= Solution
The <Cauchy-Riemann equations> determine the <harmonic conjugate>. First
$$
u_x=e^x[(x+1)\cos y-y\sin y],
$$
so $v_y=u_x$ integrates with respect to $y$ to
$$
v=e^x(x\sin y+y\cos y)+h(x).
$$
Now
$$
u_y=e^x[-(x+1)\sin y-y\cos y],
$$
and the second <Cauchy-Riemann equation>, $v_x=-u_y$, forces $h'(x)=0$. Hence
$$
f=e^x\{x\cos y-y\sin y+i(x\sin y+y\cos y)\}+iC=ze^z+iC.
$$
The normalization at zero gives $C=0$, so the required <entire function> is
$$
\boxed{f(z)=ze^z.}
$$
Back to article page
