}{2}+ixy}$$

$$f(z)=e^{D_2(y)+D_{2num}+\frac{x^2}{2}+ixy}$$

so...
$$D_1(x)+D_{1num}-\frac{y^2}{2}=D_2(y)+D_{2num}+\frac{x^2}{2}$$
$$D_1(x)=\frac{x^2}{2}$$
$$D_2(y)=-\frac{y^2}{2}$$
$$D_{1num}=D_{2num}=D$$

and analytical function is:


$$f(z)=e^{D+\frac{x^2}{2}-\frac{y^2}{2}+ixy}=Ce^{\frac{z^2}{2}}$$