Recover complex function from the real part

Question

I have $\Re f(z) = e^x(xcosy -ysiny) = e^xcosy -e^xysiny$. Therefore, $$v_x= e^xxcosy+e^xcosy-e^xysiny = e^x(xcosy+cosy-ysiny)$$ $$v_y= -e^xxsiny-e^x(siny-ycosy) = e^x(-xsiny-siny+ycosy)$$

$u_x = v_y = e^x(-xsiny-siny+ycosy)$

$u_y = -v_x = -e^x(xcosy+cosy-ysiny)$

$$u = \int u_xdx = -\int e^x(xsiny+siny-ycosy)dx = ... = -e^x(xsiny-ycosy)$$ To find $u_y$, I do:

$$u_y = -e^x(xcosy+cosy-ysiny) = -e^x(xsiny-ycosy)$$ What do I do from here?

Answer 1

First of all, we observe that $$ \mathrm{e}^x(x\cos y-y\sin y)=\mathrm{Re}\,\big(\mathrm{e}^x(\cos y+i\sin y)\cdot (x+iy)\big)=\mathrm{Re}\,(\mathrm{e}^zz). $$ Next, use the fact that, if $g$ is analytic in some region and $\,\mathrm{Re}(g)=0$, then $g\equiv ic$, where $c$ real constant.

Hence the function you are looking for is $\mathrm{e}^zz+ic$.

Answer 2

You know that$$v_x=-u_y=e^x\bigl(y\cos(y)+\sin(y)+x\sin(y)\bigr)$$and$$v_y=u_x=e^x\bigl(\cos(y)+x\cos(y)-y\sin(y)\bigr).$$Integrating, you get that$$v(x,y)=e^x\bigl(y\cos(y)+x\sin(y)\bigr)+k$$Therefore\begin{align}f(x+yi)&=u(x,y)+v(x,y)i\\&=e^x\bigl(x\cos(y)-y\sin(y)\bigr)+e^x\bigl(y\cos(y)+x\sin(y)\bigr)i+ki\\&=e^x\bigl((\cos(y)+\sin(y)i\bigr)(x+yi)+ki\\&=e^{x+yi}(x+yi)+ki\end{align}and so $f(z)=ze^z+ki$.