We have $z=x+iy \in \mathbb{C}$
is there exists entire function $f(z)$ such that $$Re(f(z))=e^x(xcos(y)-ysin(y))+2sin(x)sinh(y)+x^3-3xy^2+y$$
i already tried it like this:
$$Re(f(z))=u(x,y)\\u_x(x,y)=e^x(xcos(y)-ysin(y)+cos(y))+2cos(x)sinh(y)+3x^2-3y^2\\u_y(x,y)=e^x(-xsin(y)-sin(y)-ycos(y))+2sin(x)cosh(y)-6xy+1$$
i know from the C-R, the function is entire when $u_x=v_y$ and $u_y=-v_x$, so i did it and integrate the $v_x$ and $v_y$.
$v=2cosh(y)cos(x)+(xsin(y)+ycos(y))e^x+3yx^2-x+C$ and $v=2cosh(y)cos(x)+(xsin(y)+ycos(y))e^x+3yx^2-y^3+C$
so what is the conclusion? does the entire function $f(z)$ exists when $x=y^3$?
$$v=2cosh(y)cos(x)+(xsin(y)+ycos(y))e^x+3yx^2−y^3-x+C$$ is the answer, so the complete answer will be:
$$f(z)=e^x(xcos(y)−ysin(y))+2sin(x)sinh(y)+x^3−3xy^2+y+i(2cosh(y)cos(x)+(xsin(y)+ycos(y))e^x+3yx^2−y^3-x+C)$$