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.

ProofSupposef(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)]is analytic. Now examine^{2}[f(z)]and finish the proof.^{2}

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. Sincef(z) = u(x,y) +iv(x,y)is an analytic function then[f(z)]is also analytic. Therefore:^{2}

[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, thenu(x,y)v(x,y)is harmonic.

I feel like I'm not saying anything, though! Any insight?

Another attempt: Supposef(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)]is analytic. Thus:^{2}

[f(z)]=^{2}[u(x,y) + iv(x,y)]=^{2}[u(x,y)]Since u(x,y) is a harmonic equation then [u(x,y)]^{2}- [v(x,y)]^{2}+ i2u(x,y)(v(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?