# Harmonic functions proofs

a.) Show that if $v(x,y)$ is a harmonic conjugate of $u(x,y)$ in a domain $D$, then every harmonic conjugate of $u(x,y)$ in $D$ must be of the form $v(x,y)+a$, where $a$ is a real constant.

b.) Suppose that $f(z)$ is analytic and nonzero in a domain $D$. Prove that $ln|f(z)|$ is harmonic in $D$.

I know initutively why a is true, I also know that I must use Cauchy-Riemann equations to prove it but I do not know how to apply it correctly. For b, I am stumped.

## 1 Answer

(a)

Let $w$ be another such harmonic conjugate and so $z = u(x,y) + i w(x,y)$ where both $u$ and $w$ are real functions. $u_x = w_y$ so $u = \int w_y dx + C(y)$ where $C(y)$ is a pure real function of $y$.

Similarly, $u_y = -w_x$ so $u = -\int w_x dx + D(x)$ where $D(x)$ is a pure real function of $x$.

You know one possible antiderivative corresponding to $\int w_y dx$ and $-\int w_x dx$ already, namely $v(x,y)$ and so $C(y) = D(x)$. This is only possible when $C(y) = D(x) = a$, with $a$ a real constant.

(b)

$\ln f(z) = \ln|f(z)| + i \arg z$ and so it is the real part of an analytic function and so it is harmonic. I don't know if you're expected to do more than this

• Does $u_x = w_y$ because of the Cauchy equations? And why does $u_y = v_x$ mean that $u = \int w_x dx + D(x)$, did you mean $u_y = w_x$ instead of $v_x$? – Q.matin Mar 2 '13 at 6:36
• Sorry, I posted it without really proof-reading it. Let me fix up a few details – muzzlator Mar 2 '13 at 6:37
• And yes, $u_x = w_y$ by the Cauchy-Riemann equations for $u$ being a harmonic conjugate of $w$. – muzzlator Mar 2 '13 at 6:39
• Oh and for part b it gives a hint saying, let $\phi (x,y)=ln|f(z)|=\frac{1}{2}ln(u^2 +v^2)$, will that be expected for an alternative proof? – Q.matin Mar 2 '13 at 6:47
• Ah, then you'll be expected to do a little more than what I just said. They are probably explicitly asking you to come up with a harmonic conjugate through the integration techniques I used in part (a). The harmonic conjugate you'll get will probably be $\arctan(\frac{v}{u})$ which is $\arg z$ (Special case will need to be made for $u=0$ but it isn't hard.) – muzzlator Mar 2 '13 at 6:54