Solution
= Solution
Expanding the <derivative> <limit> along real and imaginary increments gives $u_x=v_y$ and $u_y=-v_x$. Since $\operatorname{Re}(ze^z)=e^x(x\cos y-y\sin y)$, the required <function> is $f(z)=ze^z$, 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.