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

EDIT:

$ Re ( \log [(x-1)+ iy]) = Re ( \log |(x-1)+ iy|+ i \arg z)=Re ( \log [(x-1)^2+ y^2 ]^{\frac {1}{2}} + i \arg z)= \ln [(x-1)^2 + y^2]^{\frac {1}{2}})=\frac {1}{2} \ln [(x-1)^2 + y^2]$

I would like to answer part b with the same logic but honestly i have no idea how to draw $ ln (z-1) $ but regular ln has a value of 0 at 1 this one will have a value of 0 at 0 because of the shift (in the reals anyway in C i have no idea) anyway if we look at $\frac {1}{2} \ln [(x-1)^2 + y^2 ]$ by looking at this equation we can see that when x=1 and y=0 or z=1 is the only time this will equal 0. which will not make a lot of sense cause its the origin defining an angle for it doesn't make a lot of sense.

I guess my question is how do i make an argument for why the Laplace equations hold without actually computing them?

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    Complex $\log$ is more complicated than what you just interpreted as. You first need a suitable brach cut to make it non-multivalued, and moreover then it is $\log z = \log |z| + i\arg z$. So $\Re \log z=\log |z|$ – 2017-02-18
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    but im working with the Real part only. i can toss the i arg z no? – 2017-02-18
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    Thanks though that explains where the positive sign came from. – 2017-02-18

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I assume that in the statement you assume $z=x+iy$. By changing $z-1$ into $z$ and $x-1$ into $x$, the statement becomes $$ \operatorname{Re}(\log z)=\frac{1}{2}\ln(x^2+y^2) $$ Of course $z\ne0$ or the problem is meaningless. Write $\log z=a+ib$, so $z=e^{a+ib}=e^a(\cos b+i\sin b)$ and therefore $x=e^a\cos b$, $y=e^a\sin b$. Hence $$ x^2+y^2=e^{2a} $$ and $$ 2a=\ln(x^2+y^2) $$ which means $$ a=\frac{1}{2}\ln(x^2+y^2)=\ln|z| $$

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    Wow that is really clever. – 2017-02-20