Show this function is holomorphic

Question

Let $f:ℂ→ℂ$ be a holomorphic function such for $x, y ϵ ℝ$,

$\operatorname{Im}f(x + iy) = x + y$

Find f and check that it is indeed holomorphic.

Firstly I have put $f = u+iv$ and $u = x + y$, and because f is holomorphic in $ℂ$ the Cauchy-Riemann equations hold.

So $\frac{dv}{dx} = \frac{-du}{dy} = -1$.

I'm not really sure what to do next, can anyone help?

Answer 1

If $\Im(f(x+iy))=x+y$ and $f(x+iy)=u(x,y)+iv(x,y)$, then $v(x,y)=x+y$. Take $u(x,y):=x-y$. Then \begin{align*} f(z)&=f(x+iy)=u(x,y)+iv(x,y)=x-y+i(x+y) \\ &=x+iy-y+ix=x+iy+i(x+iy) \\&=z+iz, \end{align*} so $f$ is indeed holomorphic.

Answer 2

A theorem, due to the textbook by Stein and Shakarchi (theorem 2.4, page. 13), gives a sufficient condition for a function being holomorphic:

Suppose $f = u + iv$ is a complex-valued function defined on an open set $\Omega$. If $u$ and $v$ are continuously differentiable and satisfy the Cauchy-Riemann equations on $\Omega$, then $f$ is holomorphic on $\Omega$.