# Math Help - Complex Harmonies

1. ## Complex Harmonies

I've been in trouble enough lately, I figured I would get up to speed on a few things. Pondering elementary Complex Analysis, I made it only to page 53 before I managed to get stuck.

I want a non-zero Harmonic Function that disappears on the hyperbola y = 1/x.

I'm staring at f(z) = z^2.

I created a lovely page (or 3) of algebraic manipulations that lead to interesting results and relationships, but don't solve the problem at hand. I figure I'm just missing something.

Do I get a hint? What about f(z) = z^2 is supposed to impress me?

2. I do not understand. You want an analytic function $f:\mathbb{C}\mapsto\mathbb{C}$ so that $f\not = 0$ and $f\left( x + i\frac{1}{x} \right)=0$ for all real $x\not = 0$. Is that what you are asking? Because the word "harmonic" is strange here.

Try $f(z) = z^2 - \bar z^2 - 4i$. This is not everywhere analytic but it almost works.

3. What does "disappears on the hyperbola y = 1/x" mean here?

4. Right. I seem already to have picked up the somewhat cheesy vocabulary of the author. Maybe I need a new book. Also, it seems I am assuming you can see what I'm seeing. How many students have I told not to do that?!

In any case:

1) It's u(x,y) on $\Re^{2}\mapsto\Re^{2}$
2) "disappears" or "vanishes" seems to mean "u(x,y)= 0"

So u(x,1/x) = 0, u(x,y) is not constant, and u(x,y) satisfies Laplace's Equation, second partial derivatives sum to zero.

I have a hint, "Look at f(z) = z^2". Sadly, this "look" doesn't seem to be clarifying anything.

5. Can you try $u(x,y) = 2xy - 2$? It is harmonic.
The conjugate is $v(x,y) = y^2 - x^2$.

6. Hmpf. I still don't see why the author wanted me to look at z^2 for a clue. I wasn't expecting a unique result, but it seems I haven't found the one the author was hinting at.

Time to move on.

7. Originally Posted by TKHunny
Hmpf. I still don't see why the author wanted me to look at z^2 for a clue.
Actually I did use the author’s hint.
If v is the harmonic conjugate of u then -u is the harmonic conjugate of v.
$z^2 = (x^2 - y^2 ) + i(2xy)$

8. Good call. I begin to see. I'm sure I'll be back in another 50 pages.