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How do I find a analytic function such that $\displaystyle \mathfrak{Re}(f) =u(x,y)= \frac{y}{x^{2}+y^{2}}$.

I can call the real part $u(x,y)$ and by Cauchy-Riemann I will have $u_{x}=v_{y}$ and $u_{y}=-v_{x}$. So $$v_{y}=u_{x}(x,y)= -\frac{2x}{(x^{2}+y^{2})^{2}} \ ; \qquad v_{x}=-u_{y}=\frac{1}{x^{2}+y^{2}}$$

After this what should I do? An elaborate solution will help.

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  • $\begingroup$ Check your computations for $u_x$ and $u_y$ again! $\endgroup$ – Hagen von Eitzen Jun 9 '13 at 16:12
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By eyeballing the formula, we have that $y=\Re(-iz)$ and $x^2+y^2=z\overline z$. So the function $$f(z)=\frac{-iz}{z\overline z}=\frac{-i}{\overline z}$$ would work - if only it were analytic. Now finally observe that complex conjugation does not change the real part, hence taking $$ f(z)=\overline{\frac {-i}{\overline z}}=\frac{-\overline i}{z}=\frac iz$$ does the trick.

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What about

$$f(z)=\frac{i}{z}$$

where $z=x+i y$.

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You can integrate one equation at the time. From $$v_x=\frac1{x^2+y^2}$$ you get $$v(x,y)=\frac{\arctan(x/y)}{y}+g(y)$$ for some function $g(y)$ being the integration “constant”. Now plug that into the other equation to find an ordinary differential equation for $g(y)$ (this equation should not contain any $x$, or else you made a mistake somewhere).

At some point, you will possibly find it useful to rewrite $\arctan(x/y)$ in terms of $\arg z$.

As an alternative, and much faster, is to compute the real and imaginary parts of $1/z$ and ponder the result. The above, however, is how you would do it in the general case, when such tricks aren't immediately obvious.

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  • $\begingroup$ I know you asked for an “elaborate” answer. I don't have the time to complete this for you, however. Hopefully, you can take it the rest of the way on your own. $\endgroup$ – Harald Hanche-Olsen Jun 9 '13 at 16:13

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