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Given that $ Re [ \log (z-1) ]= \frac {1}{2} \ln [(x-1)^2 + y^2 ] $

Question : Why must this equation satisfy the Laplace's equation when $z \neq 1 $? Please do not verify that the function satisfies Laplace's equation directly.

The only time this function is undefined is when x=1 and y=0 otherwise stated when z=1.

Clearly when $z \neq 1 $ considering $ Re [ \log (z-1) ] = Re [ \ln |z-1| + i arg z]= \ln |z-1|$

However $\ln |z-1|$ is defined a unique way so that $ \ln |z-1| > 0 $ by changing the $i argz $ term around.

I feel like this function is always single branch the way it has been defined but im not sure.

If it is a single branch i would like to say that $ Re [ \frac {d}{dz} \log (z-1) ]= \frac {1}{z-1}$

Once its in this form i can use a theorem from my textbook to say that $ \frac {1}{z-1}$ is a polynomial and is thus analytic whenever the denominator does not equal 0 ie $ z=1 $

Next since the derivative of $ Re [ \log (z-1) ] $ is analytic whenever $ z \neq 1 $ clearly $ Re [ \log (z-1) ] $ must be analytic whenever $ z \neq 1 $

By theorem if $ f(z) $ is analytic in a domain D then its component functions u and v are harmonic in D. Which in alot more work than just computing the partials directly but shows us that Laplace's equation holds when $z \neq 1 $

the problems i am having is that though im fairly confident that my function is single valued when $|z|< 1 $ i have a totally different function then i have when $ |z| >1 $

can it be done this way? is there a different approach that is simpler to show that $ H_{xx}(x,y) + H_{yy}(x,y)=0 $

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$f(z) = \log (z - 1)$ is analytic in a region around $z=1$. You can then make use of a result on analytic functions: if a function $f(z) = u(x,y) + iv(x,y)$ is analytic on a region $D$, then $u$ and $v$ are harmonic functions on $D$.

Since $u(x,y) = \text{Re}\{f(z)\}$, $\frac {1}{2} \ln [(x-1)^2 + y^2 ]$ is harmonic on the region.

Harmonic functions, by definition, satisfy Laplace's equation.