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Show that the mapping $w=z+\frac1z$ maps the domain outside the circle $|z| = 1$ onto the rest of the $w$ plane.

I have already showed that circles $|z|=r_0 (r_0 \ne 1)$ are mapped onto ellipses $$\frac{u^2}{(r_0+\frac{1}{r_0})^2}+\frac{v^2}{(r-\frac{1}{r_0})^2}=1$$ with $r_0 \neq 1.$ In the course of proving this I had the equalities $u = (r_0 + \frac{1}{r_0}) \cos \theta$ and $v = (r_0 - \frac{1}{r_0}) \sin \theta$ with $0 \leq \theta \leq 2\pi$. For $|z| > 1$, my intuition tells me that values of $u$ and $v$ range over the $w$ plane, but I am not sure how to formally show this. I thought about letting the radii tend to infinity, but am not sure how the argument would go.

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  • $\begingroup$ Actually, what is meant with "rest of the plane"? The image is infact the complment of $[-2,2]$ $\endgroup$ – Hagen von Eitzen May 8 '18 at 4:04
  • $\begingroup$ @HagenvonEitzen I believe that I am supposed to show that the image is the complement of the interval you stated. How would I justify that the image is simply the complement of that interval? (I can show that the image is $[-2,2]$ when $|z| = 1$, but am not sure how to formally show that the image is the complement for $|z| >1$). $\endgroup$ – Geoff Kolak May 8 '18 at 4:18
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Rewrite the equation as $wz=z^2+1$. This is a quadratic equation in $z$ and hence has two roots for each $w$, call them $\alpha$ and $\beta$. Then $\alpha\beta=1$. What does this tell you about where the roots are relative to the unit circle?

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  • $\begingroup$ The roots are at $\pm i.$ How does this show that the domain outside the circle is mapped onto "the rest of the $w$ plane?" $\endgroup$ – Geoff Kolak May 8 '18 at 5:46
  • $\begingroup$ The roots are not $\pm i$ (the left hand side is not 0). To further elaborate, note that if $z$ is a root to the equation $w=z+\frac{1}{z}$, we have $\frac{1}{z}$ is also a root. What happens if $z$ is inside the unit circle? Outside? $\endgroup$ – Bay Wei Heng May 8 '18 at 5:52
  • $\begingroup$ Inside the unit circle, $w \to 0 + \infty$ and outside the unit circle $w \to \infty + 0$ . Why did we need to consider $wz$? $\endgroup$ – Geoff Kolak May 8 '18 at 14:17
  • $\begingroup$ We're not considering $wz$, we're considering $z^2-wz+1=0$, which is a quadratic equation. Basically, we want to say that for every $w$ in the "rest of the $w$ plane", i.e. $\mathbb{C}\setminus[-2,2]$, we can find a $z$ outside the unit circle such that $z+1/z=w$. But the argument I have provided shows you can always find such a $z$. $\endgroup$ – Bay Wei Heng May 9 '18 at 5:37

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