They ask me to find the analytic function $f(z)=u(x,y)+iv(x,y)$ if we know that $$u(x,y)=\frac{e^x( (x^2-y^2)\cos y+ 2xy\sin y)}{(x^2+y^2)^2}.$$
Am I supposed to find $v(x,y)$ satisfying the Cauchy-Riemann conditions in cartesian coords.? Isn't it too cumbersome to integrate and differentiate $u$? I don't think polars make it easier since there'd be $\sin(r\sin\theta)$ type functions and a quotient...
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One method. Substitute $$ x = \frac{z+\overline{z}}{2},\qquad y=\frac{z-\overline{z}}{2i} $$ into $u(x,y)$, then simplify the expression. Result: $$ \frac{e^z}{2z^2}+\frac{e^\overline{z}}{2\overline{z}^2} . $$ Then do it backward, to conclude $$ u(x,y) = \operatorname{Re}\;\frac{e^z}{z^2} . $$
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Let $f(z), g(z)$,in short $f, g$, be two complex functions. Then we have $$\Re(fg)=\Re(f)\Re(g)-\Im(f)\Im(g).$$ The OP's function $$u(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2} e^x\cos y -\left(-\frac{2xy}{(x^2+y^2)^2}\right)e^x\sin y.$$ is $$u(x,y)=\Re(\tfrac1{z^2})\Re(e^z)-\Im(\tfrac1{z^2})\Im(e^z)=\Re( \tfrac1{z^2}e^z).$$
It's cumbersome, but not too cumbersome. On the other hand, it is a mistake to ask for the analytical function whose real part is the given function; there are infinitely many of them.
Here's a way of getting one of them. Take\begin{align}v(x,y)&=\int_0^y\frac{\partial u}{\partial x}(x,t)\,\mathrm dt\\&=\frac{e^x\left(\left(x^2-y^2\right) \sin (y)-2 x y \cos (y)\right)}{\left(x^2+y^2\right)^2}.\end{align}It is then clear that $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$. And a direct computation shows that $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$. So, you can take\begin{align}f(x+yi)&=u(x,y)+v(x,y)i\\&=\frac{e^x\left(\left(x^2-y^2\right)\cos(y)+2xy\sin(y)\right)}{\left(x^2+y^2\right)^2}+\frac{e^x\left(\left(x^2-y^2\right)\sin(y)-2xy\cos(y)\right)}{\left(x^2+y^2\right)^2}i\\&=\frac{e^x((x^2-y^2)e^{yi}-2xyie^{yi})}{(x^2+y^2)^2}\\&=\frac{e^{x+yi}(x-yi)^2}{(x+yi)^2(x-yi)^2}\\&=\frac{e^{x+yi}}{(x+yi)^2.}\end{align}In other words, you can take $f(z)=\dfrac{e^z}{z^2}$.