Find the immaginary part of $f(z)$

Question

Determine the analytic function $f(z)=u(x,y)+iv(x,y)$ knowing that: $$u(x,y)=Ref=\phi\left(\frac{y}{x}\right)$$ where $\phi\in C^2$.

Firstly, I've tried to prove that $u$ is harmonic. $$\frac{\partial^2 u}{\partial x^2}=\phi''\left(\frac{y}{x}\right)\frac{y^2}{x^4}+\phi'\left(\frac{y}{x}\right)\frac{2y}{x^3}$$ $$\frac{\partial^2 u}{\partial y^2}=\phi''\left(\frac{y}{x}\right)\frac{1}{x^2}$$ The sum must be $0$, but then? Is there another way to find it?

Answer 1

Using Cauchy-Riemann, I find

\begin{align} \frac{\partial u }{\partial x} &= \frac{\partial}{\partial x} \phi\left(\frac{y}{x}\right) \\ &= \frac{-y}{x}\phi' \\ &= \frac{\partial v}{\partial y}\\ \frac{\partial u }{\partial y } &= \frac{\partial}{\partial y} \phi\left(\frac{y}{x}\right) \\ &= \frac{1}{x}\phi' \\ &= -\frac{\partial v}{\partial x} \end{align}

Can you integrate up, remembering constants?

Answer 2

You're doing well. You have $$ (\frac{y^2}{x^4}+\frac{1}{x^2})\phi'' \left(\frac{y}{x}\right) + \frac{2y}{x^3} \phi'\left(\frac{y}{x}\right) = 0$$ that is for $t=y/x$ $$ \frac{1}{x^2} \Big((t^2+1) \phi'' \left(t\right) + 2t \phi'\left(t\right) \Big)= 0$$ $$ (t^2+1) \phi'' \left(t\right) + 2t \phi'\left(t\right) = 0 $$ Solving this equation will give you $\phi(t)$.

Another, equivalent way is to use radial coordinates. We have $$ \Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{1}{r}\frac{\partial}{\partial r} \left( r \frac{\partial u}{\partial r} \right)+ \frac{1}{r^2} \frac{\partial^2 u}{\partial \varphi^2}$$

Since $u(x,y) = \phi(\tan \varphi)$ then $\partial u/\partial r = 0$ and you get easily $$ \frac{\partial^2 u}{\partial \varphi^2} = 0$$ $$ u = a\varphi + b = a \arctan \frac{y}{x} + b$$ (at least in the first quadrant).