[SOLVED] holomorphic function

**civodul** · September 26th 2008, 07:09 AM

Hello,

I have to find all holomorphic functions F(z)=F(x+iy)

F(x+iy)=P(x,y) + i Q(x,y)

with the condition (1) that P(x,y)/Q(x,y) is a function of x only.

Combining (1) with cauchy relations I reach that the module of F depends only on y. But I can't get more precise, though I think it must be possible.

Thanks for any help.

civodul

**shawsend** · September 26th 2008, 01:30 PM

I'm no expert with this but I don't think there are any non-constant functions which satisfy this relation:

Let $f(z)=u(x,y)+iv(x,y):$

If $\frac{u}{v}=G(x)$ then $\frac{u}{v}=\frac{g(x)h(y)}{r(x)h(y)}=\frac{g}{r}= G(x)$ . Now if $f$ is holomorphic, then $u(x,y)=g(x)h(y)$ satisfies Laplace's equation: $g''h+h''g=0$ or $\frac{g''}{g}=-\frac{h''}{h}$ . The only way that's true is if $g(x)=ax+b$ and $h(y)=cy+d$ . Letting $u(x,y)=(ax+b)(cy+d)$ , then the harmonic conjugate of this is found to be $v(x,y)=\frac{acy^2}{2}+ady-\frac{cax^2}{2}-cbx+k$ and this is not a product solution of the form $v(x,y)=r(x)h(y)$ .

I hope someone here can confirm or prove this wrong.

**Opalg** · September 26th 2008, 01:51 PM

Write $F(z) = re^{i\theta}$ , where r is a function of y only, and $\theta$ is a function of x and y.

The condition that P/Q is a function of x alone tells us (implicitly) that F(z) is nonzero. Therefore (locally, at least) we can form $\ln(F(z)) = \ln r + i\theta$ , which will be an analytic function. Apply the Cauchy–Riemann equations to this function. They tell you that $\theta$ is a function of x only, and in fact $\theta(x) = \alpha x+\beta$ (where $\alpha$ and $\beta$ are real), and that $r'(y)/r(y) = -\alpha$ . That leads to the conclusion that $F(z) = \lambda e^{i\alpha z}$ , where $\lambda\in\mathbb{C}$ and $\alpha\in\mathbb{R}$ are constants.

**civodul** · September 27th 2008, 12:56 AM

Originally Posted by Opalg

The condition that P/Q is a function of x alone tells us (implicitly) that F(z) is nonzero.

Thanks for your very clear answer.
There is still one point that is not so obvious to me:
Why would the fact that P/Q is a function of x alone entail that F(z) is non zero?

civodul

**Opalg** · September 27th 2008, 01:03 AM

Originally Posted by civodul

Why would the fact that P/Q is a function of x alone entail that F(z) is non zero?

What I meant was that the question, by referring to the fraction P/Q, implicitly assumes that Q (and therefore F) is nonzero.

**civodul** · September 27th 2008, 01:15 AM

Originally Posted by Opalg

What I meant was that the question, by referring to the fraction P/Q, implicitly assumes that Q (and therefore F) is nonzero.

That is what I assumed you meant but I wasn't completely sure.

Civodul

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