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Write the Taylor series of $\text{Log}(1+w)$ with center at $w=0$ on $|w|<1$;

check that if $|z-2|<1$, then $|z|>1$. (If you have difficulties in checking this formally, try to draw a picture of the situation). Use these facts to compute

$$ \int_C \text{Log}\left(1+\frac{1}{z}\right)dz\, $$ $$ C:w(\theta) = 2 + (1/2)e^{i\theta},\quad \theta \in [0,2\pi] $$


My attempt so far:

I know the Taylor series of Log(1+w).

(I am pretty sure) I know what |z-2| < 1 looks like on the complex plane, and |z| > 1 also, but how do those two hold at the same time? |z| > 1 contains the region that |z-2| < 1 (disk centered at 2, radius 1), but that's it? They for sure aren't equal.

Writing out the Taylor series expansion of Log(1+(1/z)) seems to be reminiscent of a Laurent series with all all positive term coefficients equal to zero, but given our simple closed curve, there seems to be only singularities at z = 0, so I don't see anything resembling using residues or such.

I am sorry I am not fluent in LaTeX to formalize my question, but thanks in advance.

EDIT: I have not done any problems in a long time that simply just refer to Cauchy-Goursat holding, but IIRC, if f(z) is holomorphic within the interior of our curve of integration (and writing out several Taylor expansion terms of Log(1+(1/z))), there are no singularities in the interior of our curve, so is the integral simply zero? I doubt this, but just throwing my ideas out there.

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