Let \[ u(x,y)=\frac{a^2+b^2+x^2-y^2}{(a-x)^2+(b-y)^2}. \] Show that \(u\) is harmonic. The original problem asks for an entire function whose real part is \(u\), but the paper corrects this: explain why this is impossible, modify \(u\) slightly, and solve the corrected problem.
The problem is false as printed: the displayed \(u\) is not harmonic. On the complement of \((a,b)\), \[ \Delta u=\frac{4\left(a^2+b^2+2ax-x^2-2by+y^2\right)}{\left((a-x)^2+(b-y)^2\right)^2}, \] which is not identically zero; for \(a=1\), \(b=0\) it equals \(4\) at the origin. No entire function has real part \(u\), since \(u\) is unbounded near \((a,b)\) whenever \(a\neq0\). The source prints no modified formula. The solution below replaces \(+x^2\) by \(-x^2\); with \(c=a+ib\) and \(z=x+iy\), the modified function is \(\operatorname{Re}\frac{c+z}{c-z}\), which is harmonic on \(\CC\setminus\{c\}\).
Solution
Let \[ D(x,y)\coloneqq(a-x)^2+(b-y)^2 \] and let \[ \Omega\coloneqq\CC\setminus\{a+ib\}. \]
1. The function printed in the source is not harmonic on its naturaldomain.
For \[ u(x,y) = \frac{a^2+b^2+x^2-y^2}{D(x,y)}, \] direct differentiation gives \[ \Delta u = \frac{ 4\bigl(a^2+b^2+2ax-x^2-2by+y^2\bigr) }{ D(x,y)^2 }. \] The numerator is not the zero polynomial: its coefficient of \(x^2\) is \(-4\). Thus \(\Delta u\) is not identically zero on \(\RR^2\setminus\{(a,b)\}\), so \(u\) is not harmonic.
2. Consequently the printed \(u\) cannot be the real part of an entirefunction.
The real part of every holomorphic function is harmonic. In particular, the real part of an entire function is harmonic on all of \(\CC\). Step 1 shows that the printed \(u\) is not even harmonic on its punctured natural domain, so no entire function can have real part \(u\) there.
3. Make the single sign change\[ \widetilde u(x,y) \coloneqq \frac{a^2+b^2-x^2-y^2}{D(x,y)}. \]If\[ c\coloneqq a+ib, \]then on \(\Omega\),\[ \boxed{ \widetilde u(x,y) = \operatorname{Re} \left( \frac{c+z}{c-z} \right), \qquad z=x+iy. } \]
Since \[ \abs{c-z}^2=D(x,y), \] one has \[ \frac{c+z}{c-z} = \frac{(c+z)(\overline c-\overline z)} {\abs{c-z}^2}. \] The real part of the numerator is \[ \operatorname{Re} \bigl( (c+z)(\overline c-\overline z) \bigr) = \abs{c}^2-\abs{z}^2 = a^2+b^2-x^2-y^2, \] because \[ z\overline c-c\overline z \] is purely imaginary. Dividing by \(D(x,y)\) gives the stated identity.
4. The modified function \(\widetilde u\) is harmonic on\(\RR^2\setminus\{(a,b)\}\).
The function \[ F(z)\coloneqq\frac{c+z}{c-z} \] is holomorphic on \(\Omega\). By step 3, \[ \widetilde u=\operatorname{Re}F \] there, and the real part of a holomorphic function is harmonic.
5. The corresponding holomorphic function is\[ \boxed{ F(z)=\frac{a+ib+z}{a+ib-z} } \]on \(\CC\setminus\{a+ib\}\).
3.Thus it solves the corrected harmonic-conjugate problem on the naturaldomain of \(\widetilde u\).
6. If \((a,b)\ne(0,0)\), no entire function has real part\(\widetilde u\) on \(\Omega\).
When \(c\ne0\), \[ F(z) = -1+\frac{2c}{c-z} \] has a simple pole at \(z=c\). Equivalently, \(\widetilde u=\operatorname{Re}F\) is unbounded in every punctured neighborhood of \(c\). The real part of an entire function is continuous and finite at \(c\), so it cannot agree with \(\widetilde u\) on \(\Omega\).
7. If \((a,b)=(0,0)\), the modified function is the constant \(-1\) on\(\Omega\), and it extends across the origin.
For \(c=0\) and \(z\ne0\), \[ F(z)=\frac z{-z}=-1. \] Hence \(\widetilde u=-1\), and the constant entire function \(F(z)=-1\) gives the extension.
8. The printed \(u\) is not harmonic and is not the real part of an entirefunction. The modified function satisfies\[ \boxed{ \widetilde u = \operatorname{Re} \left( \frac{a+ib+z}{a+ib-z} \right) } \]on \(\CC\setminus\{a+ib\}\), and \(\widetilde u\) is the real part of an entirefunction if and only if \((a,b)=(0,0)\).
Steps 1 and 2 prove the claims about the printed \(u\). Step 3 gives the formula for \(\widetilde u\), and steps 6 and 7 decide when \(\widetilde u\) is the real part of an entire function.
9. Q.E.D.
Step 8 answers each part of the problem.