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Given the real function $u: A \rightarrow R$ defined by $\displaystyle u(x,y) = 100 \cdot \frac{x - y}{x^2+y^2}$ where A is an open subset of $R^2$.

  1. Determine the largest open set A where u is harmonic. Find a real function $v: R^2 \rightarrow R$ such that $f(z) = u(x,y)+iv(x,y)$ is an entire function where $x = Re(z)$ and $y = Im(z)$

  2. Write $f(z)$ down explicitly as a function of $z$

Is the following manner good or am I going in the wrong direction:

  1. Using the Cauchy-Riemann relationship $vy = ux$ and $vx = -uy$

$\displaystyle vy = ux = 100 + 100y \cdot \frac{2x+2y*y '(x)}{(x^2 + y^2) ^ 2} $

I am not sure if I have to take $2y*y '(x)$ also in the above differential with respect to x...

then integrate with respect to y to get v (the imaginary part). We then get, by integrating a constant which can be determined also by the Cauchy Riemann and thus eventually get $f(z)$.

Is the above correct?

  1. I don't really understand what is asked so can you please clarify it?

Thanks !

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  • $\begingroup$ haha I am not angry at all :p $\endgroup$ – Jantje7600 Oct 13 '14 at 11:59
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Let $f(z) =\frac{1}{z} ,$ $g(z) =\frac{-i}{z}$ then $$\mbox{Re} (f(z))= \mbox{Re}\left(\frac{x-iy}{x^2 +y^2}\right) =\frac{x}{x^2 +y^2} $$ $$\mbox{Re} (g(z))= \mbox{Re}\left(\frac{-y-ix}{x^2 +y^2}\right) =\frac{-y}{x^2 +y^2} $$ hence $$u(z) =\mbox{Re} \left(100\cdot f(z) +100\cdot g(z)\right)$$ hence $u$ is harmonic in the set $\mathbb{R}^2\setminus \{(0,0)\}.$

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