# [SOLVED]Find holomorphic function f

#### dwsmith

##### Well-known member
On $U=\mathbb{C}-\{0\}$, find a holomorphic function $f=u+iv$.
$$u(x,y) = \frac{y}{x^2+y^2}$$

u is harmonic on U

Let g be a primitive for f on U.

write $g=\varphi +i\psi$.

Then $\varphi_x = u$.

$$\varphi_{xy} = \psi_{xy} = \frac{x^2-y^2}{(x^2+y^2)^2}$$

So I can integrate the above with respect to x and find a function with the constant of integration being some h(y).

Then I would have v and I would have found my function f correct?

So I found $f$ to be
$$f(z) = u + iv = \frac{y}{x^2 + y^2} + i\left[\frac{x}{x^2 + y^2} + h(y)\right].$$

Correct?

Last edited:

#### Opalg

##### MHB Oldtimer
Staff member
On $U=\mathbb{C}-\{0\}$, find a holomorphic function $f=u+iv$.
$$u(x,y) = \frac{y}{x^2+y^2}$$

u is harmonic on U

Let g be a primitive for f on U.

write $g=\varphi +i\psi$.

Then $\varphi_x = u$.

$$\varphi_{xy} = \psi_{xy} = \frac{x^2-y^2}{(x^2+y^2)^2}$$

So I can integrate the above with respect to x and find a function with the constant of integration being some h(y).

Then I would have v and I would have found my function f correct?

So I found $f$ to be
$$f(z) = u + iv = \frac{y}{x^2 + y^2} + i\left[\frac{x}{x^2 + y^2} + h(y)\right].$$

Correct?
That is correct as far as it goes, but it would look a lot better if you pushed it a bit further. For a start, you are only asked for a holomorphic function $f$, not for all such functions. So you can ditch the term $ih(y)$ as being unnecessary.

next, you should try to simplify the terms that are left, and to express the answer by giving $f$ as a function of $z = x+iy$. In fact, $u+iv = \dfrac{y+ix}{x^2+y^2}$, and you can factorise the denominator as $(y+ix)(y-ix).$ Since $y-ix = -i(x+iy) = -iz$, that gives you a simple formula for $f(z)$ in terms of $z$ alone.