$f=u+iv$ holomorphic, $xu+yv = (x^2+y^2)e^x \cos y$, what is $f$?

Question

$f(z)=u(x,y)+iv(x,y)$ holomorphic, $xu+yv = (x^2+y^2)e^x \cos y$, what is $f$?

I tried to interprete $xu+yv$ as some part of a new function, for example, as the real part of $\overline{z}f$，but this function is no more holomorphic, so I don't know how to continue. (Maybe $\dfrac{\partial}{\partial\overline{z}}（\overline{z}f)=1$? But how to make use of it?)

Answer 1

What you need to notice is that $$ \frac{f(z)}{z} = \frac{u+iv}{x+iy} = \frac{(u+iv)(x-iy)}{x^2+y^2}, $$ of which the real part is $$ \frac{xu+yv}{x^2+y^2}, $$ which is equal to $e^x \cos{y}$ according to the problem. Now find the harmonic conjugate of $e^x \cos{y}$, which is well-known to be $e^x \sin{y}$ for obvious Euler's formula-related reasons, and rewrite the equation to include the imaginary part: $$ \frac{f(z)}{z} = e^x(\cos{y}+i\sin{y}) = e^{x+iy}=e^z. $$

Answer 2

We have $xu+yv=(x^2+y^2)e^x\cos y$. For notational efficiency, let $G(x,y) = (x^2+y^2)e^x\cos y$.

Here is a procedure for finding

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Step 1:

Take the partial derivative with respect to $x$ and multiply by $x$ to find that

$$x^2u_x+xu+xyv_x=xG_x$$

where the subscript $x$ means the first partial with respect to $x$.

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Step 2:

Take the partial derivative with respect to $y$ and multiply by $y$ to find that

$$y^2v_y+yv+xyu_y=yG_y$$

where the subscript $y$ means the first partial with respect to $y$.

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Step 3:

Now, add these last two equations together and apply the Cauchy-Riemann equations ($u_x=v_y$ and $u_y=-v_x$). This reveals that

$$(x^2+y^2)u_x+G=xG_x+yG_y$$

which implies that

$$u_x=v_y=\frac{xG_x+yG_y-G}{x^2+y^2}$$

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Step 4:

Integrate with respect to $x$ to find $u$ within and integration constant and integrate with respect to $y$ to find $v$ within and integration constant.

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Step 5:

Enforce $xu+yv=G$ for all $x$ and $y$ to find the integration constants. Then $f=u+iv$.