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Consider the function

$$f(z)=Log(\frac{z-a}{z-b})$$

where $a,b\in D(0,r)$ , the disc of radius $r$ centered at the origin, open, and $r>0$. Show that $f$ is holomophic in an open subset of complex plane, containing the complement of the disc:$\mathbb C-D(0,r)$ and compute the integral

$$ \oint_{|z|=r}z^nf(z)dz$$

where $n\geq 0$ is an integer.

I'm quite confused about this problem. Because, the $Log$ in the RHS should be the principal branch, i.e. $Log(z)=log|z|+iArg(z)$, with $Arg(z)\in (-\pi,\pi)$ and $Arg(0)=0$. Then i should prove that $\frac{z-a}{z-b}$ does not intersect the real negative axis, but how to do this? However, suppose i could prove the holomorphicity, then i'm left to compute the integral, for which i tought to use the residue formula applied to $\infty$ point, so calling $I$ the integral,

$$I=2\pi i Ind(\gamma,\infty)Res(f,\infty)$$

where $\gamma$ is the contour $|z|=1$, so that $ind(\gamma,\infty)=-1$. But then the computation of residue is very hard for me. Some help?

$$f(z)=Log(\frac{z-a}{z-b})$$

where $a,b\in D(0,r)$ , the disc of radius $r$ centered at the origin, open, and $r>0$. Show that $f$ is holomophic in an open subset of complex plane, containing the complement of the disc:$\mathbb C-D(0,r)$ and compute the integral

$$ \oint_{|z|=r}z^nf(z)dz$$

where $n\geq 0$ is an integer.

I'm quite confused about this problem. Because, the $Log$ in the RHS should be the principal branch, i.e. $Log(z)=log|z|+iArg(z)$, with $Arg(z)\in (-\pi,\pi)$ and $Arg(0)=0$. Then i should prove that $\frac{z-a}{z-b}$ does not intersect the real negative axis, but how to do this? However, suppose i could prove the holomorphicity, then i'm left to compute the integral, for which i tought to use the residue formula applied to $\infty$ point, so calling $I$ the integral,

$$I=2\pi i Ind(\gamma,\infty)Res(f,\infty)$$

where $\gamma$ is the contour $|z|=1$, so that $ind(\gamma,\infty)=-1$. But then the computation of residue is very hard for me. Some help?

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