Finding Analytic $f(z)$

Question

Let $$u(x,y)=xe^x\cos y-ye^x\sin y$$ find $$f(z)=u(x,y)+iv(x,y)$$ such that $f(z)$ is analytic

Using the theorem: if $u(x,y)$ is harmonic then there is $v(x,y)$ such that $f(z)=u(x,y)+iv(x,y)$ is analytic we first test if $u(x,y)$ is harmonic ($\Delta u=u_{xx}+u_{yy}=0$)

$$u_x=(e^x+xe^x)\cos y-ye^x\sin y$$

$$u_{xx}=(2e^x+xe^x)\cos y-ye^x\sin y$$

$$u_y=-xe^x\sin y-e^x\sin y-e^xy\cos y$$

$$u_{yy}=-xe^x\cos y-2e^x\cos y+e^xy\sin y$$

And so $u_{xx}+u_{yy}=0$ and $u(x,y)$ is harmonic.

Now let assume the that $C-R$ holds

So $$u_x=v_y=(e^x+xe^x)\cos y-ye^x\sin y$$ integrating results with $$v=e^x\cos y+xe^x\sin y+C(x)$$

so $$v_x=ye^x\cos y+e^x\sin y+xe^x\sin y+C'(x)$$

Using the second $C-R$

$$u_y=-v_x=-xe^x\sin y-e^x\sin y-e^xy\cos y$$

comparing the two $v_x$ to find $C(x)$ yields $C(x)=k$

So $$f(z)=u+iv=xe^x\cos y-ye^x\sin y+i(xe^x\sin y+e^x\sin y+ye^x\cos y)$$

Is the process correct? How should I continue to find $f(z)$ as expression of $z$ ?

Answer 1

I think you can use that $$f(z)=2u\left(\frac{z}{2}, -\frac{iz}{2}\right)-u(0,0)$$

Answer 2

Let $g(z)$ be the result of substituting $x=z$ and $y=0$ in your expression for $f$, i.e. $$ g(z) = ze^z. $$ Then $g$ is entire and so is $f$ (as it is constructed). Also $f = g$ on the real axis by construction. Hence, by the identity theorem, $f = g$ everywhere.

Answer 3

Use the Cauchy-Riemann equations

If $f(x,y) = u(x,y) + i v(x,y)$ is such that

$$ u_x = v_y\\ v_x = -u_y $$

then $f(x,y)$ is analytic.

Now we have $$u(x,y)=xe^x\cos y-ye^x\sin y$$

then

$$ v_y = u_x \Rightarrow v(x,y) = \int u_x dy + \phi_1(x)\\ v_x = -u_y \Rightarrow v(x,y) = -\int u_y dx + \phi_2(y) $$

and after those operations we get

$$ v(x,y) = e^x(y \cos y + x \sin y) $$