 Solution

Expanding the derivative limit along real and imaginary increments gives $u_x=v_y$ and $u_y=-v_x$. Since $\mathrm{Re}(z{e}^z)=e^x(x\cos y-y\sin y)$, the required function is $f(z)=z{e}^z$, and $f(0)=0$. Two such functions differ by a holomorphic function with zero real part; the open mapping theorem makes it constant, and the value at zero makes that constant zero.