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Let $$f(z) = \dfrac{1}{z-1}, z \in \mathbb{C}$$

Let $\mathcal{D}$ be a simple contour (e.g. no self loop, etc.) that encircles $z = 1$ once clockwise.

Then claim: $f(\mathcal{D})$ is a contour that encircles the origin $z = 0$ once counter-clockwise

Can someone explain why this is so?

My attempt is take $z = re^{i\theta}$, and suppose that $z \in \mathcal{D}$

Then $f(z) = \dfrac{1}{ re^{i\theta}-1} = \dfrac{1}{ re^{i\theta}-e^{i\theta}e^{-i\theta}} = \dfrac{1}{(r-e^{-i\theta})e^{i\theta}} = Ae^{-i\theta}$. Then $f(z)$ has been mapped to a point on a circle of radius $A$ encircling the origin once counter-clockwise.

However, $A$ is a function of $\theta$...

How to correctly solve this problem?

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  • $\begingroup$ A circle around $z=-1$ is $z=-1+re^{iθ}$. You are considering a circle around $z=0$. $\endgroup$ Nov 14, 2016 at 0:07
  • $\begingroup$ The function is analytical around $z=-1$ and without zeros. Are you sure you took the right point/function? Maybe $z=1$? $\endgroup$
    – A.Γ.
    Nov 14, 2016 at 0:07
  • $\begingroup$ @A.G. Ah that's right, around the pole at 1 $\endgroup$
    – Fraïssé
    Nov 14, 2016 at 2:02
  • $\begingroup$ @BeachedWhale It is easier to change variable to move the plane one unit to the left so that $z=1$ becomes the origin and the function in the new variable becomes $1/z$. $\endgroup$
    – A.Γ.
    Nov 14, 2016 at 2:09

1 Answer 1

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Suppose $\gamma$ is your contour and we parameterize it on $[a,b]$. Then we have the following.

$$2\pi n(1/(\gamma - 1; 0) = \int_{1/(\gamma - 1)} {dz/z} = \int_a^b {(1/(\gamma -1 ))'\, dt\over 1/(\gamma - 1)} = -\int_a^b {\gamma'(t)\over \gamma - 1} = -2\pi n(\gamma, 1) = 2\pi. $$

Hence $n(f\circ \gamma; 0) = 1$.

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