# Thread: [Complex Analysis] Am I heading the right direction with this proof? Feel like it's m

1. ## [Complex Analysis] Am I heading the right direction with this proof? Feel like it's m

In this problem, you are guided through the start of the proof of the proposition:
If u(x,y) is a harmonic function and v(x,y) is its harmonic conjugate, then the function φ(x,y) = u(x,y)v(x,y) is harmonic.
Proof Suppose f(z) = u(x,y) +iv(x,y) is analytic in a domain D. We saw in Section 3.2 that the product of two analytic functions is analytic. Hence [f(z)]2 is analytic. Now examine [f(z)]2 and finish the proof.

This is what I have so far: If u(x,y) is a harmonic function and v(x,y) is its harmonic conjugate, then the function φ(x,y) = u(x,y)v(x,y) is harmonic. Since f(z) = u(x,y) +iv(x,y) is an analytic function then [f(z)]2 is also analytic. Therefore:
[f(z)]2 = [u(x,y) + iv(x,y)]2 = [u(x,y)]2 - [v(x,y)]2 + i2u(x,y)(v(x,y)
Sine the product of a harmonic function is harmonic, then u(x,y)v(x,y) is harmonic.
I feel like I'm not saying anything, though! Any insight?

Another attempt: Suppose f(z) = u(x,y) +iv(x,y) is analytic in a domain D. We saw in Section 3.2 that the product of two analytic functions is analytic. Hence [f(z)]2 is analytic. Thus:
[f(z)]2 = [u(x,y) + iv(x,y)]2 = [u(x,y)]2 - [v(x,y)]2 + i2u(x,y)(v(x,y)
Since u(x,y) is a harmonic equation then [u(x,y)]2 is also harmonic. Similarly, since v(x,y) is the harmonic conjugate of u(x,y) then [v(x,y)]2 is also the harmonic conjugate of [u(x,y)]2.
Therefore, focusing on the term i2u(x,y)(v(x,y) and finding the partial derivatives with respect to x and respect to y respectively: Here I do the math.
When put into LaPlace's Formula it simplified to 4i[VxUx + UyVy]. Since the function is analytic the CR equations hold and with substitution the function is 4i[Vx(Vy) + (-Vx)Vy) = 4i(o0 =0. Showing the product of conjugate harmonic components is also harmonic.

Is this too big of a jump? Can I just assume since u(x,y) is a harmonic equation then [u(x,y)]2 is also harmonic?

2. ## Re: [Complex Analysis] Am I heading the right direction with this proof? Feel like it

just show that $\nabla^2 \varphi(x,y) = 0$

$\varphi(x,y)=uv$

$\varphi_x=\dfrac {\partial}{\partial x} \varphi = u_x v + u v_x$

$\varphi_y=\dfrac {\partial}{\partial y} \varphi = u_y v + u v_y$

$\varphi_{xx}=\dfrac {\partial}{\partial x}\left(u_x v + u v_x\right)=u_{xx}v + 2 u_x v_x + u v_{xx}$

$\varphi_{yy}=\dfrac {\partial}{\partial y}\left( u_y v + u v_y \right) = u_{yy} v + 2 u_y v_y + u v_{yy}$

$\varphi_{xx}+\varphi_{yy}=v(u_{xx}+u_{yy}) + u(v_{xx}+ v_{yy}) + 2(u_x v_x + u_y v_y) = 0 + 0 + u_x (-u_y) + u_y (u_x)=0$

the Laplacians of u and v are each 0 as they are each harmonic functions.

the last equality is from the CR equations of u and v which are satisfied as they are harmonic conjugates of one another.

3. ## Re: [Complex Analysis] Am I heading the right direction with this proof? Feel like it

Let me put your suggestions into proof form and I'll post it in a jiffy. Thanks!

4. ## Re: [Complex Analysis] Am I heading the right direction with this proof? Feel like it

Originally Posted by romsek
just show that $\nabla^2 \varphi(x,y) = 0$

$\varphi(x,y)=uv$

$\varphi_x=\dfrac {\partial}{\partial x} \varphi = u_x v + u v_x$

$\varphi_y=\dfrac {\partial}{\partial y} \varphi = u_y v + u v_y$

$\varphi_{xx}=\dfrac {\partial}{\partial x}\left(u_x v + u v_x\right)=u_{xx}v + 2 u_x v_x + u v_{xx}$

$\varphi_{yy}=\dfrac {\partial}{\partial y}\left( u_y v + u v_y \right) = u_{yy} v + 2 u_y v_y + u v_{yy}$

$\varphi_{xx}+\varphi_{yy}=v(u_{xx}+u_{yy}) + u(v_{xx}+ v_{yy}) + 2(u_x v_x + u_y v_y) = 0 + 0 + u_x (-u_y) + u_y (u_x)=0$

the Laplacians of u and v in each 0 as they are each harmonic functions.

the last equality is from the CR equations of u and v which are satisfied as they are harmonic conjugates of one another.
I'm looking through my Complex Analysis notes and I'm not seeing why $u_x (-u_y) + u_y (u_x)=0$. Is this a known theorem?

5. ## Re: [Complex Analysis] Am I heading the right direction with this proof? Feel like it

Originally Posted by WingardiumLeviOsa
I'm looking through my Complex Analysis notes and I'm not seeing why $u_x (-u_y) + u_y (u_x)=0$. Is this a known theorem?
look at it more closely

$\Large u_x (-u_y) + u_y (u_x) = u_x u_y - u_x u_y = 0$

6. ## Re: [Complex Analysis] Am I heading the right direction with this proof? Feel like it

Originally Posted by romsek
look at it more closely

$\Large u_x (-u_y) + u_y (u_x) = u_x u_y - u_x u_y = 0$
Wow... I know I'm blind but I didn't know I was that blind! Thanks! It's been a long night.

7. ## Re: [Complex Analysis] Am I heading the right direction with this proof? Feel like it

Originally Posted by romsek
look at it more closely

$\Large u_x (-u_y) + u_y (u_x) = u_x u_y - u_x u_y = 0$

Does that look great or do I need to explain myself in other parts? Thank you so much for helping me see this.

8. ## Re: [Complex Analysis] Am I heading the right direction with this proof? Feel like it

Originally Posted by WingardiumLeviOsa

Does that look great or do I need to explain myself in other parts? Thank you so much for helping me see this.
looks fine to me

9. ## Re: [Complex Analysis] Am I heading the right direction with this proof? Feel like it

Originally Posted by romsek
looks fine to me
Thanks!